RBSE Solutions for Class 10 Maths Chapter 14 Statistics Ex 14.3

Rajasthan Board RBSE Solutions for Class 10 Maths Chapter 14 Statistics Ex 14.3 Textbook Exercise Questions and Answers.

Rajasthan Board RBSE Solutions for Class 10 Maths in Hindi Medium & English Medium are part of RBSE Solutions for Class 10. Students can also read RBSE Class 10 Maths Important Questions for exam preparation. Students can also go through RBSE Class 10 Maths Notes to understand and remember the concepts easily. Make use of our handy algebraic arithmetic sequences calculator and find the Sum of n terms of the arithmetic sequence.

RBSE Class 10 Maths Solutions Chapter 14 Statistics Ex 14.3

Questions 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.

RBSE Solutions for Class 10 Maths Chapter 14 Statistics Ex 14.3 Q1

Solution:
For Median:

RBSE Solutions for Class 10 Maths Chapter 14 Statistics Ex 14.3 Q1.1

Here, Σfi = n = 68 then \(\frac{n}{2}=\frac{68}{2}\) = 34
Which lies in the class-interval 125 - 145.
∴ Median Class = 125 - 145
So, l = 125; n = 68; f = 20; cf = 22 and h = 20

RBSE Solutions for Class 10 Maths Chapter 14 Statistics Ex 14.3 Q1.2

For Mean:

RBSE Solutions for Class 10 Maths Chapter 14 Statistics Ex 14.3 Q1.3

Here, Assumed Mean (a) = 135 and Class Size (h) = 20
\(\vec{u}=\frac{\sum f_{i} u_{i}}{\sum f_{i}}=\frac{7}{68}\) = 0.103
∵ Mean = \((\overline{\mathbf{X}})=a+h \bar{u}\)
= 135 + 20(0.103)
= 135 + 2.06
= 137.06 units
For Mode:
In the given data the maximum frequency is 20 and its corresponding class is 125-145.
∴ Modal Class = 125 - 145
So l = 125; f1 = 20; f0 = 13; f2 = 14 and h = 20

RBSE Solutions for Class 10 Maths Chapter 14 Statistics Ex 14.3 Q1.4

Hence, the median, mean, and mode of the given data are 137 units, 137.06 units, and 135.77 units.

Question 2.
If the median of the distribution given below is 28.5, find the values of x and y.

RBSE Solutions for Class 10 Maths Chapter 14 Statistics Ex 14.3 Q2

Solution:

RBSE Solutions for Class 10 Maths Chapter 14 Statistics Ex 14.3 Q2.1
RBSE Solutions for Class 10 Maths Chapter 14 Statistics Ex 14.3 Q2.2

In the given datas, Σfi = n = 60
\(\frac{n}{2}=\frac{60}{2}\) = 30
Also, median of distribution = 28.5 which lies in the class interval 20 - 30.
∴ Median Class = 20 - 30
So, l = 20; f = 20; cf = 5 + x; h = 10
From Table 45 + x + y = 60
or x + y = 60 - 45 = 15
or x + y = 15 ....... (i)

RBSE Solutions for Class 10 Maths Chapter 14 Statistics Ex 14.3 Q2.3

Substituting this value of x is (i)
8 + y = 15
y = 15 - 8 = 7
Hence the values of x and y are 8 and 7.

RBSE Solutions for Class 10 Maths Chapter 14 Statistics Ex 14.3

Question 3.
A life insurgence agent found the following data for the distribution of ages of 100 policyholders. Calculate the median age, if policies are given only to persons aged 18 years onwards but less than 60 years.

RBSE Solutions for Class 10 Maths Chapter 14 Statistics Ex 14.3 Q3

Solution:

RBSE Solutions for Class 10 Maths Chapter 14 Statistics Ex 14.3 Q3.1

Here, Σfi = n = 100
Then, \(\frac{n}{2}=\frac{100}{2}\) = 50, which lies in the interval 35 - 40.
∴ Median Class = 35 - 40
So, l = 35; n = 100; f = 33; cf = 45 and h = 5

RBSE Solutions for Class 10 Maths Chapter 14 Statistics Ex 14.3 Q3.2

= 35 + 0.7575
= 35 + 0.76 (approx.)
= 35.76
Hence the median age of the given data is 35.76 years.

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:

RBSE Solutions for Class 10 Maths Chapter 14 Statistics Ex 14.3 Q4

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)
Solution:
Since the frequency distribution is not continuous, therefore we shall first convert it into a continuous frequency distribution:

RBSE Solutions for Class 10 Maths Chapter 14 Statistics Ex 14.3 Q4.1

Here, Σfi = n = 40
Then \(\frac{n}{2}=\frac{40}{2}\) = 20, which lies in the interval 144.5 - 153.5.
∴ Median Class = 144.5 - 153.5
So, l = 144.5; f = 12; cf = 17 and h = 9

RBSE Solutions for Class 10 Maths Chapter 14 Statistics Ex 14.3 Q4.2

Hence the median length of the leaves is 146.75 mm.

RBSE Solutions for Class 10 Maths Chapter 14 Statistics Ex 14.3

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

RBSE Solutions for Class 10 Maths Chapter 14 Statistics Ex 14.3 Q5

Find the median lifetime of a lamp.
Solution:

RBSE Solutions for Class 10 Maths Chapter 14 Statistics Ex 14.3 Q5.1

Here, Σfi = n = 400
\(\frac{n}{2}=\frac{400}{2}\) = 200, which lies in the interval 3000 - 3500.
∴ Median Class = 3000 - 3500
So l = 3000; n = 400; f = 86; cf = 130 and h = 500

RBSE Solutions for Class 10 Maths Chapter 14 Statistics Ex 14.3 Q5.2

Hence the median life of a lamp is 3406.98.

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:

RBSE Solutions for Class 10 Maths Chapter 14 Statistics Ex 14.3 Q6

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 Median:

RBSE Solutions for Class 10 Maths Chapter 14 Statistics Ex 14.3 Q6.1
RBSE Solutions for Class 10 Maths Chapter 14 Statistics Ex 14.3 Q6.2

Here, Σfi = n = 100
Then, \(\frac{n}{2}=\frac{100}{2}\) = 50, which lies in the interval 7-10.
∴ Median Class = 7 - 10
So l = 7; n = 100; f = 40; cf = 36 and h = 3

RBSE Solutions for Class 10 Maths Chapter 14 Statistics Ex 14.3 Q6.3

Hence, the median number of letters is 8.05.

For Mean:

RBSE Solutions for Class 10 Maths Chapter 14 Statistics Ex 14.3 Q6.4

Here Assured Mean (a) = 8.5 and Class Size (h) = 3
\(\frac{\sum f_{i} u_{i}}{\sum f_{i}}=\bar{u}=\frac{-6}{100}\) = -0.06
∵ Median \((\overline{\mathrm{X}})=a+h \bar{u}\)
= 8.5 + 3 (-0.06)
= 8.5 - 0.18
= 8.32
Hence, the median number of letters in the surnames is 8.32 letters.
For Mode:
In the given data the maximum frequency is 40 and its corresponding interval is 7-10.
∵ Median Class = 7 - 10
So l = 7; f1 = 40; f0 = 30; f2 = 16 and h = 3

RBSE Solutions for Class 10 Maths Chapter 14 Statistics Ex 14.3 Q6.5

Hence the modal size of the surnames is 7.88 letters.

RBSE Solutions for Class 10 Maths Chapter 14 Statistics Ex 14.3

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

RBSE Solutions for Class 10 Maths Chapter 14 Statistics Ex 14.3 Q7

Solution:

RBSE Solutions for Class 10 Maths Chapter 14 Statistics Ex 14.3 Q7.1

Here, Σfi = n = 30
Then, \(\frac{n}{2}=\frac{30}{2}\) = 15, which lies in the interval 55-60.
∴ Median Class = 55 - 60
So l = 55; n = 30; f = 6; cf = 13 and h = 5

RBSE Solutions for Class 10 Maths Chapter 14 Statistics Ex 14.3 Q7.2

Hence the median weight is 56.67

Raju
Last Updated on April 30, 2022, 3:16 p.m.
Published April 30, 2022