NCERT Solutions for Exercise 14.3 Class 10 Maths Chapter 14 - Statistics

Q2 If the median of the distribution given below is 28.5, find the values of x and y.

Answer:

As, $N= 60 \implies \frac{N}{2} = 30$

Given median = 28.5, which lies in the class 20-30

Therefore, Median class = 20-30

Frequency corresponding to median class, f = 20

Cumulative frequency of the class preceding the median class, c.f. = 5 + x
Lower limit, l = 20; Class height, h = 10

Thus, $Median = l + \left (\frac{\frac{n}{2}-c.f}{f} \right ).W$

After putting the values, we get:

$\\ \implies28.5= 20 + \left (\frac{30-5-x}{20} \right ).10 \\ \\ \implies8.5=\frac{25-x}{2} \\ \implies 25-x = 8.5(2) \\ \implies x = 25 - 17 = 8$

Also,

$\\ 60 = 45 + x+y \\ \implies x+y = 60-45 = 15 \\ \implies y = 15-x = 15-8 \ \ \ (\because x =8) \\ \implies y = 7$

Therefore, the required values are: x=8 and y=7

Q3 A life insurance agent found the following data for the distribution of ages of 100 policyholders. Calculate the median age, if policies are given only to persons having age 18 years onwards but less than 60 years.

Answer:

Given, $N= 100 \implies \frac{N}{2} = 50$

Therefore, Median class = 35-45

Frequency corresponding to median class, f = 21

Cumulative frequency of the class preceding the median class, c.f. = 24

Lower limit, l = 35; Class height, h = 10

Therefore, $Median = l + \left (\frac{\frac{n}{2}-c.f}{f} \right ).W$

After putting in the values, we get:

$\\ = 35 + \left (\frac{50-45}{33} \right ).5 \\ \\$

$= 35.75$

Thus, the median age is 35.75 years.

Q4 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:

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.)

Answer:

The data needs to be converted to continuous classes for finding the median, since the formula assumes continuous classes.

As,$ N= 40 \implies \frac{N}{2} = 20$

Therefore, Median class = 144.5-153.5

Lower limit, l = 144.5; Class height, h = 9

Frequency corresponding to median class, f = 12

Cumulative frequency of the class preceding the median class, c.f. = 17

$Median = l + \left (\frac{\frac{n}{2}-c.f}{f} \right ).W$

After putting in the values, we get:

$\\ = 144.5 + \left (\frac{20-17}{12} \right ).9 \\ \\$

$= 146.75$

Thus, the median length of the leaves is 146.75 mm

Q5 The following table gives the distribution of the lifetime of 400 neon lamps:

Find the median lifetime of a lamp.

Answer:

$ N= 400 \implies \frac{N}{2} = 200$

Therefore, Median class = 3000-3500

Lower limit, l = 3000; Class height, h = 500

Frequency corresponding to median class, f = 86

Cumulative frequency of the class preceding the median class, c.f. = 130

$Median = l + \left (\frac{\frac{n}{2}-c.f}{f} \right ).W$

After putting in the values, we get:

$\\ = 3000 + \left (\frac{200-130}{86} \right ).500 \\ \\ = 3000+406.97$

$= 3406.97$

Thus, the median lifetime of a lamp is 3406.97 hours

$= 146.75$

Thus, the median length of the leaves is 146.75 mm

Q6 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:

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.

Answer:

MEDIAN:

$N= 100 \implies \frac{N}{2} = 50$

$\therefore$ Median class = 7-10; Lower limit, l = 7;

Cumulative frequency of preceding class, c.f. = 36; f = 40; h = 3

$Median = l + \left (\frac{\frac{n}{2}-c.f}{f} \right ).W$

After putting in the values, we get:

$\\ = 7+ \left (\frac{50-36}{40} \right ).3 \\ \\$

$= 8.05$

Thus, the median of the data is 8.05

MODE:

The class having the maximum frequency is the modal class.

The maximum frequency is 40, and hence the modal class = 7-10

Lower limit (l) of modal class = 7, class size (h) = 3

Frequency ( $f_1$ ) of the modal class = 40

Frequency ( $f_0$ ) of class preceding the modal class = 30

Frequency ( $f_2$ ) of class succeeding the modal class = 16

$Mode = l + \left(\frac{f_1-f_0}{2f_1 - f_0 - f_2} \right).h$

After putting in the values, we get:

$\\ = 7 + \left(\frac{40-30}{2(40)-30-16} \right).3 \\ \\ = 125 + \frac{10}{34}.3$

$= 7.88$

Thus, the Mode of the data is 7.88

MEAN:

$\overline x =\frac{\sum f_ix_i}{\sum f_i}$

$= \frac{825}{100} = 8.25$

Thus, the Mean of the data is 8.25

Q7 The distribution below gives the weights of 30 students of a class. Find the median weight of the students.

Answer:

MEDIAN:

$N= 30 \implies \frac{N}{2} = 15$

Therefore, Median class = 55-60; Lower limit, l = 55;

Cumulative frequency of preceding class, c.f. = 13; f = 6; h = 5

$Median = l + \left (\frac{\frac{n}{2}-c.f}{f} \right ).W$

After putting in the values, we get:

$\\ = 55+ \left (\frac{15-13}{6} \right ).5 \\ \\ = 55+\frac{2}{6}.5$

$= 56.67$

Thus, the median weight of the student is 56.67 kg