how to find missing frequency when mode is given

NCERT Solutions for Class 10 Math Chapter 9 Mean, Average, Mode Of Grouped Information, Additive Frequency Graph And Ogive are provided here with simple step out-by-step explanations. These solutions for Mean, Median value, Musical mode Of Grouped Information, Cumulative Relative frequency Graph And Ogive are super common among Class 10 students for Math Mean, Median, Mode Of Grouped Data, Cumulative Absolute frequency Chart And Nose cone Solutions come handy for quickly complementary your prep and preparing for exams. All questions and answers from the NCERT Book of Class 10 Math Chapter 9 are provided here for you gratis. You will also love the advertising-free experience on Meritnation's NCERT Solutions. All NCERT Solutions for class Socio-economic class 10 Math are prepared by experts and are 100% accurate.

Page Nobelium 359:

Question 1:

Find the mean, using calculate method:

Class	0-10	10-20	20-30	30-40	40-50
Absolute frequency	3	5	9	5	3

Answer:

Class	Relative frequency $\left(f_i\right)$	Mid Values $\left(x_i\right)$	$(f_i\times x_i)$
0-10	3	5	15
10-20	5	15	75
20-30	9	25	225
30-40	5	35	175
40-50	3	45	135
	${\sum }_{f_i}=25$		$\sum \left(f_i\times x_i\right)=625$

$$\begin{gathered} \therefore \text{ Mean,}\overline{x}=\frac{{\displaystyle \sum (f_i\times x_i)}}{{\displaystyle \sum f_i}} \\ \text{=}\frac{625}{25} \\ \text{=25} \\ \therefore \overline{x}=25 \end{gathered}$$

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Question 2:

Find the mean, victimisation direct method:

Grade	0-10	10-20	20-30	30-40	40-50	50-60
Frequency	7	5	6	12	8	2

Answer:

Class	Frequency $\left(f_i\right)$	Mid Values $\left(x_i\right)$	$(f_i\times x_i)$
0-10	7	5	35
10-20	5	15	75
20-30	6	25	150
30-40	12	35	420
40-50	8	45	360
50-60	2	55	110
	${\sum }_{f_i}=40$		$\sum (f_i\times x_i)=1150$

$$\begin{gathered} \therefore \text{ Mean,}\overline{x}=\frac{{\displaystyle \sum (f_i\times x_i)}}{{\displaystyle \sum f_i}} \\ \text{=}\frac{1150}{40} \\ \text{=28}\text{.75} \\ \therefore \overline{x}=\text{28}\text{.75} \end{gathered}$$

Sri Frederick Handley Page Zero 359:

Question 3:

Find the meanspirited, using direct method:

Class	10-20	20-30	30-40	40-50	50-60	60-70
Frequency	11	15	20	30	14	10

Answer:

Class	Oftenness $\left(f_i\right)$	Mid Values $\left(x_i\right)$	$\left(f_i\times x_i\right)$
10-20	11	15	165
20-30	15	25	375
30-40	20	35	700
40-50	30	45	1350
50-60	14	55	770
60-70	10	65	650
	$\sum f_i=100$		$\sum (f_i\times x_i)=4010$

$$\begin{gathered} \therefore \text{ Average,}\overline{x}=\frac{{\displaystyle \sum (f_i\times x_i)}}{{\displaystyle \sum f_i}} \\ \text{=}\frac{4010}{100} \\ \text{=40}\text{.1} \\ \therefore \overline{x}=40.1 \end{gathered}$$

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Question 4:

Find the mean, using direct method:

Marks	10-20	20-30	30-40	40-50	50-60	60-70	70-80
Keep down of students	6	8	13	7	3	2	1

Answer:

Class	Frequency $\left(f_i\right)$	Middle Values $\left(x_i\right)$	$(f_i\times x_i)$
10-20	6	15	90
20-30	8	25	200
30-40	13	35	455
40-50	7	45	315
50-60	3	55	165
60-70	2	65	130
70-80	1	75	75
	$\sum f_i=40$		$\sum (f_i\times x_i)=1430$

$$\begin{gathered} \therefore \text{ Poor,}\overline{x}=\frac{{\displaystyle \sum (f_i\times x_i)}}{{\displaystyle \sum f_i}} \\ \text{=}\frac{1430}{40} \\ \text{=35}\text{.75} \\ \therefore \overline{x}=35.75 \end{gathered}$$

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Question 5:

Find the mean, using direct method:

Class	25-35	35-45	45-55	55-65	65-75
Frequency	6	10	8	12	4

Answer:

Class	Frequency $\left(f_i\right)$	Middle values $\left(x_i\right)$	$\left(f_i\times x_i\right)$
25-35	6	30	180
35-45	10	40	400
45-55	8	50	400
55-65	12	60	720
65-75	4	70	280
	$\sum f_i=40$		$\sum (f_i\times x_i)=1980$

$$\begin{gathered} \therefore \text{ Mean,}\overline{x}=\frac{{\displaystyle \sum (f_i\times x_i)}}{{\displaystyle \sum f_i}} \\ \text{=}\frac{1980}{40} \\ \text{=49}\text{.5} \\ \therefore \overline{x}=49.5 \end{gathered}$$

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Question 6:

Find the mean, using direct method:

Social class	0-100	100-200	200-300	300-400	400-500
Frequency	6	9	15	12	8

Result:

Class	Absolute frequency $\left(f_i\right)$	Mid values $\left(x_i\right)$	$(f_i\times x_i)$
0-100	6	50	300
100-200	9	150	1350
200-300	15	250	3750
300-400	12	350	4200
400-500	8	450	3600
	$\sum f_i=50$		$\sum (f_i\times x_i)=13200$

$$\begin{gathered} \therefore \text{ Mean,}\overline{x}=\frac{{\displaystyle \sum (f_i\times x_i)}}{{\displaystyle \sum f_i}} \\ \text{=}\frac{13200}{50} \\ \text{=264} \\ \therefore \overline{x}=264 \end{gathered}$$

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Question 7:

The mean of the following frequency distribution is 24. Happen the value of p.

Marks	0-10	10-20	20-30	30-40	40-50
Number of students	15	20	35	p	10

Answer:

Class	Frequency $\left(f_i\right)$	Mid Values $\left(x_i\right)$	$(f_i\times x_i)$
0-10	15	5	75
10-20	20	15	300
20-30	35	25	875
30-40	p	35	35 p
40-50	10	45	450
	$\sum f_i=80+p$		$\sum (f_i\times x_i)=1700+35p$

$$\begin{gathered} \therefore \text{ Mean,}\overline{x}=\frac{{\displaystyle \sum (f_i\times x_i)}}{{\displaystyle \sum f_i}} \\ \Rightarrow \text{24 =}\frac{1700+35p}{80+p}[\because \text{Mean=24}] \\ \Rightarrow \text{1920+24}p=1700+35p \\ \Rightarrow \text{11}p=220 \\ \Rightarrow p=20 \\ \therefore p=20 \end{gathered}$$

Foliate No 360:

Question 8:

Find the missing frequencies f ₁ and f ₂ in the table in given below, information technology is being given that the awful of the presented frequency distribution is 50.

Class	0-20	20-40	40-60	60-80	80-100	Add
Relative frequency	17	f 1	32	f 2	19	120

Answer:

Class	Relative frequency $\left(f_i\right)$	Mid values $\left(x_i\right)$	$(f_i\times x_i)$
0-20	17	10	170
20-40	f1	30	30 f1
40-60	32	50	1600
60-80	52- f1	70	3640-70 f1
80-100	19	90	1710
	$\sum f_i=120$		$\sum (f_i\times x_i)=7120-40f_1$

$$\begin{gathered} \text{ We have:} \\ \text{17+}{f}_1+32+f_2+19=120 \\ \Rightarrow f_1+f_2=52 \\ \Rightarrow f_2=52-f_1 \\ \therefore \text{ Mean,}\overline{x}=\frac{{\displaystyle \sum (f_i\times x_i)}}{{\displaystyle \sum f_i}} \\ \Rightarrow \text{50=}\frac{7120-40f_1}{120}[\because \text{Mean=50]} \\ \Rightarrow 40f_1=1120 \\ \therefore f_1=28 \\ \text{And}{f}_2=52-28 \\ \Rightarrow f_2\text{=24} \\ \therefore \text{The missing frequencies are}{f}_1=28\text{ and}{f}_2\text{ =24.} \end{gathered}$$

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Question 9:

The mean of the following frequency dispersion is 57.6 and the sum of the observations is 50

Socio-economic class	0-20	20-40	40-60	60-80	80-100	100-120
Frequency	7	f 1	12	f 2	8	5

Answer:

Class	Frequency $\left(f_i\right)$	Mid values $\left(x_i\right)$	$(f_i\times x_i)$
0-20	7	10	70
20-40	f 1	30	30 f 1
40-60	12	50	600
60-80	18- f 1	70	1260-70 f1
80-100	8	90	720
100-120	5	110	550
	$\sum f_i=50$		$\sum (f_i\times x_i)=3200-40f_1$

$$\begin{gathered} \text{ We have:} \\ \text{ 7+}{f}_1+12+f_2+8+5=50 \\ \Rightarrow f_1+f_2=18 \\ \Rightarrow f_2=18-f_1 \\ \therefore \text{ Mean,}\overline{x}=\frac{{\displaystyle \sum (f_i\times x_i)}}{{\displaystyle \sum f_i}} \\ \Rightarrow \text{57}\text{.6=}\frac{3200-40f_1}{50}[\because \text{Mean=57}\text{.6}] \\ \Rightarrow 40f_1=320 \\ \therefore f_1=8 \\ \text{And}{f}_2=18-8 \\ \Rightarrow f_2\text{=10} \\ \therefore \text{The nonexistent frequencies are}{f}_1=8\text{ and}{f}_2\text{ =10.} \end{gathered}$$

Page Zero 360:

Question 10:

Find the mean, using assumed-normal method acting:

Marks	0-10	10-20	20-30	30-40	40-50	50-60
Number of students	12	18	27	20	17	6

Answer:

Assort	Frequency $\left(f_i\right)$	Mid values$\left(x_i\right)$	Divergence $\left(d_i\right)$ $d_i=\left(x_i-25\right)$	$(f_i\times d_i)$
0-10	12	5	-20	-240
10-20	18	15	-10	-180
20-30	27	25=A	0	0
30-40	20	35	10	200
40-50	17	45	20	340
50-60	6	55	30	180
	$\sum f_i=100$			$\sum (f_i\times d_i)=300$

$$\begin{gathered} \text{Let}A=25\text{ be the taken for granted mean}\text{. Then we have:} \\ \text{Mean,}\overline{x}=A+\frac{{\displaystyle \sum (f_i\times d_i)}}{{\displaystyle \sum f_i}} \\ \text{= 25+}\frac{300}{100} \\ \text{=28} \\ \therefore \overline{x}=28 \end{gathered}$$

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Question 11:

Line up the mean, using assumed-mean method acting:

Class	0-40	40-80	80-120	120-160	160-200
Frequency	12	20	35	30	23

Answer:

Class	Oftenness $\left(f_i\right)$	Mid values$\left(x_i\right)$	Deviation $\left(d_i\right)$ $\left(d_i\right)=\left(x_i-100\right)$	$(f_i\times d_i)$
0-40	12	20	-80	-960
40-80	20	60	-40	-800
80-120	35	100=A	0	0
120-160	30	140	40	1200
160-200	23	180	80	1840
	$\sum f_i=120$			$\sum (f_i\times d_i)=1280$

$$\begin{gathered} \text{Let}A=100\text{ cost the assumed mean}\text{. Then we have:} \\ \text{Mean,}\overline{x}=A+\frac{{\displaystyle \sum (f_i\times d_i)}}{{\displaystyle \sum f_i}} \\ \text{= 100+}\frac{1280}{120} \\ \text{=100+10}\text{.67} \\ \therefore \overline{x}=110.67 \end{gathered}$$

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Question 12:

Find the mean, using assumed-mean method acting:

Class	100-120	120-140	140-160	160-180	180-200
Frequency	10	20	30	15	5

Resolve:

Class	Frequency $\left(f_i\right)$	Mid values$\left(x_i\right)$	Deviation $\left(d_i\right)$ $d_i=\left(x_i-150\right)$	$(f_i\times d_i)$
100-120	10	110	-40	-400
120-140	20	130	-20	-400
140-160	30	150=A	0	0
160-180	15	170	20	300
180-200	5	190	40	200
	$\sum f_i=80$			$\sum (f_i\times d_i)=-\text{3}00$

$$\begin{gathered} \text{Let}A=150\text{ be the assumed mean}\text{. Then we rich person:} \\ \text{Mean,}\overline{x}=A+\frac{{\displaystyle \sum (f_i\times d_i)}}{{\displaystyle \sum f_i}} \\ \text{= 150}-\frac{300}{80} \\ \text{=150}-\text{3}\text{.75} \\ \therefore \overline{x}=146.25 \end{gathered}$$

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Question 13:

Find the mean, using assumed-mean method:

Class	0-20	20-40	40-60	60-80	80-100	100-120
Frequency	20	35	52	44	38	31

Answer:

Class	Frequency $\left(f_i\right)$	Middle Values$\left(x_i\right)$	Departure $\left(d_i\right)$ $d_i=\left(x_i-50\right)$	$(f_i\times d_i)$
0-20	20	10	-40	-800
20-40	35	30	-20	-700
40-60	52	50=A	0	0
60-80	44	70	20	880
80-100	38	90	40	1520
100-120	31	110	60	1860
	$\sum f_i=220$			$\sum (f_i\times d_i)=2760$

$$\begin{gathered} \text{Rent}A=50\text{ be the assumed have in mind}\text{. Then we have:} \\ \mathrm{Like\; a\; sho},\mathrm{m}\text{ean,}\overline{x}=A+\frac{{\displaystyle \sum (f_i\times d_i)}}{{\displaystyle \sum f_i}} \\ \text{=50+}\frac{2760}{220} \\ \text{=50+12}\text{.55} \\ \therefore \overline{x}=62.55 \end{gathered}$$

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Question 14:

Find the arithmetical think of each of the following frequency distributions victimisation footprint-deviation method:

Marks	0-10	10-20	20-30	30-40	40-50	50-60
Number of students	12	18	27	20	17	6

Answer:

Class	Frequency$\left(f_i\right)$	Middle Values $\left(x_i\right)$	$u_i=\frac{(x_i-A)}{h}$ $\text{=}\frac{(x_i-25)}{10}$	$(f_i\times u_i)$
0-10	12	5	−2	−24
10-20	18	15	−1	−18
20-30	27	25=A	0	0
30-40	20	35	1	20
40-50	17	45	2	34
50-60	6	55	3	18
	$\sum f_i=100$			$\sum (f_i\times u_i)=30$

$$\begin{gathered} \text{Now,}A=25,h=10,\sum f_i=100\mathrm{and}\sum (f_{{}_i}\times u_i)=30 \\ \therefore \text{ Mean,}\overline{x}=A+\left(h\times \frac{{\displaystyle \sum (f_{{}_i}\times u_i)}}{{\displaystyle \sum f_i}}\right) \\ \text{=25+}\left(10\times \frac{30}{100}\right) \\ \text{=25+3} \\ \text{=28} \\ \therefore \overline{x}=28 \end{gathered}$$

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Question 15:

Find the arithmetical mean of all of the following absolute frequency distributions using step-deviation method acting:

Course of instruction	Number of students
4-8	2
8-12	12
12-16	15
16-20	25
20-24	18
24-28	12
28-32	13
32-36	3

Answer:

Class	Oftenness $\left(f_i\right)$	Mid values$\left(x_i\right)$	$u_i=\frac{(x_i-A)}{h}$ $\text{=}\frac{(x_i-18)}{4}$	$(f_i\times u_i)$
4-8	2	6	-3	-6
8-12	12	10	-2	-24
12-16	15	14	-1	-15
16-20	25	18=A	0	0
20-24	18	22	1	18
24-28	12	26	2	24
28-32	13	30	3	39
32-36	3	34	4	12
	$\sum f_i=100$			$\sum (f_i\times u_i)=48$

$$\begin{gathered} \text{Now,}A=18,h=4,\sum f_i=100\mathrm{and}\sum (f_{{}_i}\times u_i)=48 \\ \therefore \text{Normal,}\overline{x}=A+\{h\times \frac{{\displaystyle \sum (f_{{}_i}\times u_i)}}{{\displaystyle \sum f_i}}\} \\ \text{=18+}\{4\times \frac{48}{100}\} \\ \text{=18+1}\text{.92} \\ \text{=19}\text{.92} \\ \therefore \overline{x}=19.92 \end{gathered}$$

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Question 16:

Uncovering the expectation of each of the following frequency distributions using step-deviation method:

Class	0-30	30-60	60-90	90-120	120-150	150-180
Frequency	12	21	34	52	20	11

Answer:

Grade	Oftenness $\left(f_i\right)$	Mid values$\left(x_i\right)$	$u_i=\frac{(x_i-A)}{h}$ $\text{=}\frac{(x_i-75)}{30}$	$(f_i\times u_i)$
0-30	12	15	−2	−24
30-60	21	45	−1	−21
60-90	34	75 = A	0	0
90-120	52	105	1	52
120-150	20	135	2	40
150-180	11	165	3	33
	$\sum f_i=150$			$\sum (f_i\times u_i)=80$

$$\begin{gathered} \text{Now,}A=75,h=30,\sum f_i=150\mathrm{and}\sum (f_{{}_i}\times u_i)=80 \\ \therefore \text{Mean,}\overline{x}=A+\{h\times \frac{{\displaystyle \sum (f_{{}_i}\times u_i)}}{{\displaystyle \sum f_i}}\} \\ \text{=75+}\{30\times \frac{80}{150}\} \\ \text{=75+16} \\ \text{=91} \\ \text{∴}\overline{x}=91 \end{gathered}$$

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Question 17:

Find the arithmetical mean of each of the following frequency distributions using step-deviation method acting:

Sort	0-20	20-40	40-60	60-80	80-100	100-120	120-140
Frequency	12	18	15	25	26	15	9

Answer:

Form	Frequency $\left(f_i\right)$	Mid values$\left(x_i\right)$	$$\begin{gathered} u_i=\frac{(x_i-A)}{h} \\ \text{=}\frac{(x_i-70)}{20} \end{gathered}$$	$(f_i\times u_i)$
0-20	12	10	−3	−36
20-40	18	30	−2	−36
40-60	15	50	−1	−15
60-80	25	70 = A	0	0
80-100	26	90	1	26
100-120	15	110	2	30
120-140	9	130	3	27
	$\sum f_i=120$			$\sum (f_i\times u_i)=-4$

$$\begin{gathered} \text{In real time,}A=70,h=20,\sum f_i=120\mathrm{and}\sum (f_{{}_i}\times u_i)=-4 \\ \therefore \text{Beggarly,}\overline{x}=A+\{h\times \frac{{\displaystyle \sum (f_{{}_i}\times u_i)}}{{\displaystyle \sum f_i}}\} \\ \text{=70+}\{20\times \frac{(-4)}{120}\} \\ \text{=70-0}\text{.67} \\ \text{=69}\text{.33} \\ \text{∴}\overline{x}=69.33 \end{gathered}$$

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Question 18:

Find the arithmetic ungenerous of from each one of the following frequency distributions using step-diversion method:

Simon Marks	0-14	14-28	28-42	42-56	56-70
Number of students	7	21	35	11	16

Answer:

Class	Frequency $\left(f_i\right)$	Mid values $\left(x_i\right)$	$$\begin{gathered} u_i=\frac{(x_i-A)}{h} \\ \text{=}\frac{(x_i-35)}{14} \end{gathered}$$	$(f_i\times u_i)$
0-14	7	7	−2	−14
14-28	21	21	−1	−21
28-42	35	35 = A	0	0
42-56	11	49	1	11
56-70	16	63	2	32
	$\sum f_i=90$			$\sum (f_i\times u_i)=8$

$$\begin{gathered} \text{Now,}A=35,h=14,\sum f_i=90\mathrm{and}\sum (f_{{}_i}\times u_i)=8 \\ \therefore \text{ Mean,}\overline{x}=A+\{h\times \frac{{\displaystyle \sum (f_{{}_i}\times u_i)}}{{\displaystyle \sum f_i}}\} \\ \text{=35+}\{14\times \frac{8}{90}\} \\ \text{=35+1}\text{.24} \\ \text{=36}\text{.24} \\ \therefore \overline{x}=36.24\mathrm{Cm} \end{gathered}$$

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Interrogative 19:

Uncovering the arithmetic mean of each of the following frequency distributions using gradation-departure method:

Class	10-15	15-20	20-25	25-30	30-35	35-40
Frequency	5	6	8	12	6	3

Solution:

Socio-economic class	Frequency $\left(f_i\right)$	Mid values$\left(x_i\right)$	$$\begin{gathered} u_i=\frac{(x_i-A)}{h} \\ \text{=}\frac{(x_i-22.5)}{5} \end{gathered}$$	$(f_i\times u_i)$
10-15	5	12.5	−2	−10
15-20	6	17.5	−1	−6
20-25	8	22.5 = A	0	0
25-30	12	27.5	1	12
30-35	6	32.5	2	12
35-40	3	37.5	3	9
	$\sum f_i=40$			$\sum (f_i\times u_i)=17$

$$\begin{gathered} \text{Now,}A=22.5,h=5,\sum f_i=40\mathrm{and}\sum (f_{{}_i}\times u_i)=17 \\ \therefore \text{ Mean,}\overline{x}=A+\{h\times \frac{{\displaystyle \sum (f_{{}_i}\times u_i)}}{{\displaystyle \sum f_i}}\} \\ \text{=22}\text{.5+}\{5\times \frac{17}{40}\} \\ \text{=22}\text{.5+2}\text{.125} \\ \text{=24}\text{.625} \\ \therefore \overline{x}=24.625 \end{gathered}$$

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Wonder 20:

Find the first moment of each of the following frequency distributions using step-deviation method acting:

Age (in years)	18-24	24-30	30-36	36-42	42-48	48-54
Number of workers	6	8	12	8	4	2

Suffice:

Class	Frequence $\left(f_i\right)$	Middle values$\left(x_i\right)$	$$\begin{gathered} u_i=\frac{(x_i-A)}{h} \\ \text{=}\frac{(x_i-33)}{6} \end{gathered}$$	$(f_i\times u_i)$
18-24	6	21	−2	−12
24-30	8	27	−1	−8
30-36	12	33 = A	0	0
36-42	8	39	1	8
42-48	4	45	2	8
48-54	2	51	3	6
	$\sum f_i=40$			$\sum (f_i\times u_i)=2$

$$\begin{gathered} \text{Now,}A=33,h=6,\sum f_i=40\mathrm{and}\sum (f_{{}_i}\times u_i)=2 \\ \therefore \text{ Mean,}\overline{x}=A+\{h\times \frac{{\displaystyle \sum (f_{{}_i}\times u_i)}}{{\displaystyle \sum f_i}}\} \\ \text{=33+}\{6\times \frac{2}{40}\} \\ \text{=33+0}\text{.3} \\ \text{=33}\text{.3} \\ \therefore \overline{x}=33.3\text{years} \end{gathered}$$

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Question 21:

Find the first moment of each of the following frequency distributions using step-deviance method:

Class	84-90	90-96	96-102	102-108	108-114	114-120
Frequency	15	22	20	18	20	25

Answer:

Class	Frequency$\left(f_i\right)$	Mid values$\left(x_i\right)$	$$\begin{gathered} u_i=\frac{(x_i-A)}{h} \\ \text{=}\frac{(x_i-99)}{6} \end{gathered}$$	$(f_i\times u_i)$
84-90	15	87	−2	−30
90-96	22	93	−1	−22
96-102	20	99 = A	0	0
102-108	18	105	1	18
108-114	20	111	2	40
114-120	25	117	3	75
	$\sum f_i=120$			$\sum (f_i\times u_i)=81$

$\begin{array}{l}\text{Now,}A=99,h=6,\sum f_i=120\mathrm{and}\sum (f_{{}_i}\times u_i)=81\\ \therefore \text{ Mean,}\overline{x}=A+\{h\times \frac{{\displaystyle \sum (f_{{}_i}\times u_i)}}{{\displaystyle \sum f_i}}\}\\ \text{=99+}\{6\times \frac{81}{120}\}\\ \text{=99+4}\text{.05}\\ \text{=103}\text{.05}\\ \text{∴}\overline{x}=103.05\end{array}$

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Question 22:

Come up the arithmetic mean of each of the following frequency distributions using step-deflexion method:

Class	500-520	520-540	540-560	560-580	580-600	600-620
Absolute frequency	14	9	5	4	3	5

Answer:

Class	Frequency$\left(f_i\right)$	Mid values$\left(x_i\right)$	$$\begin{gathered} u_i=\frac{(x_i-A)}{h} \\ \text{=}\frac{(x_i-550)}{20} \end{gathered}$$	$(f_i\times u_i)$
500-520	14	510	−2	−28
520-540	9	530	−1	−9
540-560	5	550 = A	0	0
560-580	4	570	1	4
580-600	3	590	2	6
600-620	5	610	3	15
	$\sum f_i=40$			$\sum (f_i\times u_i)=-12$

$$\begin{gathered} \text{Now,}A=550,h=20,\sum f_i=40\mathrm{and}\sum (f_{{}_i}\times u_i)=-12 \\ \therefore \text{ Nasty,}\overline{x}=A+\{h\times \frac{{\displaystyle \sum (f_{{}_i}\times u_i)}}{{\displaystyle \sum f_i}}\} \\ \text{=550+}\{20\times \frac{(-12)}{40}\} \\ \text{=550-6} \\ \text{=544} \\ \therefore \overline{x}=544 \end{gathered}$$

Page No 362:

Question 23:

Find the mean senesce from the following frequency distribution:

Age (in years)	25-29	30-34	35-39	40-44	45-49	50-54	55-59
Nary. of individual	4	14	22	16	6	5	3

Answer:

Converting the series into exclusive form, we get:

Class	Relative frequency$\left(f_i\right)$	Mid values$\left(x_i\right)$	$$\begin{gathered} u_i=\frac{(x_i-A)}{h} \\ \text{=}\frac{(x_i-42)}{5} \end{gathered}$$	$(f_i\times u_i)$
24.5-29.5	4	27	−3	−12
29.5-34.5	14	32	−2	−28
34.5-39.5	22	37	−1	−22
39.5-44.5	16	42 = A	0	0
44.5-49.5	6	47	1	6
49.5-54.5	5	52	2	10
54.5-59.5	3	57	3	9
	$\sum f_i=70$			$\sum (f_i\times u_i)=-37$

$$\begin{gathered} \text{Now,}A=42,h=5,\sum f_i=70\mathrm{and}\sum (f_{{}_i}\times u_i)=-37 \\ \therefore \text{ Mean,}\overline{x}=A+\{h\times \frac{{\displaystyle \sum (f_{{}_i}\times u_i)}}{{\displaystyle \sum f_i}}\} \\ \text{=42+}\{5\times \frac{(-37)}{70}\} \\ \text{=42-2}\text{.64} \\ \text{=39}\text{.36} \\ \therefore \overline{x}=39.36 \\ \therefore \text{Mean age=39}\text{.36 years} \end{gathered}$$

Paginate No 362:

Question 24:

The following put over shows the get on distribution of patients of malaria in a village during a particular month.

Age (in geezerhood)	5-14	15-24	25-34	35-44	45-54	55-64
No. of cases	6	11	21	23	14	5

Obtain the average age of the patients.

Answer:

Converting the serial into exclusive form, we get:

Division	Frequency $\left(f_i\right)$	Middle values$\left(x_i\right)$	$$\begin{gathered} u_i=\frac{(x_i-A)}{h} \\ \text{=}\frac{(x_i-29.5)}{10} \end{gathered}$$	$(f_i\times u_i)$
4.5-14.5	6	9.5	−2	−12
14.5-24.5	11	19.5	−1	−11
24.5-34.5	21	29.5 = A	0	0
34.5-44.5	23	39.5	1	23
44.5-54.5	14	49.5	2	28
54.5-64.5	5	59.5	3	15
	$\sum f_i=80$			$\sum (f_i\times u_i)=43$

$$\begin{gathered} \text{Now,}A=29.5,h=10,\sum f_i=80\mathrm{and}\sum (f_{{}_i}\times u_i)=43 \\ \therefore \text{ Mean,}\overline{x}=A+\{h\times \frac{{\displaystyle \sum (f_{{}_i}\times u_i)}}{{\displaystyle \sum f_i}}\} \\ \text{=29}\text{.5+}\{10\times \frac{43}{80}\} \\ \text{=29}\text{.5+5}\text{.375} \\ \text{=34}\text{.875} \\ \therefore \overline{x}=\text{34}\text{.875} \\ \therefore \text{The average long time of the patients is 34}\text{.87 years}\text{.} \end{gathered}$$

Page No 368:

Question 1:

Calculate the median for the following relative frequency distribution:

Class	0-10	10-20	20-30	30-40	40-50	50-60
Frequency	3	6	8	15	10	8

Answer:

Class	Frequency (f)	Additive frequency
0-10	3	3
10-20	6	9
20-30	8	17
30-40	15	32
40-50	10	42
50-60	8	50
	N = ∑f = 50

$$\begin{gathered} N=50 \\ =>\frac{N}{2}=25 \\ \text{The cumulative frequency just greater than 25 is 32 and the corresponding class is 30-40.} \\ \text{Thus, the median class is 30-40.} \\ \therefore l=30,h=10,f=15,cf=\text{c}\text{.f}\text{. of pre-existing class = 17 and}\frac{N}{2}=25 \\ \therefore \text{Median,}M=l+\left\{h\times \frac{\left(\frac{N}{2}-cf\right)}{f}\right\} \\ =30+\left\{10\times \frac{(25-17)}{15}\right\} \\ =30+\frac{16}{3} \\ =35\text{.33} \\ \text{Hence, median=35}\text{.33} \end{gathered}$$

Page No 368:

Question 2:

Compute the median from the following data:

Marks	0-7	7-14	14-21	21-28	28-35	35-42	42-49
Number of students	3	4	7	11	0	16	9

Answer:

Class	Frequency (f)	Cumulative frequency
0-7	3	3
7-14	4	7
14-21	7	14
21-28	11	25
28-35	0	25
35-42	16	41
42-49	9	50
	N= ∑f =50

$$\begin{gathered} \begin{array}{l}N=50\\ \Rightarrow \frac{N}{2}=25\\ \text{The cumulative frequency just greater than 25 is 41 and the corresponding class is 35-42.}\\ \\ \text{Thus, the median class is 35-42.}\\ \therefore l=35,h=7,f=16,cf=\text{c}\text{.f}\text{. of preceding social class = 25 and}\frac{N}{2}=25\end{array} \\ \therefore \mathrm{Median}=l+\frac{{\displaystyle \frac{N}{2}}-c.f}{f}\times h \\ \begin{array}{l}\text{=35+}\left\{7\times \frac{(25-25)}{16}\right\}\\ \text{=35+0}\\ \text{=35}\\ \end{array} \end{gathered}$$

Page No 368:

Question 3:

The following table shows the daily reward of workers in a factory:

Day by day reward (in Rs)	0-100	100-200	200-300	300-400	400-500
Number of workers	40	32	48	22	8

Find the median daily wage income of the workers.

Answer:

Socio-economic class	Frequency(f)	Cumulative frequency
0-100	40	40
100-200	32	72
200-300	48	120
300-400	22	142
400-500	8	150
	N= ∑f =150

$$\begin{gathered} N=150 \\ \Rightarrow \frac{N}{2}=75 \\ \text{The additive frequency just greater than 75 is 120 and the corresponding class is 200-300.} \\ \text{Thence, the median class is 200-300.} \\ \therefore l=200,h=100,f=48,cf=\text{c}\text{.f}\text{. of preceding family=72 and}\frac{N}{2}=75 \\ \therefore \text{Median,}M=l+\left\{h\times \frac{\left(\frac{N}{2}-cf\right)}{f}\right\} \\ =200+\left\{100\times \frac{(75-72)}{48}\right\} \\ =200+6\text{.25} \\ =206\text{.25} \\ \text{Hence, the median daily wage income of the workers is Rs 206}\text{.25.} \end{gathered}$$

Pageboy No 368:

Question 4:

Calculate the mesial from the following frequence distribution:

Class	5-10	10-15	15-20	20-25	20-30	30-35	35-40	40-45
Frequency	5	6	15	10	5	4	2	2

Answer:

Class	Relative frequency(f)	Cumulative absolute frequency
5-10	5	5
10-15	6	11
15-20	15	26
20-25	10	36
25-30	5	41
30-35	4	45
35-40	2	47
40-45	2	49
	N= ∑f =49

$$\begin{gathered} N=49 \\ \Rightarrow \frac{N}{2}=24.5 \\ \text{The additive frequency just greater than 24}\text{.5 is 26 and the corresponding class is 15-20.} \\ \text{Thus, the median class is 15-20.} \\ \mathrm{Now},l=15,h=5,f=15,cf=\text{c}\text{.f}\text{. of preceding class = 11 and}\frac{N}{2}=24.5 \\ \therefore \text{Mesial,}M=l+\left\{h\times \frac{\left(\frac{N}{2}-cf\right)}{f}\right\} \\ \text{=15+}\left\{5\times \frac{(24.5-11)}{15}\right\} \\ \text{=15+4}\text{.5} \\ \text{=19}\text{.5} \\ \text{ Hence, median=19}\text{.5} \end{gathered}$$

Page No 369:

Question 5:

Given below is the number of units of electricity consumed in a week in a certain locality:

Consumption (in units)	65-85	85-105	105-125	125-145	145-165	165-185	195-205
Numeral of consumers	4	5	13	20	14	7	4

Forecast the average

Answer:

Separate	Relative frequency(f)	Accumulative frequency
65-85	4	4
85-105	5	9
105-125	13	22
125-145	20	42
145-165	14	56
165-185	7	63
185-205	4	67
	N= ∑f =67

$$\begin{gathered} N=67 \\ \Rightarrow \frac{N}{2}=33.5 \\ \text{The cumulative frequency reasonable greater than 33}\text{.5 is 42 and the corresponding class is 125-145.} \\ \text{Thus, the median class is 125-145.} \\ \mathrm{Now},l=125,h=20,f=20,cf=\text{c}\text{.f}\text{. of preceding class = 22 and}\frac{N}{2}=33.5 \\ \therefore \text{Average,}M=l+\left\{h\times \frac{\left(\frac{N}{2}-cf\right)}{f}\right\} \\ \text{=125+}\left\{20\times \frac{(33.5-22)}{20}\right\} \\ \text{=125+11}\text{.5} \\ \text{=136}\text{.5} \\ \text{Hence, median=136}\text{.5} \end{gathered}$$

Page No 369:

Oppugn 6:

Calculate the median from the following data:

Height (in cm)	135-140	140-145	145-150	150-155	155-160	160-165	165-170	170-175
No. of boys	6	10	18	22	20	15	6	3

Answer:

Course	Frequency(f)	Cumulative frequency
135-140	6	6
140=145	10	16
145-150	18	34
150-155	22	56
155-160	20	76
160-165	15	91
165-170	6	97
170-175	3	100
	N= ∑f =100

$$\begin{gathered} N=100 \\ \Rightarrow \frac{N}{2}=50 \\ \text{The cumulative frequency fitting greater than 50 is 56 and the corresponding year is 150-155.} \\ \text{Thence, the median sort is 150-155.} \\ \mathrm{Now},l=150,h=5,f=22,cf=\text{c}\text{.f}\text{. of preceding class=34 and}\frac{N}{2}=50 \\ \therefore \text{Median,}M=l+\left\{h\times \frac{\left(\frac{N}{2}-cf\right)}{f}\right\} \\ \text{=150+}\left\{5\times \frac{(50-34)}{22}\right\} \\ \text{=150+3}\text{.64} \\ \text{=153}\text{.64} \\ \text{Hence, average=153}\text{.64} \end{gathered}$$

Page No 369:

Head 7:

Forecast the missing frequency from the following dispersion, IT being apt that the median of the distribution is 24.

Class	0-10	10-20	20-30	30-40	40-50
Frequency	5	25	?	18	7

Answer:

Sort	Frequency (fi)	c.f
0-10	5	5
10-20	25	30
20-30	x	x+30
30-40	18	x+48
40-50	7	x+55

$$\begin{gathered} \mathrm{Central}\mathrm{is}24\mathrm{which}\mathrm{lies}\mathrm{in}20-30 \\ \therefore \mathrm{Median}\mathrm{Class}=20-30 \\ \mathrm{Let}\mathrm{the}\mathrm{unknown}\mathrm{frequency}\mathrm{be}x \\ \mathrm{Here},l=20,\frac{n}{2}=\frac{x+55}{2},\mathrm{c}.\mathrm{f}\mathrm{of}\mathrm{the}\mathrm{preceding}\mathrm{class}=c.f=30,f=x,h=10 \\ \therefore \mathrm{Average}=l+\frac{{\displaystyle \frac{n}{2}}-c.f}{f}\times h \\ \Rightarrow 24=20+\frac{{\displaystyle \frac{x+55}{2}}-30}{x}\times 10 \\ \Rightarrow 24=20+\frac{{\displaystyle \frac{x+55-60}{2}}}{x}\times 10 \\ \Rightarrow 24=20+\frac{x-5}{2x}\times 10 \\ \Rightarrow 24=20+\frac{5x-25}{x} \\ \Rightarrow 24=\frac{20x+5x-25}{x} \\ \Rightarrow 24x=25x-25 \\ \Rightarrow -x=-25 \\ \Rightarrow x=25 \\ \mathrm{Hence},\mathrm{the}\mathrm{unsuspected}\mathrm{relative\; frequency}\mathrm{is}25 \end{gathered}$$

Page No 369:

Question 8:

The medial economic value for the following relative frequency distribution is 35 and the sum of the all frequencies is 170. Using the formula for median, find the lost frequencies.

Class	0-10	10-20	20-30	30-40	40-50	50-60	60-70
Frequency	10	20	?	40	?	25	15

Answer:

Let, $f_1\mathrm{and}{f}_2$ be the frequencies of the class intervals 20-30 and 40-50, respectively.
$$\begin{gathered} \mathrm{Then}10+20+f_1+40+f_2+25+15=170 \\ \Rightarrow f_1+f_2=60 \end{gathered}$$
The median is 35 which lies in the class of 30-40. So, the median class is 30-40.
$$\begin{gathered} \mathrm{Now},l=30,h=10,f=40,N=170\mathrm{and}cf=10+20+f_1=f_1+30 \\ \therefore \mathrm{Median},M=l+\left\{h\times \frac{\left(\frac{N}{2}-cf\right)}{f}\right\} \\ \Rightarrow 30+\left[10\times \frac{85-(f_1+30)}{40}\right]=35 \\ \Rightarrow 30+\frac{55-f_1}{4}=35 \\ \Rightarrow 55-f_1=20 \\ \Rightarrow f_1=35 \\ \mathrm{Now},f_2=\left(60-35\right)=25 \\ \mathrm{Hence},f_1=35and{f}_2=25 \end{gathered}$$

Page No 369:

Question 9:

If the median of the following frequency dispersion is 32.5, find the values of f ₁ and f ₂.

Class musical interval	0-10	10-20	20-30	30-40	40-50	50-60	60-70	Total
Absolute frequency	f 1	5	9	12	f 2	3	2	40

Answer:

Course of instruction	Frequency(f)	Accumulative frequency
0-10	$f_1$	$f_1$
10-20	5	$f_1$+5
20-30	9	$f_1$+14
30-40	12	$f_1$+26
40-50	$f_2$	$f_1+f_2$+26
50-60	3	$f_1+f_2$+29
60-70	2	$f_1+f_2$+31
	N= ∑f =40

$$\begin{gathered} \begin{array}{l}\text{Straightaway,}{f}_1+f_2+31=40\\ \Rightarrow f_1+f_2=9\\ \Rightarrow f_2=9-f_1...\left(\text{i}\right)\\ \text{The median is 32}\text{.5 which lies in 30-40.}\\ \text{Thence, mesial class}=30-40\\ \text{Hera,}l=30,\frac{N}{2}=\frac{40}{2}=20,f=12\mathrm{and}c.f=14+f_1\\ \text{Now, median}=32.5\\ \Rightarrow l+\frac{\frac{N}{2}-c.f}{f}\times h=32.5\\ \Rightarrow 30+\frac{20-(14+f_1)}{12}\times 10=32.5\end{array} \\ \Rightarrow \frac{6-f_1}{12}\times 10=2.5 \\ \begin{array}{l}\Rightarrow \frac{60-10f_1}{12}=2.5\\ \Rightarrow 60-10f_1=30\\ \Rightarrow 10f_1=30\\ \Rightarrow f_1=3\\ \text{From equation}\left(\text{i}\right),\text{ we ingest:}\\ f_2=9-3\\ \Rightarrow f_2=6\end{array} \end{gathered}$$

Page No 369:

Question 10:

Calculate the median for the following information:

Geezerhoo (in years)	19-25	26-32	33-39	40-46	47-53	54-60
Frequency	35	96	68	102	35	4

Answer:

First, we testament convert the information into exclusive phase.

Socio-economic class	Relative frequency(f)	Additive relative frequency
18.5-25.5	35	35
25.5-32.5	96	131
32.5-39.5	68	199
39.5-46.5	102	301
46.5-53.5	35	336
53.5-60.5	4	340
	N= ∑f =340

$$\begin{gathered} N=340 \\ =>\frac{N}{2}=170 \\ \text{The cumulative oftenness just greater than 170 is 199 and the corresponding sort out is 32}\text{.5-39}\text{.5.} \\ \text{Thus, the median class is 32}\text{.5-39}\text{.5.} \\ \therefore l=32.5,h=7,f=68,cf=\text{c}\text{.f}\text{. of preceding class = 131 and}\frac{N}{2}=170 \\ \therefore \text{Median,}M=l+\left\{h\times \frac{\left(\frac{N}{2}-cf\right)}{f}\right\} \\ \text{= 32}\text{.5+}\left\{7\times \frac{(170-131)}{68}\right\} \\ \text{= 32}\text{.5+4}\text{.01} \\ \text{= 36}\text{.51} \\ \text{Hence, median = 36}\text{.51} \end{gathered}$$

Page No 370:

Question 11:

Find the medial wages for the following frequencies distribution:

Payoff per day (in Rs)	61-70	71-80	81-90	91-100	101-110	111-120
Zero. of women workers	5	15	20	30	20	8

Answer:

Converting the tending information into exclusive form, we get:

Class	Frequency(f)	Cumulative frequency
60.5-70.5	5	5
70.5-80.5	15	20
80.5-90.5	20	40
90.5-100.5	30	70
100.5-110.5	20	90
110.5-120.5	8	98
	N= ∑f =98

$$\begin{gathered} N=98 \\ \Rightarrow \frac{N}{2}=49 \\ \text{The cumulative frequency just greater than 49 is 70 and the corresponding class is 90}\text{.5-100}\text{.5.} \\ \text{Thus, the normal course is 90}\text{.5-100}\text{.5.} \\ \mathrm{Now},l=90.5,h=10,f=30,cf=\text{c}\text{.f}\text{. of outgoing class = 40 and}\frac{N}{2}=49 \\ \therefore \text{Median,}M=l+\{h\times \frac{(\frac{N}{2}-cf)}{f}\} \\ \text{=90}\text{.5+}\{10\times \frac{(49-40)}{30}\} \\ \text{=90}\text{.5+3} \\ \text{=93}\text{.5} \\ \text{Hence, median wages = Rs 93}\text{.50} \end{gathered}$$

Sri Frederick Handley Page No 370:

Doubt 12:

The following table gives the marks obtained by 50 students in a family test:

Marks	11-15	16-20	21-25	26-30	31-35	36-40	41-45	46-50
No. of students	2	3	6	7	14	12	4	2

Find the median.

Answer:

First of every last, we will convert the given data into exclusive descriptor.

Class	Frequence(f)	Cumulative frequency
10.5-15.5	2	2
15.5-20.5	3	5
20.5-25.5	6	11
25.5-30.5	7	18
30.5-35.5	14	32
35.5-40.5	12	44
40.5-45.5	4	48
45.5-50.5	2	50
	N= ∑f =50

$$\begin{gathered} N=50 \\ \Rightarrow \frac{N}{2}=25 \\ \text{The accumulative frequence just greater than 25 is 32 and the corresponding course of instruction is 30}\text{.5-35}\text{.5.} \\ \text{Thence, the median class is 30}\text{.5-35}\text{.5.} \\ \therefore l=30.5,h=5,f=14,cf=\text{c}\text{.f}\text{. of preceding class = 18 and}\frac{N}{2}=25 \\ \therefore \text{Median,}M=l+\left\{h\times \frac{\left(\frac{N}{2}-cf\right)}{f}\right\} \\ \text{=30}\text{.5+}\left\{5\times \frac{(25-18)}{14}\right\} \\ \text{=30}\text{.5+2}\text{.5} \\ \text{=33} \\ \text{ Hence, median=33} \end{gathered}$$

Page No 370:

Question 13:

Find the median from the following data:

Class	1-5	6-10	11-15	16-20	21-25	26-30	31-35	35-40	41-45
Frequency	7	10	16	32	24	16	11	5	2

Answer:

Converting into exclusive form, we get:

Class	Oftenness(f)	Cumulative frequency
0.5-5.5	7	7
5.5-10.5	10	17
10.5-15.5	16	33
15.5-20.5	32	65
20.5-25.5	24	89
25.5-30.5	16	105
30.5-35.5	11	116
35.5-40.5	5	121
40.5-45.5	2	123
	N= ∑f=123

$$\begin{gathered} N=123 \\ \Rightarrow \frac{N}{2}=61.5 \\ \text{The cumulative frequency sporting greater than 61}\text{.5 is 65 and the corresponding class is 15}\text{.5-20}\text{.5.} \\ \text{Olibanum, the median grade is 15}\text{.5-20}\text{.5.} \\ \therefore l=15.5,h=5,f=32,cf=\text{c}\text{.f}\text{. of preceding class = 33 and}\frac{N}{2}=61.5 \\ \therefore \text{Median,}M=l+\left\{h\times \frac{\left(\frac{N}{2}-cf\right)}{f}\right\} \\ \text{=15}\text{.5+}\left\{5\times \frac{(61.5-33)}{32}\right\} \\ \text{=15}\text{.5+4}\text{.45} \\ \text{=19}\text{.95} \\ \text{Hence, median=19}\text{.95} \end{gathered}$$

Page No 370:

Question 14:

Find the medial from the following data:

Marks	No. of students
Below 10	12
Below 20	32
Below 30	57
Downstairs 40	80
Below 50	92
Below 60	116
To a lower place 70	164
Below 80	200

Solution:

Class	Additive frequency	Frequency (f)
0-10	12	12
10-20	32	20
20-30	57	25
30-40	80	23
40-50	92	12
50-60	116	24
60-70	164	48
70-80	200	36
		N = ∑f =200

$\begin{array}{l}N=200\\ \Rightarrow \frac{N}{2}=100\\ \text{The cumulative frequency but greater than 100 is 116 and}\mathrm{the corresponding class is 50-60}.\\ \\ \text{Thus, the median class is 50-60.}\\ \therefore l=50,h=10,f=24,cf=\text{c}\text{.f}\text{. of preceding class = 92 and}\frac{N}{2}=100\\ \\ \therefore \text{Normal,}M=l+\left\{h\times \frac{\left(\frac{N}{2}-cf\right)}{f}\right\}\\ \text{= 50 +}\left\{10\times \frac{(100-92)}{24}\right\}\\ \text{= 50 + 3.33}\\ \text{= 53.33}\\ \text{Hence, normal = 53.33}\end{array}$

Page Nobelium 375:

Question 1:

Find the mode of the Simon Marks obtained by 80 students in a year try out in West Germanic language as precondition on a lower floor:

Marks	0-10	10-20	20-3	30-40	40-50	50-60	60-70	70-80
Number of students	3	5	16	12	13	20	6	5

Result:

As the class 50-60 has the maximum frequency, it is the normal grade.$$\begin{gathered} \mathrm{Now},x_k=50,h=10,f_k=20,f_{k-1}=13,f_{k+1}=6 \\ \therefore \mathrm{Mode},M_0=x_k+\left\{h\times \frac{\left(f_k-f_{k-1}\right)}{\left(2f_k-f_{k-1}-f_{k+1}\right)}\right\} \\ =50+\left\{10\times \frac{\left(20-13\right)}{\left(2\times 20-13-6\right)}\right\} \\ =50+\left\{10\times \frac{7}{21}\right\} \\ =\left(50+\frac{10}{3}\right) \\ =\left(50+3.33\right) \\ =53.33 \end{gathered}$$

Hence, mode=53.33

Page No 375:

Question 2:

Find the mode of the ages of 181 workers of a factory from the following frequency distribution:

Age (in years)	20-30	30-40	40-50	50-60	60-70
Number of workers	25	47	62	37	10

Answer:

American Samoa the class 40-50 has the maximum frequency, it is the diatonic scale class.

$$\begin{gathered} \mathrm{Now},x_k=40,h=10,f_k=62,f_{k-1}=47\mathrm{and}{f}_{k+1}=37 \\ \therefore \mathrm{Mode},M_0=x_k+\left\{h\times \frac{\left(f_k-f_{k-1}\right)}{\left(2f_k-f_{k-1}-f_{k+1}\right)}\right\} \\ =40+\left\{10\times \frac{\left(62-47\right)}{\left(2\times 62-47-37\right)}\right\} \\ =40+\left\{10\times \frac{15}{40}\right\} \\ =\left(40+\frac{15}{4}\right) \\ =\left(40+3.75\right) \\ =43.75 \end{gathered}$$

Hence, mode = 43.75 years

Page No 375:

Interrogative 3:

Find the mode of the following distribution:

Class interval	10-14	14-18	18-22	22-26	26-30	30-34	34-38	38-42
Frequency	8	6	11	20	25	22	10	4

Answer:

As the course 26-30 has the maximum relative frequency, IT is the modal class.

$$\begin{gathered} \mathrm{Now},x_k=26,h=4,f_k=25,f_{k-1}=20,f_{k+1}=22 \\ \therefore \mathrm{Manner},M_0=x_k+\left\{h\times \frac{\left(f_k-f_{k-1}\right)}{\left(2f_k-f_{k-1}-f_{k+1}\right)}\right\} \\ =26+\left\{4\times \frac{\left(25-20\right)}{\left(2\times 25-20-22\right)}\right\} \\ =26+\left\{4\times \frac{5}{8}\right\} \\ =\left(26+2.5\right) \\ =28.5 \end{gathered}$$

Page No 375:

Question 4:

Given below is the distribution of tot up household expenditure of 200 manual workers in a city:

Expenditure (in Rs)	No. of manual workers
1000-1500	24
1500-2000	40
2000-2500	31
2500-3000	28
3000-3500	32
3500-4000	23
4000-4500	17
4500-5000	5

Bump the expenditure cooked away maximum act of manual workers.

Solvent:

As the class 1500-2000 has the maximum frequency, it is the modal class.

$$\begin{gathered} \mathrm{Now},x_k=1500,h=500,f_k=40,f_{k-1}=24\mathrm{and}{f}_{k+1}=31 \\ \therefore \mathrm{Modality},M_0=x_k+\left\{h\times \frac{\left(f_k-f_{k-1}\right)}{\left(2f_k-f_{k-1}-f_{k+1}\right)}\right\} \\ =1500+\left\{500\times \frac{\left(40-24\right)}{\left(2\times 40-24-31\right)}\right\} \\ =1500+\left\{500\times \frac{16}{25}\right\} \\ =\left(1500+320\right) \\ =1820 \end{gathered}$$

Hence, mode = Rs 1820

Page No 376:

Question 5:

Estimate the mode from the pursual data:

Monthly salary (in Rs)	No. of employees
0-5000	90
5000-1000	150
10000-15000	100
15000-20000	80
20000-25000	70
25000-30000	10

Answer:

As the class 5000-10000 has the maximum frequency, it is the modal class.$$\begin{gathered} \mathrm{Now},x_k=5000,h=5000,f_k=150,f_{k-1}=90\mathrm{and}{f}_{k+1}=100 \\ \therefore \mathrm{Modality},M_0=x_k+\left\{h\times \frac{\left(f_k-f_{k-1}\right)}{\left(2f_k-f_{k-1}-f_{k+1}\right)}\right\} \\ =5000+\left\{5000\times \frac{\left(150-90\right)}{\left(2\times 150-90-100\right)}\right\} \\ =5000+\left\{5000\times \frac{60}{110}\right\} \\ =\left(5000+2727.27\right) \\ =7727.27 \end{gathered}$$

Hence, mode = Rs 7727.27

Page Nary 376:

Question 6:

Compute the modality from the chase data:

Age (in years)	0-5	5-10	10-15	15-20	20-25	25-30	30-35
Number of patients	6	11	18	24	17	13	5

Answer:

As the grade 15-20 has the uttermost frequency, it is the modal classify.

$$\begin{gathered} \mathrm{Now},x_k=15,h=5,f_k=24,f_{k-1}=18\mathrm{and}{f}_{k+1}=17 \\ \therefore \mathrm{Mode},M_0=x_k+\left\{h\times \frac{\left(f_k-f_{k-1}\right)}{\left(2f_k-f_{k-1}-f_{k+1}\right)}\right\} \\ =15+\left\{5\times \frac{\left(24-18\right)}{\left(2\times 24-18-17\right)}\right\} \\ =15+\left\{5\times \frac{6}{13}\right\} \\ =\left(15+2.3\right) \\ =17.3 \end{gathered}$$

Hence, mode=17.3 years

Page No 376:

Question 7:

Compute the mode from the following series:

Size	45-55	55-65	65-75	75-85	85-95	95-105	105-115
Frequency	7	12	17	30	32	6	10

Answer:

As the class 85-95 has the maximum frequency, it is the modal class.$$\begin{gathered} \mathrm{Immediately},x_k=85,h=10,f_k=32,f_{k-1}=30\mathrm{and}{f}_{k+1}=6 \\ \therefore \mathrm{Mode},M_0=x_k+\left\{h\times \frac{\left(f_k-f_{k-1}\right)}{\left(2f_k-f_{k-1}-f_{k+1}\right)}\right\} \\ =85+\left\{10\times \frac{\left(32-30\right)}{\left(2\times 32-30-6\right)}\right\} \\ =85+\left\{10\times \frac{2}{28}\right\} \\ =\left(85+.71\right) \\ =85.71 \end{gathered}$$

Hence, mode=85.71

Page No 376:

Enquiry 8:

Compute the musical mode from the following data:

Class musical interval	1-5	6-10	11-15	16-20	21-25	26-30	31-35	36-40	41-45	46-50
Frequency	3	8	13	18	28	20	13	8	6	4

Answer:

Clearly, we have to find the mode of the data. The bestowed data is an inclusive serial. So, we will convert it to an exclusive figure as given on a lower floor:

Class interval	0.5-5.5	5.5-10.5	10.5-15.5	15.5-20.5	20.5-25.5	25.5-30.5	30.5-35.5	35.5-40.5	40.5-45.5	45.5-50.5
Absolute frequency	3	8	13	18	28	20	13	8	6	4

As the class 20.5-25.5 has the maximal frequency, it is the modal class.

$$\begin{gathered} \mathrm{Now},x_k=20.5,h=5,f_k=28,f_{k-1}=18\mathrm{and}{f}_{k+1}=20 \\ \therefore \mathrm{Mode},M_0=x_k+\left\{h\times \frac{\left(f_k-f_{k-1}\right)}{\left(2f_k-f_{k-1}-f_{k+1}\right)}\right\} \\ =20.5+\left\{5\times \frac{\left(28-18\right)}{\left(2\times 28-18-20\right)}\right\} \\ =20.5+\left\{5\times \frac{10}{18}\right\} \\ =\left(20.5+2.78\right) \\ =23.28 \end{gathered}$$

Hence, mode=23.28

Page No 377:

Question 1:

Rule the average, mode and median of the following information:

Course	0-10	10-20	20-30	30-40	40-50	50-60	60-70
Frequence	5	10	18	30	20	12	5

Resolution:

We induce the following:

Class	Middle value $\left(x_i\right)$	Absolute frequency $\left(f_i\right)$	Cumulative relative frequency	$f_i{x}_i$
0-10	5	5	5	25
10-20	15	10	15	150
20-30	25	18	33	450
30-40	35	30	63	1050
40-50	45	20	83	900
50-60	55	12	95	660
60-70	65	5	100	325
		$\sum f_i=100$		$f_i{x}_i=3560$

Mean, $\overline{x}$=$\frac{\sum \left(f_i\times x_i\right)}{\sum f_i}$
=$\frac{3560}{100}$
= 35.6.
Hera, $\frac{N}{2}=\frac{100}{2}=50$
The additive frequency just greater than 50 is 63 and the in proportion to class is 30-40.
Thus, the median class is 30-40.

$\therefore l=30,h=10,f=35,c=cf$

of preceding class = 33 and $\frac{N}{2}=50$
Median, $M_e=l+\left\{h\times \frac{\left(\frac{N}{2}-c\right)}{f}\right\}$
$$\begin{gathered} =30+\left\{10\times \frac{\left(50-33\right)}{30}\right\} \\ =\left(30+10\times \frac{17}{30}\right) \\ =\left(30+\frac{17}{3}\right) \\ =30+5.67 \\ =35.67 \end{gathered}$$

∴ Mode = 3(Median value) $-$ 2(Mean)
= (3

$\times 35.67-2\times 35.6$

)
= 35.8

Pageboy No 378:

Interrogation 2:

100 surnames were randomly picked up from a local phonebook and the distribution of number of letters of the English language alphabet in the surnames was obtained as follows:

Number of letters	1-4	4-7	7-10	10-13	13-16	16-19
Number of surnames	6	30	40	16	4	4

Determine the median and mean number of letters in the surname. Also, find the modal size of surnames.

Answer:

We have the following:

Number of letters	Middle treasure$\left(x_i\right)$	Frequence$\left(f_i\right)$	Cumulative frequency	$f_i\times x_i$
1-4	2.5	6	6	15
4-7	5.5	30	36	165
7-10	8.5	40	76	340
10-13	11.5	16	92	184
13-16	14.5	4	96	58
16-19	17.5	4	100	70
		$f_i=100$		$\left(f_i\times x_i\right)=832$

Mean, $\overline{x}$=$\frac{\sum \left(f_i\times x_i\right)}{\sum f_i}$
=$\frac{832}{100}$
=8.32

$$\begin{gathered} \mathrm{Here},N=100 \\ \Rightarrow \frac{N}{2}=50 \end{gathered}$$
The cumulative frequency conscionable greater than 50 is 76 and the related to class is 7-10.
Thusly, the median value class is 7-10.

$\therefore l=7,h=3,f=40,c=cf$

of preceding class = 36 and $\frac{N}{2}=50$
Median, $M_e=l+\left\{h\times \frac{\left(\frac{N}{2}-c\right)}{f}\right\}$
$$\begin{gathered} =7+\left\{3\times \frac{\left(50-36\right)}{40}\right\} \\ =\left(7+3\times \frac{14}{40}\right) \\ =\left(7+\frac{42}{40}\right) \\ =7+1.05 \\ =8.05 \end{gathered}$$

∴ Mode = 3(Median)

$-$

2(Mean)
= (3$\times 8.05-2\times 8.32$)
= 7.51

Page None 378:

Question 3:

A appraise regarding the high (in cm) of 50 girls of a class was conducted and the following information was obtained:

Height in cm	120-130	130-140	140-150	150-160	160-170	Total
Number of girls	2	8	12	20	8	50

Find the mean, median and mode of the above data.

Answer:

We have the following:

Pinnacle in centimeter	Mid value $\left(x_i\right)$	Relative frequency $\left(f_i\right)$	Cumulative frequency	$f_i\times x_i$
120-130	125	2	2	250
130-140	135	8	10	1080
140-150	145	12	22	1740
150-160	155	20	42	3100
160-170	165	8	50	1320
		∑ $f_i=50$		∑ $\left(f_i\times x_i\right)=$ 7490

Mean, $\overline{x}$=$\frac{\sum \left(f_i\times x_i\right)}{\sum f_i}$
=$\frac{7490}{50}$
=149.8

$$\begin{gathered} \mathrm{At\; present},N=50 \\ \Rightarrow \frac{N}{2}=25 \end{gathered}$$
The cumulative frequency just greater than 25 is 42 and the related class is 150-160.
Thus, the median class is 150-160.

$\mathrm{Directly},l=150,h=10,f=20,c=cf$

of outgoing class = 22 and $\frac{N}{2}=25$
$$\begin{gathered} \mathrm{Median},M_e=l+\left\{h\times \frac{\left(\frac{N}{2}-c\right)}{f}\right\} \\ =150+\left\{10\times \frac{\left(25-22\right)}{20}\right\} \\ =\left(150+10\times \frac{3}{20}\right) \end{gathered}$$
=151.5
∴ Musical mode = 3(Median) $-2$(Mean)
= 3$\times 151.5-2\times 149.8$
= 154.9

Page No 378:

Question 4:

The following table gives the daily income of 50 workers of a manufacturing plant:

Daily income (in Rs)	100-120	120-140	140-160	160-180	180-200
Issue of workers	12	14	8	6	10

Bump the mean, mode and median of the above information.

Answer:

We have the following:

Daily income	Mid value $\left(x_i\right)$	Frequency $\left(f_i\right)$	Cumulative absolute frequency	$f_i\times x_i$
100-120	110	12	12	1320
120-140	130	14	26	1820
140-160	150	8	34	1200
160-180	170	6	40	1020
180-200	190	10	50	1900
		∑ $f_i=50$		∑$\left(f_i\times x_i\right)=$7260

Mean, $\overline{x}$=$\frac{\sum \left(f_i\times x_i\right)}{\sum f_i}$
=$\frac{7260}{50}$
=145.2

$$\begin{gathered} \mathrm{Hera},N=50 \\ \Rightarrow \frac{N}{2}=25 \end{gathered}$$
The accumulative frequency just greater than 25 is 26 and the corresponding class is 120-140.
Thus, the median sort is 120-140.

$Now,l=120,h=20,f=14,c=cf$

of prevenient class = 12 and$\frac{N}{2}=25$
Median, $M_e=l+\left\{h\times \frac{\left(\frac{N}{2}-c\right)}{f}\right\}$
$$\begin{gathered} =120+\left\{20\times \frac{\left(25-12\right)}{14}\right\} \\ =\left(120+20\times \frac{13}{14}\right) \end{gathered}$$
=138.57
Mood = 3(Mesial) $-$ 2(mean)
$$\begin{gathered} =3\times 138.57-2\times 145.2 \\ =125.31 \end{gathered}$$

Pageboy Atomic number 102 378:

Interrogative sentence 5:

The table below shows the day-to-day expenditure of food of 30 households in a locality:

Daily expenditure (in Rs)	Number of households
100-150	6
150-200	7
200-250	12
250-300	3
300-350	2

Find the mean and median daily expenditare on food.

Answer:

We have the succeeding:

Daily income	Mid prise	Frequency$\left(f_i\right)$	Cumulative frequency	$f_i\times x_i$
100-150	125	6	6	750
150-200	175	7	13	1225
200-250	225	12	25	2700
250-300	275	3	28	825
300-350	325	2	30	650
		∑ $f_i=30$		∑ $\left(f_i\times x_i\right)=$ 6150

Average, $\overline{x}$=$\frac{\sum \left(f_i\times x_i\right)}{\sum f_i}$
=$\frac{6150}{30}$
=205

$$\begin{gathered} \mathrm{Now},N=30 \\ \Rightarrow \frac{N}{2}=15 \end{gathered}$$

The additive frequency but greater than 15 is 25 and the corresponding class is 200-250.
Thus, the average class is 200-250.

$\mathrm{Now},l=200,h=50,f=12,c=cf$

of preceding class = 13 and $\frac{N}{2}=15$
Medial, $M_e=l+\left\{h\times \frac{\left(\frac{N}{2}-c\right)}{f}\right\}$
$$\begin{gathered} =200+\left\{50\times \frac{\left(15-13\right)}{12}\right\} \\ =\left(200+50\times \frac{2}{12}\right) \end{gathered}$$
=200 + 8.33
=208.33

Sri Frederick Handley Page No 387:

Question 1:

Draw a cumulative frequency curve (to a lesser degree type) for the following data and find the median from it:

Class interval	200-220	220-240	240-260	260-280	280-300	300-320
Frequency	7	3	6	8	2	4

Answer:

From the conferred table, we Crataegus oxycantha prepare the 'less than' frequency table as shown below:

Sort	c.f
Less than 220	7
Less than 240	10
Little than 260	16
Less than 280	24
Less than 300	26
To a lesser degree 320	30

We plot the points A(220,7), B(240,10), C(260,16), D(300,24), E(300,26) and F(320,30).
Join AB, BC, Cadmium, DE, EF and FA with a free hand to get the curve representing the 'less than type' series.

Here, N=30
$\Rightarrow$ $\frac{N}{2}=15$
From P(0,15), draw$PQ$ $\parallel x-\mathrm{axis}$ meeting the arch at Q. Draw$QM$ $\perp OX$ meeting$x-\mathrm{axis}$ at M.
Clear, OM= 256 units
Hence, Median=256

Page No 387:

Question 2:

Favourable is the distribution of marks of 70 students in a periodical test:

Marks	Numeral of students
Marks less than 10 Marks less than 20 Marks less than 30 Marks less than 40 Marks inferior than 50	3 11 28 48 70

Draw a cumulative frequency curve for the above data and find the median.

Result:

$$\begin{gathered} \mathrm{We}\mathrm{plot}\mathrm{the}\mathrm{points}\mathrm{A}(10,3),\mathrm{B(20,11),}\mathrm{C(30,28),}\mathrm{D(40,48)}\mathrm{and}\mathrm{E(50,70)\; .} \\ \mathrm{Conjoin}\mathrm{Av,}\mathrm{BC,}\mathrm{CD,}\mathrm{DE}\mathrm{and}\mathrm{EA}\mathrm{with}\mathrm{a}\mathrm{free}\mathrm{hand}\mathrm{to}\mathrm{have}\mathrm{the}\mathrm{bend}\mathrm{representing}\mathrm{the}\text{'}\mathrm{to\; a\; lesser\; extent}\mathrm{than}\mathrm{type}\text{'}\mathrm{series}. \end{gathered}$$

​

$$\begin{gathered} \mathrm{From}\mathrm{P(0,35)},\mathrm{soak\; up}\mathrm{PQ\; \parallel x}-\mathrm{axis}\mathrm{meeting}\mathrm{the}\mathrm{curve}\mathrm{at}\mathrm{Q.} \\ \mathrm{Draw}\mathrm{QM\perp Wild\; ox}\mathrm{meeting}\mathrm{the}\mathrm{x-axis\; vertebra}\mathrm{a}\mathrm{t}\mathrm{M.} \\ \mathrm{C}\mathrm{learly},\mathrm{OM=\; 33.5}\mathrm{u}\mathrm{nits.} \\ \mathrm{Hence},Median=33.5 \end{gathered}$$

Varlet No 388:

Head 3:

The following table gives the high (in metres) of 360 trees:

Peak	Number of trees
Fewer than 7 m To a lesser degree 14 m To a lesser degree 21 m Less than 28 m Less than 35 m Less than 42 m Less than 49 m To a lesser degree 56 m	25 45 95 140 235 275 320 360

From the above information, attraction an nose cone and observe the median.

Respond:

$$\begin{gathered} \mathrm{We}\mathrm{plot}\mathrm{the}\mathrm{points}\mathrm{A(7,25),}\mathrm{B(14,45),}\mathrm{C(21,95),}\mathrm{D(28,140),}\mathrm{E(35,235)},\mathrm{F(42,275),}\mathrm{G(49,320)}\mathrm{and}\mathrm{H(56,360)\; .} \\ \mathrm{Juncture}\mathrm{AB,}\mathrm{BC,}\mathrm{Cardinal,}\mathrm{DE,}\mathrm{EF,}\mathrm{FG,}\mathrm{GH}\mathrm{and}\mathrm{HA}\mathrm{with}a\mathrm{costless}\mathrm{hand}\mathrm{to}\mathrm{get}\mathrm{the}\mathrm{curve}\mathrm{representing}\mathrm{the}\text{'}\mathrm{less}\mathrm{than}\mathrm{type}\text{'}\mathrm{series}. \end{gathered}$$

$$\begin{gathered} \mathrm{Here,}\mathrm{N=360} \\ \Rightarrow \mathrm{N/}2=180 \\ \mathrm{From}\mathrm{P(0,180)},\mathrm{draw}\mathrm{PQ\; \parallel x}-\mathrm{axis}\mathrm{coming\; together}\mathrm{the}\mathrm{curve}\mathrm{at}\mathrm{Q.} \\ \mathrm{Draw}\mathrm{QM\perp OX}\mathrm{meeting}\mathrm{the}\mathrm{x-axis}\mathrm{at}\mathrm{M.} \\ \mathrm{Clear},\mathrm{OM=\; 32.}5\mathrm{m} \\ \mathrm{Hence},\mathrm{median=32.}5\mathrm{m} \end{gathered}$$

Page No 388:

Question 4:

From the following relative frequency statistical distribution, prepare the 'To a lesser degree Ogive'.

Capital (in Lak of Rs)	List of Companies
0-10	2
10-20	3
20-30	7
30-40	11
40-50	15
50-60	7
60-70	2
70-80	3

Also, Line up the median.

Answer:

From the given table, we may educate the 'less than' frequency table as shown below:

Capital (in Lakhs)	Number of companies
Less than 10	2
To a lesser degree 20	5
Less than 30	12
To a lesser degree 40	23
To a lesser degree 50	38
To a lesser degree 60	45
To a lesser degree 70	47
To a lesser degree 80	50

We secret plan the points A(10,2), B(20,5), C(30,12), D(40,23), E(50,38), F(60,45), G(70,47) and H(80,50).
Join Abdominal, BC, CD, DE, EF, FG, GH and HA with a free hand to get the curve representing the 'to a lesser extent than type' series.

Here, N=50

$\Rightarrow \frac{N}{2}=25$

From P(0,25), draw PQ meeting the curve at Q. Draw QM meeting at M.
Clearly, OM = Rs 42 Lakh
Hence, median = Rs 42 Lakh

Page Nobelium 389:

Question 5:

From the next absolute frequency distribution, prepare the 'More Then Nose cone'.

Score	Numerate of candidates
400-450	20
450-500	35
500-550	40
550-600	32
600-650	24
650-700	27
700-750	18
750-800	24
Total	230

Also find the median.

Answer:

From the precondition table, we may prepare the 'more than' relative frequency postpone as shown down the stairs:

Score	Number of candidates
More than 750	34
Sir Thomas More than 700	52
More than than 650	79
More than 600	103
More than 550	135
More than 500	175
More than 450	210
To a higher degree 400	230

We plot the points A(750,34), B(700,52), C(650,79), D(600,103), E(550,135), F(500,175), G(450,210) and H(400,230).
Join AB, BC, CD, DE, EF, FG, GH and HA with a free paw to commence the wind representing the 'more than typecast' series.

Here, N=230
⇒

$\frac{N}{2}=115$

From P(0,115), suck PQ meeting the curve at Q. Force QM meeting at M.
Clearly, OM = 590 units
Hence, median = 590 units

Page No 389:

Question 6:

The marks obtained by 100 students of a class in an examination are given below:

Marks	Number of students
0-5	2
5-10	5
10-15	6
15-20	8
20-25	10
25-30	25
30-35	20
35-40	18
40-45	4
45-50	2

Draw cumulative oftenness curves by using (i) 'to a lesser degree' series and (cardinal) 'more than' series
Hence, find the median.

Answer:

(i) From the given postpone, we may prepare the 'to a lesser degree' frequency table as shown down the stairs:

Marks	No. of students
Less than 5	2
Less than 10	7
Less than 15	13
Less than 20	21
To a lesser degree 25	31
Less than 30	56
Less than 35	76
Less than 40	94
To a lesser degree 45	98
Less than 50	100

We game the points A(5,2), B(10,7), C(15,13), D(20,21), E(25,31), F(30,56), G(35,76), H(40,94), I(45,98) and J(50,100).
Link Group AB, BC, CD, DE, EF, FG, GH, HI, IJ and JA with a free hand to get the curve ball representing the 'to a lesser degree type' series.

(cardinal) Much series:

Marks	No. of student
More than 0	100
More than 5	98
More than 10	93
More 15	87
More than 20	79
More than 25	69
More than 30	44
More than 35	24
More than 40	6
More than 45	2

Now, on the same chart paper, we plot the points (0,100), (5,98), (10,94), (15,76), (20,56), (25,31), (30,21), (35,13), (40,6) and (45,2).
Join , with a free hand to get the 'more than type' serial.

The deuce curves cross at taper L. Draw $\mathrm{LM}\perp \mathrm{OX}$ edged the $x-axis$at M.
Clearly, M = 29.5
Thu, Median = 29.5

Varlet No 390:

Question 7:

From the following data, draw the two types of additive frequency curves and find out the median:

Height (in cm)	Frequency
140-144	3
144-148	9
148-152	24
152-156	31
156-160	42
160-164	64
164-168	75
168-172	82
172-176	86
176-180	34

Answer:

(i) Inferior than series:

Mark s	No. of students
Less than 144	3
Less than 148	12
To a lesser degree 152	36
Less than 156	67
Less than 160	109
Less than 164	173
Less than 168	248
Less than 172	330
Little than 176	416
Little than 180	450

We plot the points A(144,3), B(148,12), C(152,36), D(156,67), E(160,109) F(164,173), G(168,248), H(172,330), I(176,416) and J(180,450). Join AB, BC, CD, DE, EF, FG, GH, HI, IJ and JA with a free hand to go the curve representing  the 'less than type' series.

(ii) More than series:

Marks	No. of students
More than 140	450
More than 144	447
More than 148	438
More than 152	414
More than 156	383
More than 160	341
Thomas More than 164	277
Sir Thomas More than 168	202
More than 172	120
More than 176	34

Now on the duplicate graphical record wallpaper, we plot the points $A_1$(140,450), $B_1$(144,447), $C_1$(148,438), $D_1$(152,414), $E_1$(156,383), $F_1$(160,341), $G_1$(164,277), $H_1$(168,202), $I_1$(172,120) and $J_1($176,34).
Join $A_1{B}_1,B_1{C}_1,C_1{D}_1,D_1{E}_1,E_1{F}_1,F_1{G}_1,G_1{H}_1,H_1{I}_1\mathrm{and}{I}_1{J}_1$ with a autonomous hand to get the 'much than type' series.

The two curves intersect at point L. Trace $\mathrm{Lumen}\perp \mathrm{OX}$  cutting the $x-axis$at M.
Distinctly, M = 166 centimeter
Hence, Median = 166 cm

Page No 391:

Question 1:

Which of the following is not a measure of key trend?
(a) Mean
(b) Mode
(c) Median
(d) Standard deviation

Solution:

(d) Standard divergence

The standard deviation is a measure of dispersion. Information technology is the action or process of distributing things ended a wide area (nothing about central location).

Page No more 391:

Question 2:

Which of the following cannot be determined diagrammatically?
(a) Mean
(b) Median
(c) Mode
(d) No of these

Reply:

(a) Think of

The base can not be ascertained diagrammatically because the values cannot personify summed.

Page No 392:

Question 3:

The mode of a frequency distribution is obtained diagrammatically from
(a) a frequency curve
(b) a frequency polygon
(c) a histogram
(d) an nose cone

Answer:

The correct option is (c).

The mode of a frequence statistical distribution can be obtained graphically from a histogram.

Page No 392:

Question 4:

The median value of a frequency statistical distribution is found graphically with the help of
(a) a histogram
(b) a frequency curve
(c) a frequency polygon
(d) ogives

Answer:

(d) ogives

​This is because mesial of a oftenness distribution is establish graphically with the help of ogives.

Page No 392:

Wonder 5:

The cumulative frequency table is usable in deciding the
(a) mingy
(b) median
(c) mode
(d) altogether of these

Answer:

The additive frequency table is useful in determining the (b) median.

Page No 392:

Enquiry 6:

The abscissa of the intersection point of the Less Than Type and of the More Than Type accumulative frequency curves of a grouped data gives its
(a) mean
(b) median
(c) mode
(d) no of these

Answer:

The abscissa of the point of intersection of the 'less than case' and that of the 'more than type' cumulative frequency curves of a grouped information gives its (b) normal.

Page No 392:

Question 7:

If x _i' s are the midpoints of the form intervals of a sorted data, f _i' s are the corresponding frequency and$\overline{x}$ is the intend, then $\sum f_i(x_i-\overline{x})=?$
(a) 1
(b) 0
(c) −1
(d) 2

Reply:

$\begin{array}{l}\text{We know that}\overline{x}=\frac{{\displaystyle \sum f_i{x}_i}}{{\displaystyle \sum f_i}}\\ \Rightarrow \overline{x}\sum f_i=\sum f_i{x}_i...\left(\text{i}\right)\\ \text{Now,}\sum f_i(x_i-\overline{x})=\sum f_i{x}_i-\overline{x}\sum f_i\\ \Rightarrow \sum f_i(x_i-\overline{x})=\sum f_i{x}_i-\sum f_i{x}_i[\mathrm{U}\text{sing}\left(\text{i}\right)]\\ \Rightarrow \sum f_i(x_i-\overline{x})=0\\ \text{Hence, the correct option is}\left(\text{b}\right).\end{array}$

Paginate No 392:

Question 8:

For finding the normal by using the formula, $\overline{x}=A+h\left(\frac{\sum f_i{u}_i}{\sum f_i}\right)$, we have u_i=?
(a) $\frac{(A-x_i)}{h}$
(b) $\frac{(x_i-A)}{h}$
(c) $\frac{(A+x_i)}{h}$
(d) $h(x_i-A)$

Page No 392:

Question 9:

In the formula,$\overline{x}=\left\{A+\frac{\sum f_i{d}_i}{\sum f_i}\right\}$ for finding the nasty of the grouped data, the d _i's are the deviations from A of
(a) lower limits of the classes
(b) upper berth limits of the classes
(c) midpoints of the classes
(d) no of these

Answer:

The _{$d_i$'s} are the deviations from $A$ of (c) midpoints of the classes.

Page No 392:

Question 10:

While computing the have in mind of the grouped data, we assume that the frequencies are
(a) evenly distributed over the classes
(b) century at the class marks of the classes
(c) centred at the lower limits of the classes
(d) centred at the pep pill limits of the classes

Solvent:

While calculation the mean of the group data, we wear that the frequencies are (b) centred at the class marks of the classes.

Thomas Nelson Page Nary 392:

Question 11:

The relation between base, mode and median is
(a) mode = (3 × mean) − (2 × median)
(b) mode = (3 × medial) − (2 × stand for)
(c) way = (3 × mean) − (2 × mode)
(d) mode = (3 × median) − (2 × mode)

Answer:

(b) $\mathrm{mode}=\left(3\times \mathrm{median}\right)-\left(2\times \mathrm{mean}\right)$

Page No 393:

Enquiry 12:

Consider the frequency distribution of the high of 60 students of a class

Summit (in cm)	No more. of Students	Cumulative Absolute frequency
150-155	16	16
155-160	12	28
160-165	9	37
165-170	7	44
170-175	10	54
175-180	6	60

The sum of the lower limit of the modal class and the high limit of the median class is
(a) 310
(b) 315
(c) 320
(d) 330

Resolution:

(b) 315
The course having the maximum frequency is the modal class.

$$\begin{gathered} \mathrm{So},\mathrm{}\mathrm{the}\mathrm{}\mathrm{modal}\mathrm{}\mathrm{class}\mathrm{}\mathrm{is}\mathrm{}150-155\mathrm{and}\mathrm{}\mathrm{its}\mathrm{}\mathrm{get\; down}\mathrm{}\mathrm{limit}\mathrm{}\mathrm{is}\mathrm{}150. \\ \mathrm{Also},\mathrm{}\mathrm{N}=60 \\ \Rightarrow \frac{\mathrm{N}}{2}=30\mathrm{} \\ \mathrm{The}\mathrm{}\mathrm{cumulative}\mathrm{}\mathrm{freequency}\mathrm{just}\mathrm{}\mathrm{more}\mathrm{}\mathrm{than}\mathrm{}30\mathrm{}\mathrm{is}\mathrm{}37\mathrm{and}\mathrm{}\mathrm{its}\mathrm{}\mathrm{class}\mathrm{}\mathrm{is}\mathrm{}160-165,\mathrm{whose}\mathrm{}\mathrm{upper}\mathrm{}\mathrm{limit}\mathrm{}\mathrm{is}\mathrm{}165. \\ \therefore \mathrm{Required}\mathrm{}\mathrm{sum}=\left(150+165\right)=315 \end{gathered}$$

Page No 393:

Question 13:

Consider the favorable frequency distribution

Class	0-10	10-20	20-30	30-40	40-50	50-60
Frequency	3	9	15	30	18	5

The modal grade is
(a) 10-20
(b) 20-30
(c) 30-40
(d) 50-60

Answer:

$\left(\mathrm{c}\right)30-40.\mathrm{The}\mathrm{}\mathrm{separate}\mathrm{}30-40\mathrm{}\mathrm{has}\mathrm{}\mathrm{the}\mathrm{utmost}\mathrm{}\mathrm{freequency},\mathrm{i}.\mathrm{e}.,30.\mathrm{So},\mathrm{}\mathrm{the}\mathrm{}\mathrm{modal}\mathrm{}\mathrm{class}\mathrm{}\mathrm{is}\mathrm{}30-40.$

Page No 393:

Call into question 14:

Style = ?
(a) $x_k+h·\left\{\frac{(f_{k-1}-f_k)}{(2f_k-f_{k-1}-f_{k+1})}\right\}$
(b) $x_k+h·\left\{\frac{(f_k-f_{k-1})}{(2f_k-f_{k-1}-f_{k+1})}\right\}$
(c) $x_k+h·\left\{\frac{(f_k-f_{k-1})}{(f_k-2f_{k-1}-f_{k+1})}\right\}$
(d) $x_k+h·\left\{\frac{(f_k-f_{k-1})}{(f_k-f_{k-1}-2f_{k+1})}\right\}$

Answer:

$\left(b\right)x_k+h\left\{\frac{f_k-f_{k-1}}{2f_k-f_{k-1}-f_{k+1}}\right\}$

Page No 393:

Question 15:

Median = ?
(a) $l+\left\{h\times \frac{\left({\displaystyle \frac{\mathrm{N}}{2}}-cf\right)}{f}\right\}$
(b) $l+\left\{h\times \frac{\left(cf-{\displaystyle \frac{\mathrm{N}}{2}}\right)}{f}\right\}$
(c) $l-\left\{h\times \frac{\left({\displaystyle \frac{\mathrm{N}}{2}}-cf\right)}{f}\right\}$
(d) None of these

Page No 394:

Question 16:

If the mean and median of a set of number are 8.9 and 9 respectively, and then the mode will be
(a) 7.2
(b) 8.2
(c) 9.2
(d) 10.2

Answer:

$$\begin{gathered} \left(\mathrm{c}\right)9.2 \\ \mathrm{It}\mathrm{is}\mathrm{relinquished}\mathrm{that}\mathrm{the}\mathrm{}\mathrm{mean}\mathrm{}\mathrm{and}\mathrm{}\mathrm{median}\mathrm{}\mathrm{are}\mathrm{}8.9\mathrm{}\mathrm{a}\mathrm{n}\mathrm{d}\mathrm{}9,\mathrm{}\mathrm{r}\mathrm{e}\mathrm{s}\mathrm{p}\mathrm{e}\mathrm{c}\mathrm{t}\mathrm{i}\mathrm{v}\mathrm{e}\mathrm{l}\mathrm{y}. \\ \therefore \mathrm{M}\mathrm{o}\mathrm{d}\mathrm{e}=\left(3\times \mathrm{M}\mathrm{e}\mathrm{d}\mathrm{i}\mathrm{a}\mathrm{n}\right)-\left(2\times \mathrm{M}\mathrm{e}\mathrm{a}\mathrm{n}\right) \\ \Rightarrow \mathrm{M}\mathrm{o}\mathrm{d}\mathrm{e}=\left(3\times 9\right)-\left(2\times 8.9\right) \\ =27-17.8 \\ =9.2 \end{gathered}$$

Page No 394:

Question 17:

Take the frequency distribution table given below:

Class musical interval	35-45	45-55	55-65	65-75
Frequency	8	12	20	10

The median of the above distribution is
(a) 56.5
(b) 57.5
(c) 58.5
(d) 59

Answer:

(b) 57.5

$\mathrm{Class}\mathrm{}\mathrm{interval}$	$35-45$	$45-55$	$55-65$	$65-75$
$\mathrm{Frequency}$	$8$	$12$	$20$	$10$
$\mathrm{Cumulative}\mathrm{}\mathrm{frequency}$	$8$	$20$	$40$	$50$

$$\begin{gathered} \mathrm{Hera},N=50 \\ \Rightarrow \frac{N}{2}=25,whichliesintheclassintervalof55-65. \\ \mathrm{Now},cf=55,f=20\mathrm{and}l=50 \\ \therefore \mathrm{Median}=l+\left\{h\times \frac{\left(\frac{N}{2}-cf\right)}{f}\right\} \\ =50+\frac{65-55}{20}\times \left(25-20\right) \\ =57.5 \end{gathered}$$

Page No 394:

Oppugn 18:

Weigh the following table:

Class interval	10-14	14-18	18-22	22-26	26-30
Frequence	5	11	16	25	19

The modal value of the above data is
(a) 23.5
(b) 24
(c) 24.4
(d) 25

Response:

(c) 24.4
The maximum frequency is 25 and the modal sort out is 22-26.$$\begin{gathered} \mathrm{Now},x_k=22,f_k=25,f_{k-1}=16,f_{k+1}=19andh=4 \\ \therefore \mathrm{Style}=x_k+h\times \left\{\frac{f_k-f_{k-1}}{2f_k-f_{k-1}-f_{k+!}}\right\} \\ =22+4\times \frac{\left(25-16\right)}{\left(2\times 25-16-19\right)} \\ =22+4\times \frac{\left(25-16\right)}{\left(50-16-19\right)} \\ =22+4\times \frac{9}{15} \\ =22+\frac{12}{5} \\ =22+2.4 \\ =24.4 \end{gathered}$$

Foliate No more 394:

Question 19:

The mean and mode of a frequency dispersion are 28 and 16 respectively. The median is
(a) 22
(b) 23.5
(c) 24
(d) 24.5

Answer:

(c) 24
$$\begin{gathered} \mathrm{Mode}=\left(3\times \mathrm{median}\right)-\left(2\times \mathrm{intend}\right) \\ \Rightarrow \left(3\times \mathrm{Average}\right)=\left(\mathrm{modality}+2\mathrm{think\; of}\right) \\ \Rightarrow \left(3\times \mathrm{Median}\right)=16+56 \\ \Rightarrow \left(3\times \mathrm{Median}\right)=72 \\ \Rightarrow \mathrm{Median}=\frac{72}{3} \\ \therefore \mathrm{Mesial}=24 \end{gathered}$$

Page No 394:

Interrogative sentence 20:

The central and mode of a frequency dispersion are 26 and 29 severally. Then, the entail is
(a) 27.5
(b) 24.5
(c) 28.4
(d) 25.8

Answer:

(b) 24.5
$$\begin{gathered} \mathrm{Modality}=\left(3\times \mathrm{median}\right)-\left(2\times \mathrm{mean}\right) \\ \Rightarrow \left(2\times \mathrm{Mean}\right)=\left(3\times \mathrm{median}\right)-\mathrm{mode} \\ \Rightarrow \left(2\times \mathrm{Imply}\right)=3\times 26-29 \\ \Rightarrow \left(2\times \mathrm{Mean}\right)=49 \\ \Rightarrow \mathrm{Meanspirited}=\frac{49}{2} \\ \therefore \mathrm{Mean}=24.5 \end{gathered}$$

Page No 394:

Question 21:

For a symmetrical frequency distribution, we have:
(a) mean < mode < median
(b) skilled > mode > median
(c) mean = way = median
(d) musical mode = $\frac{1}{2}$ (nasty + median)

Answer:

$\left(c\right)\mathrm{mean}=\mathrm{style}=\mathrm{median}$
A symmetric distribution is peerless where the left and right hand sides of the dispersion are roughly equally balanced just about the mean.

Page No 394:

Interrogation 22:

Look at the cumulative frequence distribution shelve given below:

Monthly income	Phone number of families
More that Rs 10000	100
More that Rs 14000	85
More that Rs 18000	69
More that Rs 20000	50
More that Rs 25000	37
More that Rs 30000	15

Bi of families having income range Rs 20000 to Rs 25000 is
(a) 19
(b) 16
(c) 13
(d) 22

Solution:

Converting the given data into a frequency table, we get:

Monthly income	No. of families	Frequency
30,000 and In a higher place	15	15
25,000-30,000	37	(37 − 15) = 22
20,000-25,000	50	(50 − 37) = 13
18,000-20,000	69	(69 − 50) = 19
14,000-18,000	85	(85 − 69) = 16
10,000-14,000	100	(100 − 85) = 15

Hence, the number of families having an income range of Rs 20,000-Rs 25,000 is 13.
​The correct pick is (c).

Page No 395:

Question 23:

Match the following columns:

Column I	Column II
(a) The virtually frequent value in a data is known as ........ .	(p) standard deviation
(b) Which of the following cannot be determined graphically out of mean, mode and median?	(q) median
(c) An nose cone is exploited to determine ....... .	(r) mean
(d) Out of skilled, mode, median and standard deviation, which is not a measure of focal tendency?	(s) mode

Answer:

Column I	Column II
(a) The most shop value in a data is titled ........ .	(s) mode
(b) Which of the following cannot be determined diagrammatically out of mean, mode and median?	(r) mean
(c) An ogive is used to determine ....... .	(q) median
(d) Out of mean, mode, median and standard deviation, which is not a measure of central disposition?	(p) standard deviation

Pageboy No 395:

Question 24:

Assertion (A)
If the median and fashion of a frequency distribution are 150 and 154 respectively, then its mean is 148.
Reason (R)
Mean, central and mode of a frequency distribution are related as:

(a) Both Asseveration (A) and Reason (R) are even and Reason (R) is a correct account of Statement (A).
(b) Both Assertion  (A) and Reason (R) are true but Reason (R) is a non a correct account of Assertion (A).
(c) Assertion (A) is true and Understanding (R) is false.
(d) Assertion (A) is false and Reason (R) is true.

Answer:

(a) Some Assertion (A) and Reason (R) are geographic and Reason (R) is a even out explanation of Assertion (A).
Distinctly, reason (R) is true.
Using the relation given in reason (R), we have:
$$\begin{gathered} 2\mathrm{mean}=\left(3\times \mathrm{median}\right)-\mathrm{mode}=\left(3\times 150\right)-154=450-154=296 \\ \Rightarrow \mathrm{Mean}=148,\mathrm{}\mathrm{which}\mathrm{}\mathrm{is}\mathrm{}\mathrm{genuine}. \\ \therefore \mathrm{This}\mathrm{}\mathrm{assertion}\mathrm{}\left(\mathrm{A}\right)\mathrm{}\mathrm{and}\mathrm{}\mathrm{ground}\left(\mathrm{R}\right)\mathrm{}\mathrm{are}\mathrm{}\mathrm{both}\mathrm{}\mathrm{true}\mathrm{}\mathrm{and}\mathrm{}\mathrm{reason}\mathrm{}\left(\mathrm{R}\right)\mathrm{is}\mathrm{}\mathrm{the}\mathrm{}\mathrm{correct}\mathrm{}\mathrm{account}\mathrm{}\mathrm{of}\mathrm{}\mathrm{assertion}\mathrm{}\left(\mathrm{A}\right). \end{gathered}$$

Page No 395:

Question 25:

Statement (A)
Consider the following frequency distribution:

Class musical interval	3-6	6-9	9-12	12-15	15-18	18-21
Frequency	2	5	21	23	10	12

The modality of the to a higher place data is 12.4.

Reason (R)
The value of the variable which occurs nigh often is the style.

(a) Both Assertion (A) and Reason (R) are geographic and Reason (R) is a chasten explanation of Assertion (A).
(b) Both Assertion  (A) and Reason (R) are rightful merely Reason (R) is a non a correct explanation of Assertion (A).
(c) Assertion (A) is dead on target and Understanding (R) is dishonest.
(d) Assertion (A) is unharmonious and Argue (R) is straight.

Answer:

(b) Both Statement (A) and Reason (R) are true, merely Reason (R) is a not a correct explanation of Assertion (A).
Clearly, ground (R) is true.
The maximum frequency is 23 and the modal class is 12-15.
$$\begin{gathered} \mathrm{Now},x_k=12,f_k=23,\mathrm{}{f}_{k\mathit{-}\mathit{1}}=21,f_{k\mathit{+}\mathit{1}}=23\mathrm{and}\mathrm{}\mathrm{h}=3 \\ \therefore \mathrm{Mode}=\left\{12+3\times \frac{\left(23-21\right)}{\left(2\times 23-21-10\right)}\right\} \\ =\left(12+3\times \frac{2}{15}\right) \\ =\left(12+0.4\right) \\ =12 \\ \therefore \mathrm{A}\mathrm{s}\mathrm{s}\mathrm{e}\mathrm{r}\mathrm{t}\mathrm{i}\mathrm{o}\mathrm{n}\mathrm{}\left(\mathrm{A}\right)\mathrm{i}\mathrm{s}\mathrm{}\mathrm{t}\mathrm{r}\mathrm{u}\mathrm{e}. \\ \mathrm{Yet},\mathrm{r}\mathrm{e}\mathrm{a}\mathrm{s}\mathrm{o}\mathrm{n}\mathrm{}\left(\mathrm{R}\right)\mathrm{i}\mathrm{s}\mathrm{}\mathrm{n}\mathrm{o}\mathrm{t}\mathrm{}\mathrm{a}\mathrm{}\mathrm{c}\mathrm{o}\mathrm{r}\mathrm{r}\mathrm{e}\mathrm{c}\mathrm{t}\mathrm{}\mathrm{e}\mathrm{x}\mathrm{p}\mathrm{l}\mathrm{a}\mathrm{n}\mathrm{a}\mathrm{t}\mathrm{i}\mathrm{o}\mathrm{n}\mathrm{}\mathrm{o}\mathrm{f}\mathrm{}\mathrm{a}\mathrm{s}\mathrm{s}\mathrm{e}\mathrm{r}\mathrm{t}\mathrm{i}\mathrm{o}\mathrm{n}\mathrm{}\left(\mathrm{A}\right). \end{gathered}$$

Page No 398:

Question 1:

If the think of a data is 27 and its medial is 33. Then, the mode is
(a) 30
(b) 43
(c) 45
(d) 47

Do:

(c) 45
$$\begin{gathered} \mathrm{Here},\mathrm{mean}=27\mathrm{and}\mathrm{mode}=33 \\ \therefore \mathrm{Modality}=3\mathrm{Median}-2\mathrm{Stingy} \\ =3\times 33-2\times 27 \\ =99-54 \\ =45 \end{gathered}$$

Page Atomic number 102 398:

Question 2:

Which measure of central tendency is obtained graphically every bit the x coordinate of the point of crossing of the two ogives?
(a) Skilled
(b) Median
(c) Mode
(d) None of these

Pageboy No 398:

Question 3:

For the following distribution:

Year	0-5	5-10	10-15	15-20	20-25
Absolute frequency	10	15	12	20	9

The sum of the lower limits of the median social class and the modal class is
(a) 15
(b) 25
(c) 30
(d) 35

Answer:

(b) 25

Class	Frequency	Cumulative frequency
0-5	10	10
5-10	15	25
10-15	12	37
15-20	20	57
20-25	9	63

Now, N=63
$\therefore \frac{N}{2}=32.5$
The additive frequency just greater than 32.5 is 37 and the corresponding class is 10-15.
$\therefore \mathrm{Median}\mathrm{class}=10-15$
Here, the highest frequency is 20 and its corresponding class is 15-20.
$\therefore \mathrm{Average}\mathrm{Class}=15-20$
Sum of the lower limits of the median value class and modal class = 10 + 15 = 25

Page No 398:

Question 4:

Consider the chase frequence distribution:

Class	0-5	6-11	12-17	18-23	24-29
Absolute frequency	13	10	15	8	11

The maximum of the median class is
(a) 16.5
(b) 18.5
(c) 18
(d) 17.5

Answer:

(d) 17.5
Converting the tending serial publication into uninterrupted serial publication, we get:

Class	Frequency	Cumulative oftenness
0.5-5.5	13	13
5.5-11.5	10	23
11.5-17.5	15	38
17.5-23.5	8	46
23.5-29.5	11	57

Now, N =57
$\therefore \frac{N}{2}=28.5$

The cumulative absolute frequency just greater than 28.5 is 38 and its corresponding class is 11.5-17.5.
∴ The median class is 11.5-17.5 and the related to maximum is 17.5.

Pageboy No 398:

Question 5:

Write down the formula showing the relation between think, median and mode.

Do:

The formula that shows the relation between mean, median and mode is given below:
$\mathrm{Manner}=3\mathrm{Median}-2\mathrm{Mean}$

Page No 398:

Question 6:

If the mean and mode of a oftenness distribution be 53.4 and 55.2 respectively, find the median.

Answer:

Surrendered:
$$\begin{gathered} \mathrm{Tight}=53.4 \\ \mathrm{Style}=55.2 \\ \mathrm{We}\mathrm{know}\mathrm{that}\mathrm{fashion}=3\mathrm{median}-2\mathrm{imply} \\ \Rightarrow 55.2=3\mathrm{median}-2\times 53.4 \\ \Rightarrow 3\mathrm{Median}=55.2+106.8 \\ \Rightarrow \mathrm{Median}=\frac{162}{3}=54 \end{gathered}$$

Page No 398:

Question 7:

In the table given below, the multiplication stolen by 120 athletes to melt down a 100 m hurdles are given:

Family	13.8-14	14-14.2	14.2-14.4	14.4-14.6	14.6-14.8	14.8-15
Frequency	2	4	15	54	25	20

Find the number of athletes who completed the race in to a lesser degree 14.6 seconds.

Answer:

Conferred distribution table behind be scripted as following:

Class	Frequency	Additive frequency
To a lesser degree 14	2	2
Less than 14.2	4	6
To a lesser degree 14.4	15	21
Less than 14.6	54	75
Less than 14.8	25	100
To a lesser degree 15	20	120

Number of athletes who completed the race in less than 14.6 sec = 75

Page No 399:

Question 8:

Conceive the following frequency distribution:

Classify	0-5	6-11	12-17	18-23	24-29
Absolute frequency	13	10	15	8	11

Find the maximum of the mesial assort.

Answer:

Class	Frequency	Additive frequence
0.5-5.5	13	13
5.5-11.5	10	23
11.5-17.5	15	38
17.5-23.5	8	46
23.5-29.5	11	57

 N=57
$\therefore \frac{N}{2}=28.5$

The additive frequency just greater than 28.5 is 38 and its corresponding class is 11.5-17.5.

$$\begin{gathered} \mathrm{Now},\mathrm{we}\mathrm{have}: \\ \mathrm{Median}\mathrm{class}=11.5-17.5 \\ \mathrm{Upper}\mathrm{limit}=17.5 \end{gathered}$$

Page No 399:

Question 9:

Find the mean of the pursuing frequency distribution:

Class	1-3	3-5	5-7	7-9
Frequency	9	22	27	18

Answer:

We have the followers table:

Class	Mid value $\left(x_i\right)$	Frequency $\left(f_i\right)$	$f_i{x}_i$
1-3	2	9	18
3-5	4	22	88
5-7	6	27	162
7-9	8	18	144
		$\sum f_i=76$	$\sum f_i{x}_i=412$

$$\begin{gathered} \therefore \mathrm{Mean},\overline{x}=\frac{\sum f_i{x}_i}{\sum f_i} \\ =\frac{412}{76} \\ =5.42 \end{gathered}$$

Page No 399:

Call into question 10:

The maximum bowling speeds (in kilometres per hour) of 33 players at a cricket coaching job centre are given below:

Accelerate in km/h	85-100	100-115	115-130	130-145
No. of players	10	4	7	9

Calculate the median bowling speed.

Response:

Speed (km/h)	No. of players $\left(f_i\right)$	c.f
85-100	10	10
100-115	4	14
115-130	7	21
130-145	9	30

N =30
$\therefore \frac{N}{2}=15$
The cumulative oftenness just greater than 15 is 21.
$\therefore \mathrm{Average}\mathrm{Grade}=115-130$
$\mathrm{i}.\mathrm{e}.,{\mathrm{x}}_{\mathrm{k}}=115,\mathrm{h}=15,\mathrm{f}=7,\mathrm{c}=\mathrm{c}.\mathrm{f}\mathrm{of}\mathrm{the}\mathrm{preceeding}\mathrm{course}=14,\frac{\mathrm{N}}{2}=15$
$$\begin{gathered} \therefore M_e=x_k+\left\{h\times \frac{\left(\frac{N}{2}-c\right)}{f}\right\} \\ =115+\left(15\times \frac{15-14}{7}\right) \\ =115+2.1 \\ =117.1 \end{gathered}$$
Hence, the required speed is 117.1 km/h.

Page No 399:

Enquiry 11:

The expected value of the following frequency distribution is 50.

Sort out	0-10	10-20	20-30	30-40	40-50
Frequency	16	p	30	32	14

Find the economic value of p.

Answer:

Class	Mid value $\left(x_i\right)$	Frequency $\left(f_i\right)$	$\left(x_i\times f_i\right)$
0-10	5	16	80
10-20	15	$p$	15 $p$
20-30	25	30	750
30-40	35	32	1120
40-50	45	14	630
		$\sum f_i=p+92$	$\sum f_i{x}_i=15p+2580$

$$\begin{gathered} \therefore \mathrm{Mean},\overline{\mathrm{x}}=\frac{\sum f_i{x}_i}{\sum f_i} \\ =\frac{15p+2580}{p+92} \\ \mathrm{Today},\frac{15p+2580}{p+92}=50 \\ \Rightarrow 50p+4600=15p+2580 \\ \Rightarrow 35p=-2020 \\ \Rightarrow p=-\frac{404}{7}\left(\mathrm{absolute\; frequency}\mathrm{}\mathrm{cannot}\mathrm{}\mathrm{represent}\mathrm{}\mathrm{negative}\right) \end{gathered}$$

So, there is an misplay dubious.

Thomas Nelson Page No 399:

Question 12:

Find the median of the following relative frequency distribution:

Course of study	0-10	10-20	20-30	30-40	40-50
Frequency	2	12	22	8	6

Answer:

Class	Frequency $\left(f_i\right)$	Cumulative frequency $\left(c.f\right)$
0-10	2	2
10-20	12	14
20-30	22	36
30-40	8	44
40-50	6	50

$$\begin{gathered} N=50 \\ \therefore \frac{N}{2}=25 \end{gathered}$$

The accumulative frequency just greater than 25 is 36.

$$\begin{gathered} \therefore \mathrm{Median}\mathrm{assort}=20-30 \\ \mathrm{i}.\mathrm{e}.,x_k=20,h=10,f=22,\mathrm{c}=\mathrm{c}.\mathrm{f}\mathrm{of}\mathrm{the}\mathrm{preceeding}\mathrm{class}=14,\frac{N}{2}=25 \\ \therefore M_e=x_k+h\times \frac{\frac{N}{2}-c}{f} \\ =20+(10\times \frac{25-14}{22}) \\ =20+5 \\ =25 \end{gathered}$$

Varlet No 399:

Interrogation 13:

Calculate the mode of the following frequency distribution:

Marks	0-10	10-20	20-30	30-40	40-50
Number of students	6	16	30	9	4

Answer:

As the class 20-30 has the maximum frequency, it is the modal auxiliary verb class.$$\begin{gathered} \therefore x_k=20,h=10,f_k=30,f_{k-1}=16\mathrm{and}{f}_{k+1}=9 \\ \mathrm{Mode},M_0=x_k+\left\{h\times \frac{\left(f_k-f_{k-1}\right)}{\left(2f_k-f_{k-1}-f_{k+1}\right)}\right\} \\ =20+\left\{10\times \frac{\left(30-16\right)}{\left(2\times 30-16-9\right)}\right\} \\ =20+\left\{10\times \frac{14}{35}\right\} \\ =\left(20+4\right) \\ =24 \end{gathered}$$

Foliate No 399:

Question 14:

Pursuing is the distribution of marks of 70 students in a periodical test:

Marks	To a lesser extent than 10	To a lesser degree 20	Less than 30	To a lesser degree 40	Less than 50
Routine of students	3	11	28	48	70

Draw a cumulative relative frequency for the above data.

Answer:

 Army of the Pure us diagram the points A(10,3), B(20,11), C(30,28), D(40,48) and E(50,70).
Now, let the States joint Group AB, BC, Cadmium and DE with a free hand to get the twist representing the 'to a lesser degree character' series.

Page No 400:

Question 15:

The following distribution gives the daily income of 50 workers of a factory:

Daily income (in Rs)	100-120	120-140	140-160	160-180	180-200
Number of workers	12	14	8	6	10

Write the above distribution as less than typecast cumulative frequency distribution.

Answer:

Daily income (in Rs.)	No. of workers (f)	Additive frequency (c.f)
Less than 120	12	12
To a lesser extent than 140	14	26
Less than 160	8	34
Less than 180	6	40
Less than 200	10	50

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Question 16:

Find the mode of the following distribution of marks obtained by 80 students:

Marks obtained	0-10	10-20	20-30	30-40	40-50
Number of students	6	10	12	32	20

Answer:

Every bit the class 30-40 has the maximum frequency, it is the modal class.$$\begin{gathered} \therefore x_k=30,h=10,f_k=32,f_{k-1}=12,f_{k+1}=20 \\ \mathrm{Mode},M_0=x_k+\left\{h\times \frac{\left(f_k-f_{k-1}\right)}{\left(2f_k-f_{k-1}-f_{k+1}\right)}\right\} \\ =30+\left\{10\times \frac{\left(32-12\right)}{\left(2\times 32-12-20\right)}\right\} \\ =30+\left\{10\times \frac{20}{32}\right\} \\ =\left(30+6.25\right) \\ =36.25 \end{gathered}$$

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Doubtfulness 17:

Find the mean of the pursual data victimisation step deviation method:

Separate	0-10	10-20	20-30	30-40	40-50
Frequency	7	12	13	10	8

Respond:

Paginate No 400:

Question 18:

Find the central of the following information:

Class	0-10	10-20	20-30	30-40	40-50	50-60	60-70	70-80	80-90	90-100
frequency	5	3	4	3	3	4	7	9	7	8

Answer:

Class	Frequency$\left(f_i\right)$	Cumulative relative frequency$\left(c.f\right)$
0-10	5	5
10-20	3	8
20-30	4	12
30-40	3	15
40-50	3	18
50-60	4	22
60-70	7	29
70-80	9	38
80-90	7	45
90-100	8	53

$$\begin{gathered} \mathrm{Here},N=53 \\ \Rightarrow \frac{N}{2}=26.5 \end{gathered}$$
The accumulative frequency just greater than 26.5 is 29 and the corresponding sort out is 60-70.
Thus, the median social class is 60-70.
$\therefore l=60,h=10,f=7,\mathrm{c}=c.f\mathrm{}\mathrm{of}\mathrm{preceeding}\mathrm{course}=22\mathrm{and}\frac{N}{2}=26.5$
$$\begin{gathered} \mathrm{Straight\; off},\mathrm{median},M_e=l+\left\{h\times \frac{\left(\frac{N}{2}-c\right)}{f}\right\} \\ =60+\left\{10\times \frac{\left(26.5-22\right)}{7}\right\} \\ =\left(60+10\times \frac{4.5}{7}\right) \\ =\left(60+\frac{45}{7}\right) \\ =60+6.43 \\ =66.43 \end{gathered}$$

Foliate No 400:

Question 19:

The following prorogue gives the production yield per hectare of wheat berry of 100 farms of a village.

Production yield in kg/hectare	50-55	55-60	60-65	65-70	70-75	75-80
Issue of farms	2	8	12	24	38	16

Change the above distribution to more than typecast distribution and draw its ogive .

Answer:

Production yield in kg/hectare	Number of farms
Much than 50	100
More than 55	98
More 60	90
More 65	78
More than 70	54
More than 75	16

We plot the points A(50,100), B(55,98), C(60,90), D(65,78), E(70,54) and F(75,16).
Join AB, Before Christ, Cadmium, DE, EF and FA with a loos hand to get the curve representing the 'more than' curve.

$$\begin{gathered} \mathrm{Here},N=100 \\ \Rightarrow \frac{N}{2}=50 \end{gathered}$$

From (0,50) draw PQ meeting the curve at Q. Take out QMmeeting at M.
Clearly, OM=70 kilo/hectare
∴ Mesial=70 kg/hectare

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Question 20:

Find the mood of the following frequency dispersion:

Class interval	0-4	4-8	8-12	12-16
Absolute frequency	4	8	5	6

Answer:

A the class interval 4-8 has the maximum frequency, IT is the average separate.$$\begin{gathered} \therefore x_k=4,h=4,f_k=8,f_{k-1}=4,f_{k+1}=5 \\ \mathrm{Mode},M_0=x_k+\left\{h\times \frac{\left(f_k-f_{k-1}\right)}{\left(2f_k-f_{k-1}-f_{k+1}\right)}\right\} \\ =4+\left\{4\times \frac{\left(8-4\right)}{\left(2\times 8-4-5\right)}\right\} \\ =4+\left\{4\times \frac{4}{7}\right\} \\ =\left(4+2.29\right) \\ =6.29 \end{gathered}$$

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