# NCERT solution class 10 chapter 15 Statistics exercise 15.3 mathematics

## EXERCISE 15.3

#### Question 1:

The following frequency distribution gives the monthly consumption of electricity of 68 consumers of a locality. Find the median, mean and mode of the data and compare them.

 Monthly consumption (in units) Number of consumers 65 − 85 4 85 − 105 5 105 − 125 13 125 − 145 20 145 − 165 14 165 − 185 8 185 − 205 4

To find the class marks, the following relation is used.

Taking 135 as assumed mean (a), diui, fiui are calculated according to step deviation method as follows.

 Monthly consumption (in units) Number of consumers (f i) xi class mark di= xi− 135 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 Total 68 7

From the table, we obtain

From the table, it can be observed that the maximum class frequency is 20, belonging to class interval 125 − 145.

Modal class = 125 − 145

Lower limit (l) of modal class = 125

Class size (h) = 20

Frequency (f1) of modal class = 20

Frequency (f0) of class preceding modal class = 13

Frequency (f2) of class succeeding the modal class = 14

To find the median of the given data, cumulative frequency is calculated as follows.

 Monthly consumption (in units) Number of consumers Cumulative frequency 65 − 85 4 4 85 − 105 5 4 + 5 = 9 105 − 125 13 9 + 13 = 22 125 − 145 20 22 + 20 = 42 145 − 165 14 42 + 14 = 56 165 − 185 8 56 + 8 = 64 185 − 205 4 64 + 4 = 68

From the table, we obtain

n = 68

Cumulative frequency (cf) just greater than is 42, belonging to interval 125 − 145.

Therefore, median class = 125 − 145

Lower limit (l) of median class = 125

Class size (h) = 20

Frequency (f) of median class = 20

Cumulative frequency (cf) of class preceding median class = 22

Therefore, median, mode, mean of the given data is 137, 135.76, and 137.05 respectively.

The three measures are approximately the same in this case.

Video Solution

#### Question 2:

If the median of the distribution is given below is 28.5, find the values of x and y.

 Class interval Frequency 0 − 10 5 10 − 20 x 20 − 30 20 30 − 40 15 40 − 50 y 50 − 60 5 Total 60

The cumulative frequency for the given data is calculated as follows.

 Class interval Frequency Cumulative frequency 0 − 10 5 5 10 − 20 x 5+ x 20 − 30 20 25 + x 30 − 40 15 40 + x 40 − 50 y 40+ x + y 50 − 60 5 45 + x + y Total (n) 60

From the table, it can be observed that = 60

45 + x + y = 60

x + y = 15 (1)

Median of the data is given as 28.5 which lies in interval 20 − 30.

Therefore, median class = 20 − 30

Lower limit (l) of median class = 20

Cumulative frequency (cf) of class preceding the median class = 5 + x

Frequency (f) of median class = 20

Class size (h) = 10

From equation (1),

8 + y = 15

y = 7

Hence, the values of x and y are 8 and 7 respectively.

#### Question 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 holders 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

Here, class width is not the same. There is no requirement of adjusting the frequencies according to class intervals. The given frequency table is of less than type represented with upper class limits. The policies were given only to persons with age 18 years onwards but less than 60 years. Therefore, class intervals with their respective cumulative frequency can be defined as below.

 Age (in years) Number of policy holders (fi) Cumulative frequency (cf) 18 − 20 2 2 20 − 25 6 − 2 = 4 6 25 − 30 24 − 6 = 18 24 30 − 35 45 − 24 = 21 45 35 − 40 78 − 45 = 33 78 40 − 45 89 − 78 = 11 89 45 − 50 92 − 89 = 3 92 50 − 55 98 − 92 = 6 98 55 − 60 100 − 98 = 2 100 Total (n)

From the table, it can be observed that n = 100.

Cumulative frequency (cf) just greater than is 78, belonging to interval 35 − 40.

Therefore, median class = 35 − 40

Lower limit (l) of median class = 35

Class size (h) = 5

Frequency (f) of median class = 33

Cumulative frequency (cf) of class preceding median class = 45

Therefore, median age is 35.76 years.

Video Solution

#### Question 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 or leaves fi 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 the leaves.

(Hint: The data needs to be converted to continuous classes for finding the median, since the formula assumes continuous classes. The classes then change to 117.5 − 126.5, 126.5 − 135.5… 171.5 − 180.5)

The given data does not have continuous class intervals. It can be observed that the difference between two class intervals is 1. Therefore,  has to be added and subtracted to upper class limits and lower class limits respectively.

Continuous class intervals with respective cumulative frequencies can be represented as follows.

 Length (in mm) Number or leaves fi Cumulative frequency 117.5 − 126.5 3 3 126.5 − 135.5 5 3 + 5 = 8 135.5 − 144.5 9 8 + 9 = 17 144.5 − 153.5 12 17 + 12 = 29 153.5 − 162.5 5 29 + 5 = 34 162.5 − 171.5 4 34 + 4 = 38 171.5 − 180.5 2 38 + 2 = 40

From the table, it can be observed that the cumulative frequency just greater than  is 29, belonging to class interval 144.5 − 153.5.

Median class = 144.5 − 153.5

Lower limit (l) of median class = 144.5

Class size (h) = 9

Frequency (f) of median class = 12

Cumulative frequency (cf) of class preceding median class = 17

Median

Therefore, median length of leaves is 146.75 mm.

Video Solution

#### Question 5:

Find the following table gives the distribution of the life time of 400 neon lamps:

 Life time (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 life time of a lamp.

Thecumulative frequencies with their respective class intervals are as follows.

 Life time Number of lamps (fi) Cumulative frequency 1500 − 2000 14 14 2000 − 2500 56 14 + 56 = 70 2500 − 3000 60 70 + 60 = 130 3000 − 3500 86 130 + 86 = 216 3500 − 4000 74 216 + 74 = 290 4000 − 4500 62 290 + 62 = 352 4500 − 5000 48 352 + 48 = 400 Total (n) 400

It can be observed that the cumulative frequency just greater than  is 216, belonging to class interval 3000 − 3500.

Median class = 3000 − 3500

Lower limit (l) of median class = 3000

Frequency (f) of median class = 86

Cumulative frequency (cf) of class preceding median class = 130

Class size (h) = 500

Median

= 3406.976

Therefore, median life time of lamps is 3406.98 hours.

Video Solution

#### Question 6:

100 surnames were randomly picked up from a local telephone directory and the frequency distribution of the number of letters in the 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 6 4 4

Determine the median number of letters in the surnames. Find the mean number of letters in the surnames? Also, find the modal size of the surnames.

The cumulative frequencies with their respective class intervals are as follows.

 Number of letters Frequency (fi) Cumulative frequency 1 − 4 6 6 4 − 7 30 30 + 6 = 36 7 − 10 40 36 + 40 = 76 10 − 13 16 76 + 16 = 92 13 − 16 4 92 + 4 = 96 16 − 19 4 96 + 4 = 100 Total (n) 100

It can be observed that the cumulative frequency just greater than is 76, belonging to class interval 7 − 10.

Median class = 7 − 10

Lower limit (l) of median class = 7

Cumulative frequency (cf) of class preceding median class = 36

Frequency (f) of median class = 40

Class size (h) = 3

Median

= 8.05

To find the class marks of the given class intervals, the following relation is used.

Taking 11.5 as assumed mean (a), diui, and fiui are calculated according to step deviation method as follows.

 Number of letters Number of surnamesfi xi di = xi− 11.5 fiui 1 − 4 6 2.5 − 9 − 3 − 18 4 − 7 30 5.5 − 6 − 2 − 60 7 − 10 40 8.5 − 3 − 1 − 40 10 − 13 16 11.5 0 0 0 13 − 16 4 14.5 3 1 4 16 − 19 4 17.5 6 2 8 Total 100 − 106

From the table, we obtain

fiui = −106

fi = 100

Mean,

= 11.5 − 3.18 = 8.32

The data in the given table can be written as

 Number of letters Frequency (fi) 1 − 4 6 4 − 7 30 7 − 10 40 10 − 13 16 13 − 16 4 16 − 19 4 Total (n) 100

From the table, it can be observed that the maximum class frequency is 40 belonging to class interval 7 − 10.

Modal class = 7 − 10

Lower limit (l) of modal class = 7

Class size (h) = 3

Frequency (f1) of modal class = 40

Frequency (f0) of class preceding the modal class = 30

Frequency (f2) of class succeeding the modal class = 16

Therefore, median number and mean number of letters in surnames is 8.05 and 8.32 respectively while modal size of surnames is 7.88.

Video Solution

#### Question 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

The cumulative frequencies with their respective class intervals are as follows.

 Weight (in kg) Frequency (fi) Cumulative frequency 40 − 45 2 2 45 − 50 3 2 + 3 = 5 50 − 55 8 5 + 8 = 13 55 − 60 6 13 + 6 = 19 60 − 65 6 19 + 6 = 25 65 − 70 3 25 + 3 = 28 70 − 75 2 28 + 2 = 30 Total (n) 30

Cumulative frequency just greater than  is 19, belonging to class interval 55 − 60.

Median class = 55 − 60

Lower limit (l) of median class = 55

Frequency (f) of median class = 6

Cumulative frequency (cf) of median class = 13

Class size (h) = 5

Median

= 56.67

Therefore, median weight is 56.67 kg.

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