Haryana State Board HBSE 10th Class Maths Solutions Chapter 14 Statistics Ex 14.3 Textbook Exercise Questions and Answers.

## Haryana Board 10th Class Maths Solutions Chapter 14 Statistics Ex 14.3

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.

Solution:

For calculating the median we prepare cumulative frequency distribution table as given below :

n = 68

â‡’ \(\frac{n}{2}=\frac{68}{2}\) = 34

But 34 comes under the cumulative frequency 42 and the class interval against the cumulative frequency 42 is 125 – 145. So, it is the median class.

âˆ´ l = 125, cf = 22, f = 20 and h = 20

Median = l + \(\left(\frac{\frac{n}{2}-c f}{f}\right)\) Ã— h

= 125 + \(\left(\frac{34-22}{20}\right)\) Ã— 20

= 125 + 12 = 137.

For calculating mean, we prepare the table as given below :

From the table, we have

Î£f_{i} = 68, Î£f_{i}u_{i} = 7, a = 135, h = 20

Mean = a + \(\left(\frac{\Sigma f_i u_i}{\Sigma f_i}\right)\) Ã— h

= 135 + \(\frac{7}{68}\) Ã— 20

= 135 + \(\frac{140}{68}\)

= 135 + 2.05 = 137.05.

For calculating mode, the class 125 – 145 has maximum frequency. So, it is the modal class.

âˆ´ l = 125, f_{0} = 13, f_{1} = 20, f_{2} = 14, and h = 20

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

= 125 + \(\left(\frac{20-13}{2 \times 20-13-14}\right)\) Ã— 20

= 125 + \(\frac{7 \times 20}{13}\)

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

= 125 + 10.76 = 135.76.

Hence, median, mean and mode are 137, 137.05 and 135.76 reapectively.

The three measures are approximately same in this case.

Question 2.

If the median of the distribution given below is 1. 285, find the values of x and y.

Solution:

For calculating the median, we prepare the cumulative frequency distribution table as given below :

n = 60

â‡’ 45 + x + y = 60

â‡’ x + y = 60 – 45

â‡’ x + y = 15

Median is 28.5, which lies in the class 20 – 30.

So, it is the median class.

âˆ´ l = 20, f = 20, cf = 5 + x and h = 10.

âˆ´ Median = l + \(-\left(\frac{\frac{n}{2}-c f}{f}\right)\) Ã— h

28.5 = 20 + \(\frac{\left[\frac{60}{2}-(5+x)\right]}{20}\) Ã— 10

28.5 = 20 + \(\frac{(30-5-x)}{20}\) Ã— 10

28.5 = 20 + \(\frac{(25-x)}{2}\)

28.5 – 20 = \(\frac{25-x}{2}\)

8.5 = \(\frac{25-x}{2}\)

â‡’ 17 = 25 – x

â‡’ x = 25 – 17

â‡’ x = 8

Putting the value of x in equation (1), we get

8 + y = 15

â‡’ y = 15 – 8 = 7.

Hence, x = 8, y = 7.

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.

Solution:

Let us prepare the table depicting class intervals with their respective frequencies and cumulative frequencies from the given data as below:

n = 100

â‡’ \(\frac{n}{2}=\frac{100}{2}\) = 50

But 50 comes under the cumulative frequency 78 and the class interval against the cumulative frequency 78 is 35 – 40. So, it is the median class.

âˆ´ l = 35, cf = 45, f = 33 and h = 5

âˆ´ Median = l + \(\left(\frac{\frac{n}{2}-c f}{f}\right)\) Ã— h

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

= 35 + \(\frac{25}{33}\)

= 35 + 0.76

= 35.76

Hence, median age = 35.76 years.

Question 4.

The lengths of 40 leaves of a plant are measured correct to the nearest millimetre, and data obtained is represented in the following table :

Find the median length of the leaves.

Solution:

The series is in inclusive form. We convert it into exclusive form and prepare the cumulative frequency table as given below :

n = 40

â‡’ \(\frac{n}{2}\) = 20

But 20 comes under the cumulative frequency 29 and the class interval against the cumulative frequency 29 is 144.5 – 153.5. So, it is the median class.

âˆ´ l = 144.5, cf = 17, f = 12, and h = 9

Median = l + \(\left(\frac{\frac{n}{2}-c f}{f}\right)\) Ã— h

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

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

= 144.5 + \(\frac{9}{4}\)

= 144.5 + 2.25

= 146.75 mm.

Hence, median length of leaves = 146.75 mm.

Question 5.

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

Find the median life time of a lamp.

Solution:

We prepare the cumulative frequency distribution table as given below:

n = 400

â‡’ \(\frac{n}{2}=\frac{400}{2}\) = 200

But 200 comes under the cumulative frequency 216 and the class interval against the cumulative frequency 216 is 3000 – 3500. So, it is the median class.

âˆ´ l = 3000, cf =130, f = 86, and h = 500

âˆ´ Median = l + \(\left(\frac{\frac{n}{2}-c f}{f}\right)\) Ã— h

= 3000 + \(\left(\frac{200-130}{86}\right)\) Ã— 500

= 3000 + \(\frac{35000}{86}\)

= 3000 + 406.98 = 3406.98

Hence, median life time of a lamp = 3406.98 hours.

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 :

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.

Solution:

For calculating the median, we prepare the cumulative frequency distribution table as given below :

Total n = 100

â‡’ \(\frac{n}{2}=\frac{100}{2}\) = 50

But 50 comes under the cumulative frequency 76. The class interval against cumulative frequency 76 is 7 – 10.

So, it is the median class.

âˆ´ l = 7, cf = 36, f = 40 and h = 3

âˆ´ Median = l + \(\left(\frac{\frac{n}{2}-c f}{f}\right)\) Ã— h

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

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

= 7 + \(\frac{42}{40}\)

= 7 + 1.05 = 8.05

For calculating the mean we prepare table as given below :

From the table, we have Î£f_{i} = 100, Î£f_{i}d_{i} = – 18, a = 8.5

âˆ´ Mean = a + \(\left(\frac{\Sigma f_i d_i}{\Sigma f_i}\right)\)

= 8.5 + \(\left(\frac{-18}{100}\right)\)

= 8.5 + \(\frac{18}{100}\)

= 8.5 – 0.18 = 8.32.

For calculating mode, the class 7 – 10 has maximum frequency.

So, it is the modal class.

âˆ´ l = 7, f_{1} = 40, f_{0} = 30, f_{2} = 16 and h = 3

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

= 7 + \(\left(\frac{40-30}{2 \times 40-30-16}\right)\) Ã— 3

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

= 7 + \(\frac{30}{34}\)

= 7 + 0.88 = 7.88

Hence, median = 8.05, Mean = 8.32 and modal size = 7.88.

Question 7.

The distribution below gives the weights of 30 students of a class. Find the median weight of the students. â€™

Solution:

We prepare the cumulative frequency distribution table as given below.

n = 30

â‡’ \(\frac{n}{2}=\frac{30}{2}\) = 15

But 15 comes under the cumulative frequency 19 and the class interval against the cumulative frequency 19 is 55 – 60. So, it is the median class.

âˆ´ l = 55, cf = 13, f = 6 and h = 5

Median = 55 + \(\left(\frac{\frac{n}{2}-c f}{f}\right)\) Ã— h

= 55 + \(\left(\frac{15-13}{6}\right)\) Ã— 5

= 55 + \(\frac{10}{6}\)

= 55 + 1.67 = 56.67

Hence, median weight of the students = 56.67 kg.