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.

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.

Solution:
For Median:

Here, Σf_i = 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

For Mean:

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

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.

Solution:

In the given datas, Σf_i = 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)

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.

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.

Solution:

Here, Σf_i = 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

= 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:

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:

Here, Σf_i = 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

Hence the median length of the leaves is 146.75 mm.

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

Find the median lifetime of a lamp.
Solution:

Here, Σf_i = 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

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:

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:

Here, Σf_i = 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

Hence, the median number of letters is 8.05.

For Mean:

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

Hence the modal size of the surnames is 7.88 letters.

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

Solution:

Here, Σf_i = 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

Hence the median weight is 56.67