NCERT solution for class 10 maths statistics ( Chapter 14)

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Solution for Exercise - 14.1

1. A survey was conducted by a group of students as a part of their environment awareness program, in which they collected the following data regarding the number of plants in 20 houses in a locality. Find the mean number of plants per house.
Number of Plants 0-2 2-4 4-6 6-8 8-10 10-12 12-14
Number of Houses 1 2 1 5 6 2 3

Which method did you use for finding the mean, and why?

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Answer :


In order to find the mean value, we will use direct method because the numerical value of \(f_i \) and \( x_i \) are small.

The midpoint of the given interval is found by formula: \(Midpoint \ (x_i) \ = \ \frac{upper \ limit \ + \ lower \ limit}{2} \)

No. of plants (Class interval) No. of houses (Frequency \( (f_i) \) ) Mid-point \((x_i)\) \(f_i x_i \)
0-2 1 1 1
2-4 2 3 6
4-6 1 5 5
6-8 5 7 35
8-10 6 9 54
10-12 2 11 22
12-14 3 13 39
\( \sum \ f_i \ = \ 20 \) \( \sum \ f_i x_i \ = \ 162 \)

The formula to find the mean is:    \( \overline{x} \ = \ \frac{ \sum \ f_i x_i}{ \sum f_i} \)

\( = \ \frac{162}{20} \ = \ 8.1 \)

∴ The mean number of plants per house is 8.1

2. Consider the following distribution of daily wages of 50 workers of a factory.
Daily wages (in Rs.) 100-120 120-140 140-160 160-180 180-200
Number of workers 12 14 8 6 10
Find the mean daily wages of the workers of the factory by using an appropriate method.

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Answer :


The midpoint of the given interval is found by formula: \(Midpoint \ (x_i) \ = \ \frac{upper \ limit \ + \ lower \ limit}{2} \)

In this case, the value of mid-point ( \(x_i\)) is very large,so let us assume the mean value, A = 150 and class interval is h = 20.

So, \(u_i \ = \ \frac{x_i \ - \ A}{h} \ => \ u_i \ = \ \frac{x_i \ - \ 150}{20} \)

Daily wages (Class interval) Number of workers (frequency (\(f_i\)) ) Mid-point ( \(x_i \) ) \( u_i \ = \ \frac{x_i \ - \ 150}{20} \) \(f_i u_i \)
100-120 12 110 -2 -24
120-140 14 130 -1 -14
140-160 8 150 0 0
160-180 6 170 1 6
180-200 10 190 2 20
Total \( \sum f_i \ = \ 50 \) \( \sum f_i u_i \ = \ -12 \)

So, the formula to find out the mean is : \( \ \overline{x} \ = \ A \ + h \frac{ \sum \ f_i x_i}{ \sum f_i} \)
\( = \ 150 \ + \ 20 \ × \ \frac{-12}{50} \)
\(= \ 150 \ - \ 4.8 \)
\(= \ 145.20 \)
∴ Mean daily wage of the workers = Rs. 145.20

3. The following distribution shows the daily pocket allowance of children of a locality. The mean pocket allowance is Rs 18. Find the missing frequency f.
Daily Pocket Allowence(in Rs) 11-13 13-15 15-17 17-19 19-21 21-23 23-35
Number of children 7 6 9 13 f 5 4

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Answer :

Given, \( \overline{x} \ = \ 18 \)

Class interval Number of children (frequency (\(f_i\)) ) Mid-point \( (x_i) \) \(f_i u_i \)
11-13 7 12 84
13-15 6 14 84
15-17 9 16 144
17-19 13 18 234
19-21 f 20 20f
21-23 5 22 110
23-25 4 24 96
Total \( \sum f_i \ = \ 44 \ + \ f \) \( \sum f_i x_i \ = \ 752 \ + \ 20f \)

Thus, \( \ \overline{x} \ = \ \frac{ \sum \ f_i x_i}{ \sum f_i} \)

\( => \ 18 \ = \ \frac{752 \ + \ 20f}{44 \ + \ f} \)

\(=> \ 18(44 \ + \ f) \ = \ (752 \ + \ 20f) \)

\(=> \ 792 \ + \ 18f \ = \ 752 \ + \ 20f \)

\(=> \ 792 \ - \ 752 \ = \ 20f \ - \ 18f \ => \ 40 \ = \ 2f \)

\( => \ f \ = \ 20 \)

So, the missing frequency, f = 20.

4. Thirty women were examined in a hospital by a doctor and the number of heartbeats per minute were recorded and summarized as follows. Find the mean heart beats per minute for these women, choosing a suitable method.
Number of heartbeats per minute 65-68 68-71 71-74 74-77 77-80 80-83 83-86
Number of women 2 4 3 8 7 4 2

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Answer :

From the given data, let us assume the mean as A = 75.5
Class size (h) = 3

So, \(u_i \ = \ \frac{x_i \ - \ A}{h} \ => \ u_i \ = \ \frac{x_i \ - \ 75.5}{h} \)

Class Interval Number of women ( \( f_i \) ) Mid-point ( \(x_i \) ) \(u_i \ = \ \frac{x_i \ - \ 75.5}{h} \) \( f_i u_i \)
65-68 2 66.5 -3 -6
68-71 4 69.5 -2 -8
71-74 3 72.5 -1 -3
74-77 8 75.5 0 0
77-80 7 78.5 1 7
80-83 4 81.5 3 8
83-86 2 84.5 3 6
\( \sum f_i \ = 30 \) \( \sum f_i u_i \ = 4 \)

\( \overline{x} \ = \ A \ + h \frac{ \sum \ f_i x_i}{ \sum f_i} \)

\(= 75.5 \ + \ 3 \ × \ \frac{4}{30} \ = \ 75.5 \ + \ \frac{4}{10} \ = \ 75.9 \)

∴ The mean heart beats per minute for these women is 75.9

5. In a retail market, fruit vendors were selling mangoes kept in packing boxes. These boxes contained varying number of mangoes. The following was the distribution of mangoes according to the number of boxes.
Number of mangoes 50-52 53-55 56-58 59-61 62-64
Number of boxes 15 110 135 115 25
Find the mean number of mangoes kept in a packing box. Which method of finding the mean did you choose?

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Answer :

Since, the given data is not continuous so we add 0.5 to the upper limit and subtract 0.5 from the lower limit as the gap between two intervals are 1

Here, assumed mean (A) = 57, Class size (h) = 3

Here, the step deviation is used because the frequency values are big.

Class Interval Number of boxes \( (f_i) \) Mid-point \( (x_i) \) \( d_i \ = \ x_i \ - \ A \) \( f_i d_i \)
49.5-52.5 15 51 -6 90
52.5-55.5 110 54 -3 -330
55.5-58.5 135 57 0 0
58.5-61.5 115 60 3 345
61.5-64.5 25 63 6 150
\( \sum f_i \ = \ 400 \) \( \sum f_i d_i \ = \ 75 \)

The formula to find out the Mean is:

\( \overline{x} \ = \ A \ + h \frac{ \sum \ f_i d_i}{ \sum f_i} \ = \ 57 \ + \ 3 × \frac{75}{400} \)

\(= \ 57 \ + \ 0.1875 \ = \ 57.19 \)

∴ The mean number of mangoes kept in a packing box is 57.19

6. The table below shows the daily expenditure on food of 25 households in a locality. Find the mean daily expenditure on food by a suitable method.
Daily expenditure(in Rs) 100-150 150-200 200-250 250-300 300-350
Number of households 4 5 12 2 2

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Answer :

Let us assume the mean (A) = 225, Class size (h) = 50

Class Interval Number of households \( ( f_i ) \) Mid-point \( ( x_i ) \) \( d_i \ = \ x_i \ - \ A \) \( u_i \ = \ \frac{d_i}{50} \) \( f_i u_i \)
100-150 4 125 -100 -2 -8
150-200 5 175 -50 -1 -5
200-250 12 225 0 0 0
250-300 2 275 50 1 2
300-350 2 325 100 2 4
\( \sum f_i \ = \ 25 \) \( \sum f_i u_i \ = \ -7 \)

\( \overline{x} \ = \ A \ + h \frac{ \sum \ f_i u_i}{ \sum f_i} \)

\( = \ 225 \ + \ 50 × \frac{-7}{25} \)

\(= \ 225 \ - \ 14 \ = \ 211 \)

∴ The mean daily expenditure on food is 211

7. To find out the concentration of SO2 in the air (in parts per million, i.e., ppm), the data was collected for 30 localities in a certain city and is presented below:
Concentration of SO2 ( in ppm) Frequency
0.00 – 0.04 4
0.04 – 0.08 9
0.08 – 0.12 9
0.12 – 0.16 2
0.16 – 0.20 4
0.20 – 0.24 2
Find the mean concentration of SO2 in the air.

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Answer :

Calculate mean by direct method:

Concentration of SO2 (in ppm) Frequency \( ( f_i ) \) Mid-point \( (x_i) \) \( f_i x_i \)
0.00 - 0.04 4 0.02 0.08
0.04 - 0.08 9 0.06 0.54
0.08 - 0.12 9 0.10 0.90
0.12 - 0.16 2 0.14 0.28
0.16 - 0.20 4 0.18 0.72
0.20 - 0.24 2 0.20 0.40
Total \( \sum f_i \ = \ 30 \) \( \sum f_i x_i \ = \ 2.96 \)

The formula to find out the mean is

\( \overline{x} \ = \ \frac{ \sum \ f_i x_i}{ \sum f_i} \)

\( = \ \frac{2.96}{30} \)

\( = \ 0.099 \) ppm

∴ The mean concentration of SO2 in air is 0.099 ppm.

8. A class teacher has the following absentee record of 40 students of a class for the whole term. Find the mean number of days a student was absent.
Number of days 0-6 6-10 10-14 14-20 20-28 28-38 38-40
Number of students 11 10 7 4 4 3 1

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Answer :

Class interval Frequency \( (f_i) \) Mid-point \( (x_i) \) \( f_i x_i \)
0-6 11 3 33
6-10 10 8 80
10-14 7 12 84
14-20 4 17 68
20-28 4 24 96
28-38 3 33 99
38-40 1 39 39
\( \sum f_i \ = \ 40 \) \( \sum f_i x_i \ = \ 499 \)

The mean formula is,

\( \overline{x} \ = \ \frac{ \sum \ f_i x_i}{ \sum f_i} \)

\( = \ \frac{499}{40} \)

\(= \ 12.48 \) days

∴ The mean number of days a student was absent = 12.48.

9. The following table gives the literacy rate (in percentage) of 35 cities. Find the mean literacy rate.
Literacy rate (in %) 45-55 55-65 65-75 75-85 85-98
Number of cities 3 10 11 8 3

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Answer :

In this case, the value of mid-point (xi) is very large,

so let us assume the mean value, A = 70 and class interval is h = 10.

So, \( u_i \ = \ \frac{x_i \ - \ A}{h} \)

\(=> \ u_i \ = \ \frac{x_i \ - \ 70}{10} \)

Class Interval Frequency \( (f_i) \) \( (x_i) \) \( d_i \ = \ x_i \ - \ A \) \(u_i \ = \ \frac{d_i}{h} \) \( f_i u_i \)
45-55 3 50 -20 -2 -6
55-65 10 60 -10 -1 -10
65-75 11 70 0 0 0
75-85 8 80 10 1 8
85-95 3 90 20 2 6
\( \sum f_i \ = \ 35 \) \( \sum f_i u_i \ = \ -2 \)

So,

\( \overline{x} \ = \ A \ + h \frac{ \sum \ f_i x_i}{ \sum f_i} \)

\( = \ 70 \ + \ \frac{-2}{35} \ × \ 10 \)

\(= \ 69.42 \)

∴ The mean literacy part is 69.42

Solution for Exercise - 14.2

1. The following table shows the ages of the patients admitted in a hospital during a year:
Age (in years) 5-15 15-25 25-35 35-45 45-55 55-65
Number of patients 6 11 21 23 14 5

Find the mode and the mean of the data given above. Compare and interpret the two measures of central tendency.

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Answer :


The class 35-45 has the maximum frequency, therefore, this is the modal class.

Here l = 35 , h = 10 , f1 = 23 , f0 = 21 , f2 =14

Now, let us substitute these values in the formula

\(Mode \ = \ l \ + \ ( \frac{f_1 \ - \ f_0}{2f_1 \ - \ f_0 \ - \ f_2}) \ × \ h \)

\(= \ 35 \ + \ \frac{23 \ - \ 21}{2 \ × \ 23 \ - \ 21 \ - \ 14} \ × \ 10 \)

\(= \ 35 \ + \ \frac{2}{46 \ - \ 21 \ - \ 14} \ × \ 10 \)

\(= \ 35 \ + \ \frac{2}{11} \ × \ 10 \)

\(= \ 35 \ + \ 1.8 \ = \ 36.8 \)

Now, to calculate the Mean,

Class Interval Frequency ( \(f_i\) ) Mid-point ( \(x_i\) ) \( f_i x_i \)
5-15 6 10 60
15-25 11 20 220
25-35 21 30 630
35-45 23 40 920
45-55 14 50 700
55-65 5 60 300
\( \sum f_i \ = \ 80 \) \( \sum f_i x_i \ = \ 2830 \)

The formula for mean is :

\( \overline{x} \ = \ \frac{ \sum \ f_i x_i}{ \sum f_i} \)

\(= \ \frac{2830}{80} \ = \ 35.37 \) years

∴ The mean of the given data = 35.37 years

2. The following data gives the information on the observed lifetimes (in hours) of 225 electrical components:
Lifetime (in hours) 0-20 20-40 40-60 60-80 80-100 100-120
Frequency 10 35 52 61 38 29

Determine the modal lifetimes of the components.

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Answer :


The class 60-80 has the maximum frequency, therefore, this is the modal class.

Here l = 60 , h = 20 , f1 = 61 , f0 = 52 , and f2 = 38

Now, let us substitute these values in the formula

\(Mode \ = \ l \ + \ ( \frac{f_1 \ - \ f_0}{2f_1 \ - \ f_0 \ - \ f_2}) \ × \ h \)

\( = \ 60 \ + \ ( \frac{61 \ - \ 52}{2 \ × \ 61 \ - \ 52 \ - \ 38}) \ × \ 20 \)

\(= \ 60 \ + \ ( \frac{61 \ - \ 52}{122 \ - \ 52 \ - \ 38}) \ × \ 20 \)

\(= \ 60 \ + \ \frac{9}{32} \ × \ 20 \)

\( = \ 60 \ + \ 5.625 \ = \ 65.625 \)

∴ The modal lifetimes of the components is 65.625 hours.

3. The following data gives the distribution of total monthly household expenditure of 200 families of a village. Find the modal monthly expenditure of the families. Also, find the mean monthly expenditure :
Expenditure Number of families
1000-1500 24
1500-2000 40
2000-2500 33
2500-3000 28
3000-3500 30
3500-4000 22
4000-4500 16
4500-5000 7

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Answer :


The class 1500 – 2000 has the maximum frequency. Therefore, it is the modal class.

Here l = 1500, h = 500, f1 =40 , f0 =24 and f2 =33

Now, let us substitute these values in the formula

Mode \(= \ l \ + \ ( \frac{f_1 \ - \ f_0}{2f_1 \ - \ f_0 \ - \ f_2}) \ × \ h \)

\(= \ 1500 \ + \ ( \frac{40 \ - \ 24}{2 \ × \ 40 \ - \ 24 \ - \ 33}) \ × \ 500\)

\(= \ 1500 \ + \ \frac{40 \ -\ 24}{80 \ - \ 24 \ - \ 33} \ × \ 500 \)

\(= \ 1500 \ + \ \frac{16}{23} \ × \ 500 \)

\(= \ 1500 \ + \ 347.83 \ = \ 1847.83 \)

∴ Modal monthly expenditure = Rs. 1847.83

Let the assumed mean be A = 2750 and h = 500.

Class Interval \(f_i \) \( x_i \) \(d_i \ = \ x_i \ - \ A \) \(u_i \ = \ \frac{d_i}{h} \) \(f_i u_i \)
1000-1500 24 1250 -1500 -3 -72
1500-2000 40 1750 -1000 -2 -80
2000-2500 33 2250 -500 -1 -33
2500-3000 28 2750 0 0 0
3000-3500 30 3250 500 1 30
3500-4000 22 3750 1000 2 44
4000-4500 16 4250 1500 3 48
4500-5000 7 4750 2000 4 28
\( \sum f_i \ = \ 200 \) \( \sum f_i u_i \ = \ -35 \)

The formula to calculate the mean,

\( \overline{x} \ = \ A \ + h \frac{ \sum \ f_i x_i}{ \sum f_i} \)

\(= \ 2750 \ + \ ( \frac{-35}{200}) \ × \ 500 \)

\( = \ 2750 \ - \ 87.50 \ = \ 2662.50 \)

So, the mean monthly expenditure of the families is Rs 2662.50

4. The following distribution gives the state-wise teacher-student ratio in higher secondary schools of India. Find the mode and mean of this data. Interpret the two measures
No of Students per teacher Number of states / U.T
15-20 3
20-25 8
25-30 9
30-35 10
35-40 3
40-45 0
45-50 0
50-55 2

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Answer :


The class 30-35 has the maximum frequency, therefore, this is the modal class.

Here l = 30, h = 5, f1 = 10, f0 = 9 and f2 = 3

Now, let us substitute these values in the formula

Mode \(= \ l \ + \ ( \frac{f_1 \ - \ f_0}{2f_1 \ - \ f_0 \ - \ f_2}) \ × \ h \)

\( = \ 30 \ + \ ( \frac{10 \ - \ 9}{2 \ × \ 10 \ - \ 9 \ - \ 3}) \ × \ 5 \)

\(= \ 30 \ + \ \frac{10 \ - \ 9}{20 \ - \ 9 \ - \ 3} \ × \ 5 \)

\(= \ 30 \ + \ \frac{1}{8} \ × \ 5 \ = \ 30 \ + \ 0.625 \)

\( = \ 30.625 \ = \ 30.6 \)

Calculation of mean :

Class Interval Frequency \( (f_i) \) Mid-point \( (x_i) \) \( f_i x_i \)
15-20 3 17.5 52.5
20-25 8 22.5 180.0
25-30 9 27.5 247.5
30-35 10 32.5 325.0
35-40 3 37.5 112.5
40-45 0 42.5 0
45-50 0 47.5 0
50-55 2 52.5 105.5
\( \sum f_i \ = \ 35 \) \( \sum f_i x_i \ = \ 1022.5 \)

\( \overline{x} \ = \ \frac{ \sum \ f_i x_i}{ \sum f_i} \)

\(= \ \frac{1022.5}{35} \ = \ 29.2 \)

∴ mean = 29.2

5. The given distribution shows the number of runs scored by some top batsmen of the world in one- day international cricket matches.
Run Scored Number of Batsman
3000-4000 4
4000-5000 18
5000-6000 9
6000-7000 7
7000-8000 6
8000-9000 3
9000-10000 1
10000-11000 1
Find the mode of the data.

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Answer :


The class 4000-5000 has the maximum frequency ,
therefore, this is the modal class.

Here l = 4000, h = 1000, f1 = 18, f0 = 4 and f2 = 9

Now, let us substitute these values in the formula

Mode \(= l \ + \ ( \frac{f_1 \ - \ f_0}{2f_1 \ - \ f_0 \ - \ f_2}) \ × \ h \)

\(= \ 4000 \ + \ ( \frac{18 \ - \ 4}{2 \ × \ 18 \ - \ 4 \ - \ 9}) \ × \ 1000 \)

\(= \ 4000 \ + \ \frac{18 \ - \ 4}{36 \ - \ 4 \ - \ 9} \ × \ 1000 \)

\(= \ 4000 \ + \ \frac{14}{23} \ × \ 1000 \)

\(= \ 4000 \ + \ 608.695 \ = 4608.7 \)

∴ The mode of the given data is 4608.7 runs.

6. A student noted the number of cars passing through a spot on a road for 100 periods each of 3 minutes and summarised it in the table given below. Find the mode of the data:
Number of cars Frequency
0-10 7
10-20 14
20-30 13
30-40 12
40-50 20
50-60 11
60-70 15
70-80 8

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Answer :


The class 40-50 has the maximum frequency, therefore, this is the modal class.

Here l = 40, h =10, f1 = 20, f0 = 12 and f2 = 11

Now, let us substitute these values in the formula
Mode \( = \ l \ + \ ( \frac{f_1 \ - \ f_0}{2f_1 \ - \ f_0 \ - \ f_2}) \ × \ h \)

\(= \ 40 \ + \ ( \frac{20 \ - \ 12}{40 \ - \ 12 \ - \ 11}) \ × \ 10 \)

\( = \ 40 \ + \ \frac{8}{17} \ × \ 10 \)

\( = \ 40+4.7 \ = \ 44.7 \)

∴ Mode = 44.7 cars

Solution for Exercise - 14.3

1. The following frequency distribution gives the monthly consumption of an electricity of 68 consumers in a locality. Find the median, mean and mode of the data and compare them.
Monthly consumption(in units) No. of customers
65-85 4
85-105 5
105-125 13
125-145 20
145-165 14
165-185 8
185-205 4

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Answer :

Class Interval Frequency Cumulative frequency
65-85 4 4
85-105 5 9
105-125 13 22
125-145 20 42
145-165 14 56
165-185 8 64
185-205 4 68
N=68

We have n = 68 \( => \ \frac{n}{2} \ = \ \frac{68}{2} \ = \ 34 \)

The cumulative frequency just greater than \( \frac{n}{2} \) , is 42 and the corresponding class is 125-145.

Thus, 125-145 is the median class such that \( \frac{n}{2} \ = \ 34 , l = 125 , c.f. = 22 , f = 20 , h = 20. \)

Median \(= \ l \ + \ ( \frac{ \frac{n}{2} \ - \ cf}{f}) \ × \ h \),

Median \( = \ 125 \ + \ ( \frac{34 \ - \ 22}{20}) \ × \ 20 \ = \ 125 \ + \ 12 \ = \ 137 \) units

To calculate the mode :

Modal class = \(125 - 145 , f_1 = 20 , f_0 = 13 , f_2 = 14, and \ h = 20 \)

Mode formula :

Mode = \( l + \frac{(f_1- f_0)}{(2f_1- f_0 - f_2)} × h \)

Mode = \( 125 + \frac{(20 -13)}{(40 -13 -14)} × 20 \)

\( = 125 + \frac{140}{13} \)

\( = 125 + 10.77 = 135.77 \)

Therefore, mode = 135.77

Calculation of mean :

Let the assumed mean, A = 135, class interval, h = 20

so \(u_i \ = \ \frac{x_i \ - \ A}{h} \ = \ \frac{x_i \ - \ 135}{20} \)

Class Interval \( f_i \) \( x_i \) \( d_i \ = \ x_i \ - \ A \) \( u_i \ = \ \frac{d_i}{h} \) \( f_i u_i \)
65-85 4 75 -60 -3 -12
85-105 5 95 -40 -2 -10
105-125 13 115 -20 -1 -13
125-145 20 135 0 0 0
145-165 14 155 20 1 14
165-185 8 175 40 2 16
185-205 4 195 60 3 12
\( \sum f_i \ = \ 68 \) \( \sum f_i u_i \ = \ 7 \)

\( \overline{x} \ = \ A \ + h \frac{ \sum \ f_i x_i}{ \sum f_i} \)

\( = \ 135 \ + \ 20( \frac{7}{68}) \)

Mean = 137.05

In this case, mean, median and mode are more/less equal in this distribution.

2. If the median of a distribution given below is 28.5 then, find the value of an x &y.
Class Interval Frequency
0-10 5
10-20 x
20-30 20
30-40 15
40-50 y
50-60 5
Total 60

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Answer :


Given that median is 28.5 and \(n \ = \ \sum f_i \ = \ 60 \)

Here n = 60 \( => \ \frac{n}{2} \ = \ 30 \)

Since the median is given to be 28.5, thus the median class is 20-30.

∴ l = 20, h = 10, f = 20 and cf = 5 + x

∴ median \(= \ l \ + \ ( \frac{ \frac{n}{2} \ - \ cf}{f} ) \ × \ h \)

\(=> \ 28.5 \ = \ 20 \ + \ \frac{30 \ - \ 5 \ - \ x}{20} \ × \ 10 \)

\( => \ 57 \ = \ 40 \ + \ 25 \ - \ x \)

\( => \ x \ = \ 65 \ - \ 57 \ => \ x \ = \ 8 \)

Also, 45 + x + y = 60      [ As n = 60 ]

\(=> \ 45 \ + \ 8 \ + \ y \ = \ 60 \)

\(=> \ y \ = \ 60 \ - \ 53 \ => \ y \ = \ 7 \)

∴ x = 8, y = 7

Q.3 A life insurance agent found the following data for distribution of ages of 100 policy holders. Calculate the median age, if policies are given only to persons having age 18 years onwards but less than 60 year.
Age(in years) Number of policy holder
Below 20 2
Below 25 6
Below 30 24
Below 35 45
Below 40 78
Below 45 89
Below 50 92
Below 55 98
Below 60 100

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Answer :

We are given the cumulative frequency distribution. So, we first construct a frequency table from the given cumulative frequency distribution and then we will make necessary computations to compute median .

Class interval Frequency Cumulative frequency
15-20 2 2
20-25 4 6
25-30 18 24
30-35 21 45
35-40 33 78
40-45 11 89
45-50 3 92
50-55 6 98
55-60 2 100

Here \( n \ = \ \sum f_i \ = \ 100 \ => \ \frac{n}{2} \ = \ 50 \)

We see that the cumulative frequency just greater than \( \frac{n}{2} \), i.e., 50 is 78 and the corresponding class is 35 - 40.

So, 35 - 40 is the median class.

∴ \( \frac{n}{2} \ = \ 50 , l = 35, c.f = 45 , f = 33, h = 5 \)

Now, let us substitute these values in the formula

Median \( = \ l \ + \ ( \frac{ \frac{n}{2} \ - \ cf}{f} ) \ × \ h \)

\(= \ 35 \ + \ \frac{50 \ - \ 45}{33} \ × \ 5 \)

\( = \ 35 \ + \ \frac{5}{33} \ × \ 5 \)

\( = \ 35 \ + \ 0.76 \ = \ 35.76 \)

∴ The median age = 35.76 years

Q.4 The lengths of 40 leaves of a plant are measured correct to the nearest millimeter, and the data obtained is represented in the following table:
Length(in mm) Number of leaves
118-126 3
127-135 5
136-144 9
145-153 12
154-162 5
163-171 4
172-180 2
Find the median length of leaves.

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Answer :

Since the data are not continuous reduce 0.5 in the lower limit and add 0.5 in the upper limit.

Class Interval Frequency Cumulative frequency
117.5-126.5 3 3
126.5-135.5 5 8
135.5-144.5 9 17
144.5-153.5 12 29
153.5-162.5 5 34
162.5-171.5 4 38
171.5-180.5 2 40

So, n = 40 and \( \frac{n}{2} \ = \ 20 \)

Median class = 144.5 -153.5

then, \( l = 144.5, cf = 17, f = 12, h = 9 \)

Now we know that, Median \(= \ l \ + \ ( \frac{ \frac{n}{2} \ - \ cf}{f} ) \ × \ h \)

=> Median \( = \ 144.5 \ + \ ( \frac{20 \ - \ 17}{12}) \ × \ 9 \)

\( = \ 144.5 \ + \ \frac{3}{12} \ × \ 9 \)

\( = \ 144.5 \ + \ 2.25 \ = \ 146.75 \)

∴ The median length of the leaves = 146.75 mm.

Q.5 The following table gives the distribution of the life time of 400 neon lamps :
Lifetime(in hours) Number of lamps
1500-2000 14
2000-2500 56
2500-3000 60
3000-3500 86
3500-4000 74
4000-4500 62
4500-5000 48
Find the median lifetime of a lamp.

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Answer :

Class Interval Frequency Cumulative
1500-2000 14 14
2000-2500 56 70
2500-3000 60 130
3000-3500 86 216
3500-4000 74 290
4000-4500 62 352
4500-5000 48 400

n = 400 => \( \frac{n}{2} \ = \ 200 \)

Median class = 3000 - 3500

∴ \( l = 3000, cf = 130, f = 86, h = 500 \)

Median \( = \ l \ + \ ( \frac{ \frac{n}{2} \ - \ cf}{f} ) \ × \ h \)

Median \(= \ 3000 \ + \ ( \frac{200 \ - \ 130}{86}) \ × \ 500 \)

\(= \ 3000 \ + \ \frac{70}{86} \ × \ 500 \)

\(= \ 3000 \ + \ 406.98 \)

\(= \ 3406.98 \)

∴ median lifetime of lamps = 3406.98 hours

6. In this 100 surnames were randomly picked up from a local telephone directory and the frequency distribution of the number of letters in English alphabets 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 number of median letters in the surnames. Find the number of mean letters in the surnames and also, find the size of modal in the surnames.

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Answer :

Class Interval Frequency Cumulative Frequency
1-4 6 6
4-7 30 36
7-10 40 76
10-13 16 92
13-16 4 96
16-19 4 100

We have n = 100 => \( \frac{n}{2} \ = \ 50 \)

The cumulative frequency just greater than \( \frac{n}{2} \) is 76 and the corresponding class is 7-10 .

Thus 7-10 is the median class such that

\( \frac{n}{2} \ = \ 50 , l = 7, cf = 36 , f = 40 and h = 3 \)

Median \( = \ l \ + \ ( \frac{ \frac{n}{2} \ - \ cf}{f} ) \ × \ h \)

\( = \ 7 \ + \ ( \frac{50 \ - \ 36}{40}) \ × \ 3 \)

\( = \ 7 \ + \ \frac{14}{40} \ × \ 3 \)

\(= \ 7 \ + \ 1.05 \ = \ 8.05 \)

Now, to calculate the mean,
Class Interval \( f_i \) \( x_i \) \( f_i x_i \)
1-4 6 2.5 15
4-7 30 5.5 165
7-10 40 8.5 340
10-13 16 11.5 184
13-16 4 14.5 51
16-19 4 17.5 70
\( \sum f_i \ = \ 100 \) \( \sum f_i x_i \ = \ 825 \)

∴ Mean = \( \overline{x} \ = \ \frac{ \sum f_i x_i}{ \sum f_i} \)

\( = \ \frac{832}{100} \)

Therefore, Mean = \( 8.32 \)

Calculation of mode :

The class 7-10 has the maximum frequency therefore, this is the modal class.

Here \( l = 7 , h = 3, f1 = 40 , f0 = 30 and f2 = 16 \)

Mode \( = \ l \ + \ ( \frac{f_1 \ - \ f_0}{2f_1 \ - \ f_0 \ - \ f_2}) \ × \ h \)

\(= \ 7 \ + \ \frac{40 \ - \ 30}{80 \ - \ 30 \ - \ 16} \ × \ 3 \)

\(= \ 7 \ + \ \frac{10}{34} \ × \ 3 \)

\( = \ 7 \ + \ 0.88 \ = \ 7.88 \)

Median = 8.05 , Mean = 8.32 and Mode = 7.88

Q.7 The distribution below gives the weights of 30 students of a class. Find the median weight of the students.
Weight(in kg) 40-45 45-50 50-55 55-60 60-65 65-70 70-75
Number of students 2 3 8 6 6 3 2

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Answer :

Class Interval Frequency Cumulative frequency
40-45 2 2
45-50 3 5
50-55 8 13
55-60 6 19
60-65 6 25
65-70 3 28
70-75 2 30

We have n = 30 \( => \ \frac{n}{2} \ = \ 15 \)

The cumulative frequency just greater than \( \frac{n}{2} \) is 19 and the corresponding class is 55 - 60.

Thus, 55 - 60 is the median class such that :

\( \frac{n}{2} = 15 , l = 55 , f = 6 , cf = 13 \ and \ h = 5 \)

Median \(= \ l \ + \ ( \frac{ \frac{n}{2} \ - \ cf}{f} ) \ × \ h \)

\(= \ 55 \ + \ ( \frac{15 \ - \ 13}{6}) \ × \ 5 \)

\(= \ 55 \ + \ \frac{2}{6} \ × \ 5 \ = \)

\( 55 \ + \ 1.67 \ = \ 56.67 \)

∴ The median weight = 56.67 kg .

Solution for Exercise - 14.4

1. The following distribution gives the daily income of 50 workers of a factory.
Daily income in Rupees 100-120 120-140 140-160 160-180 180-200
Number of workers 12 14 8 6 10
Convert the distribution above to a less than type cumulative frequency distribution and draw its ogive.

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Answer :

Daily income Frequency Cumulative Frequency
Less than 120 12 12
Less than 140 14 26
Less than 160 8 34
Less than 180 6 40
Less than 200 10 50

From the table plot the points corresponding to the ordered pairs such as (120, 12), (140, 26), (160, 34), (180, 40) and (200, 50) on graph paper and the plotted points are joined to get a smooth curve and the obtained curve is known as less than type ogive curve


2. During the medical check-up of 35 students of a class, their weights were recorded as follows:
Weight in kg Number of students
Less than 38 0
Less than 40 3
Less than 42 5
Less than 44 9
Less than 46 14
Less than 48 28
Less than 50 32
Less than 52 35


Draw a less than type ogive for the given data. Hence obtain the median weight from the graph and verify the result by using the formula

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Answer :


From the given data, to represent the table in the form of graph, choose the upper limits of the class intervals are in x-axis and frequencies on y-axis by choosing the convenient scale.

Now plot the points corresponding to the ordered pairs given by (38, 0), (40, 3), (42, 5), (44, 9),(46, 14), (48, 28), (50, 32) and (52, 35) on a graph paper an join them to get a smooth curve. The curve obtained is known as less than type ogive.



Locate the point 17.5 on the y-axis and draw a line parallel to the x-axis cutting the curve at a point. From the point, draw a perpendicular line to the x-axis. The intersection point perpendicular to x-axis is the median of the given data. Now, to find the mode by making a table.

Class interval Number of students(Frequency) Cumulative Frequency
Less than 38 0 0
Less than 40 3-0=3 3
Less than 42 5-3=2 8
Less than 44 9-5=4 9
Less than 46 14-9=5 14
Less than 48 28-14=14 28
Less than 50 32-28=4 32
Less than 52 35-22=3 35

The class 46-48 has the maximum frequency, therefore, this is modal class

Here, l = 46, h = 2, f1 = 14, f0 = 5 and f2 = 4

Now, Mode \( = \ l \ + \ ( \frac{f_1 \ - \ f_0}{2f_1 \ - \ f_0 \ - \ f_2}) \ × \ h \)

\(= \ 46 \ + \ \frac{14 \ - \ 5}{28 \ - \ 5 \ - \ 4} \ × \ 2 \ = \ 46 \ + \ \frac{9}{19} \ × \ 2 \)

\(= \ 46 \ + \ 0.95 \ = \ 46.95 \)

Thus, mode is verified.

3. The following tables gives production yield per hectare of wheat of 100 farms of a village.
Production Yield 50-55 55-60 60-65 65-70 70-75 75-80
Number of farms 2 8 12 24 38 16
Change the distribution to a more than type distribution and draw its ogive.

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Answer :

Converting the given distribution to a more than type distribution, we get

Production Yield (kg/ha) Number of farms
More than or equal to 50 100
More than or equal to 55 100-2 = 98
More than or equal to 60 98-8= 90
More than or equal to 65 90-12=78
More than or equal to 70 78-24=54
More than or equal to 75 54-38 =16
From the table obtained draw the ogive by plotting the corresponding points where the upper limits in x-axis and the frequencies obtained in the y-axis are (50, 100), (55, 98), (60, 90), (65, 78), (70, 54) and (75, 16) on this graph paper. The graph obtained is known as more than type ogive curve.