HBSE 10th Class Maths Solutions Chapter 14 Statistics Ex 14.3

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.

Haryana Board 10th Class Maths Solutions Chapter 14 Statistics Ex 14.3 1

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

Haryana Board 10th Class Maths Solutions Chapter 14 Statistics Ex 14.3 2

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

n = 68
⇒ \(\frac{n}{2}=\frac{68}{2}\) = 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.
∴ l = 125, cf = 22, f = 20 and h = 20
Median = l + \(\left(\frac{\frac{n}{2}-c f}{f}\right)\) × h
= 125 + \(\left(\frac{34-22}{20}\right)\) × 20
= 125 + 12 = 137.
For calculating mean, we prepare the table as given below :

Haryana Board 10th Class Maths Solutions Chapter 14 Statistics Ex 14.3 3

From the table, we have
Σfi = 68, Σfiui = 7, a = 135, h = 20
Mean = a + \(\left(\frac{\Sigma f_i u_i}{\Sigma f_i}\right)\) × h
= 135 + \(\frac{7}{68}\) × 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.
∴ l = 125, f0 = 13, f1 = 20, f2 = 14, and h = 20
Mode = l + \(\left(\frac{f_1-f_0}{2 f_1-f_0-f_2}\right)\) × h
= 125 + \(\left(\frac{20-13}{2 \times 20-13-14}\right)\) × 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.

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

Question 2.
If the median of the distribution given below is 1. 285, find the values of x and y.

Haryana Board 10th Class Maths Solutions Chapter 14 Statistics Ex 14.3 4

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

Haryana Board 10th Class Maths Solutions Chapter 14 Statistics Ex 14.3 5

n = 60
⇒ 45 + x + y = 60
⇒ x + y = 60 – 45
⇒ x + y = 15
Median is 28.5, which lies in the class 20 – 30.
So, it is the median class.
∴ l = 20, f = 20, cf = 5 + x and h = 10.
∴ Median = l + \(-\left(\frac{\frac{n}{2}-c f}{f}\right)\) × h
28.5 = 20 + \(\frac{\left[\frac{60}{2}-(5+x)\right]}{20}\) × 10
28.5 = 20 + \(\frac{(30-5-x)}{20}\) × 10
28.5 = 20 + \(\frac{(25-x)}{2}\)
28.5 – 20 = \(\frac{25-x}{2}\)
8.5 = \(\frac{25-x}{2}\)
⇒ 17 = 25 – x
⇒ x = 25 – 17
⇒ x = 8
Putting the value of x in equation (1), we get
8 + y = 15
⇒ y = 15 – 8 = 7.
Hence, x = 8, y = 7.

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

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.

Haryana Board 10th Class Maths Solutions Chapter 14 Statistics Ex 14.3 6

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

Haryana Board 10th Class Maths Solutions Chapter 14 Statistics Ex 14.3 7

n = 100
⇒ \(\frac{n}{2}=\frac{100}{2}\) = 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.
∴ l = 35, cf = 45, f = 33 and h = 5
∴ Median = l + \(\left(\frac{\frac{n}{2}-c f}{f}\right)\) × h
= 35 + \(\left(\frac{50-45}{33}\right)\) × 5
= 35 + \(\frac{25}{33}\)
= 35 + 0.76
= 35.76
Hence, median age = 35.76 years.

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

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 :

Haryana Board 10th Class Maths Solutions Chapter 14 Statistics Ex 14.3 8

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 :

Haryana Board 10th Class Maths Solutions Chapter 14 Statistics Ex 14.3 9

n = 40
⇒ \(\frac{n}{2}\) = 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.
∴ l = 144.5, cf = 17, f = 12, and h = 9
Median = l + \(\left(\frac{\frac{n}{2}-c f}{f}\right)\) × h
= 144.5 + \(\left(\frac{20-17}{12}\right)\) × 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.

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

Question 5.
The following table gives the distribution of the life time of 400 neon lamps :

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

Find the median life time of a lamp.
Solution:
We prepare the cumulative frequency distribution table as given below:

Haryana Board 10th Class Maths Solutions Chapter 14 Statistics Ex 14.3 11

n = 400
⇒ \(\frac{n}{2}=\frac{400}{2}\) = 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.
∴ l = 3000, cf =130, f = 86, and h = 500
∴ Median = l + \(\left(\frac{\frac{n}{2}-c f}{f}\right)\) × h
= 3000 + \(\left(\frac{200-130}{86}\right)\) × 500
= 3000 + \(\frac{35000}{86}\)
= 3000 + 406.98 = 3406.98
Hence, median life time of a lamp = 3406.98 hours.

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

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 :

Haryana Board 10th Class Maths Solutions Chapter 14 Statistics Ex 14.3 12

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 :

Haryana Board 10th Class Maths Solutions Chapter 14 Statistics Ex 14.3 13

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

Total n = 100
⇒ \(\frac{n}{2}=\frac{100}{2}\) = 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.
∴ l = 7, cf = 36, f = 40 and h = 3
∴ Median = l + \(\left(\frac{\frac{n}{2}-c f}{f}\right)\) × h
= 7 + \(\left(\frac{50-36}{40}\right)\) × 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 :

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

From the table, we have Σfi = 100, Σfidi = – 18, a = 8.5
∴ 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.
∴ l = 7, f1 = 40, f0 = 30, f2 = 16 and h = 3
Mode = l + \(\left(\frac{f_1-f_0}{2 f_1-f_0-f_2}\right)\) × h
= 7 + \(\left(\frac{40-30}{2 \times 40-30-16}\right)\) × 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.

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

Question 7.
The distribution below gives the weights of 30 students of a class. Find the median weight of the students. ’

Haryana Board 10th Class Maths Solutions Chapter 14 Statistics Ex 14.3 15

Solution:
We prepare the cumulative frequency distribution table as given below.

Haryana Board 10th Class Maths Solutions Chapter 14 Statistics Ex 14.3 16

n = 30
⇒ \(\frac{n}{2}=\frac{30}{2}\) = 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.
∴ l = 55, cf = 13, f = 6 and h = 5
Median = 55 + \(\left(\frac{\frac{n}{2}-c f}{f}\right)\) × h
= 55 + \(\left(\frac{15-13}{6}\right)\) × 5
= 55 + \(\frac{10}{6}\)
= 55 + 1.67 = 56.67
Hence, median weight of the students = 56.67 kg.

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