PSEB 10th Class Maths Solutions Chapter 14 Statistics Ex 14.3

Punjab State Board PSEB 10th Class Maths Book Solutions Chapter Statistics Ex 14.3 Textbook Exercise Questions and Answers.

PSEB Solutions for Class 10 Maths 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:

Here, \(\Sigma f_{i}\) = 68 then \(\frac{n}{2}=\frac{68}{2}\) = 34
Which lies in interval 125 – 145
Median class = 125 – 145
So, l = 125; n = 68; f = 20; çf = 22 and h = 20

Using formula, Median = l + \(\left[\frac{\frac{n}{2}-c f}{f}\right]\) × h

= 125 + \(\left\{\frac{\frac{68}{2}-22}{20}\right\}\) × 20

=125+ \(\frac{34-22}{20}\) × 20

= 125 + 12 = 137

For mean:

From above data, assumed mean (a) = 135
Width of class (h) = 20
∴ \(\bar{u}=\frac{\Sigma f_{i} u_{i}}{\Sigma f_{i}}=\frac{7}{68}\) = 0.102
Using formula, Mean \((\overline{\mathrm{X}})=a+h \bar{u}\)
\(\overline{\mathrm{X}}\) = 135 + 20 (0.102)
= 135 + 2.04 = 137.04.

For Mode:
In the given data,
Maximum frequency is 20 and it correspond to 125 – 145.
∴ Modal class = 125 – 145
So l = 125; f₁ = 20; f₀ = 13; f₂ = 14and h = 20
Using formula, Mode = l + \(\left(\frac{f_{1}-f_{0}}{2 f_{1}-f_{0}-f_{2}}\right)\) × h

Mode = 125 + \(\left(\frac{20-13}{2(20)-13-14}\right)\) × 20

= 125 + \(\frac{7}{40-27}\) × 20

= 125 + \(\frac{140}{13}\)
= 125 + 10.76923
= 125 + 10.77 = 135.77.
Hence. median, mean and mode of given data is 137 units: 137.04 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 thegiven data, \(\Sigma f_{i}\) = n = 60
∴ \(\frac{n}{2}=\frac{60}{2}\) = 30
Also, median of the distribution = 28.5 ………….(Given)
which lies in the class interval 20 – 30
Median class = 20 – 30
So, l = 20; f = 20; cf = 5 + x; h = 10
From table, it is clear that 45 + x + y = 60
x + y = 60 – 45 = 15
or x + y = 15 ……………….(1)
Now, using formula, Median = l + {\(\left\{\frac{\frac{n}{2}-c f}{f}\right\}\)

28.5 = 2o + \(\left\{\frac{30-(5+x)}{20}\right\}\)

or 28.5 = 20 + \(\frac{30-5-x}{2}\)

or 28.5 = \(\)

or 2(28.5) = 65 – x
or 57.0 = 65 – x
or x = 65 – 57 = 8
∴ x = 8
Substitute this value of x in (1), we get
8 + y = 15
Hence, values of x and y is 8 and 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 only given to persons having age 18 years onwards but less than 60 years.

Solution:

Here, \(\Sigma f_{i}\) = 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
Using formula, Median = l + \(\left\{\frac{\frac{n}{2}-c f}{f}\right\}\) × h

= 35 + \(\left\{\frac{\frac{100}{2}-45}{33}\right\}\) × 5

= 35 + \(\frac{50-45}{33}\) × 5

= 35 + \(\frac{25}{33}\)
= 35 + 0.7575 = 35 + 0.76 (approx.) = 35.76
Hence, median age of given data is 35.76 years.

Question 4.
The lengths of 40 leaves of a plant are measured correct to the nearest millimetre, and the data obtained is represented in the following table:

Find the median length of the leaves.
Solution:
Since the frequency distribution is not continuous, so firstly we shall make it continuous.

Here, \(\Sigma f_{i}\) = 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; h = 9
Using formula, Median = l + \(\left\{\frac{\frac{n}{2}-c f}{f}\right\}\) × h

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

= 144.5 + \(\frac{3 \times 9}{12}\)
= 144.5 + 225 = 146.75
Hence, median length of the leaves is 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:

Here, \(\Sigma f_{i}\) = 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
Using formula, Median = l + \(\left\{\frac{\frac{n}{2}-c f}{f}\right\}\) × h

Median = 3000 + \(\left\{\frac{\frac{400}{2}-130}{86}\right\}\) × 500

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

= 3000 + \(\frac{70 \times 500}{86}\) + 406.9767441

= 3000 + 406.98 (approx.) = 3406.98
Hence, median life time of a lamp is 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 Median:

Here, Here, \(\Sigma f_{i}\) = n = 100
∴ \(\frac{n}{2}=\frac{100}{2}\) = 50, which lies in interval 7 – 10.
∴ Median class = 7 – 10
So, l = 7; n = 100; f = 40; cf = 36 and h = 3
Using formula, Median = l + \(\left\{\frac{\frac{n}{2}-c f}{f}\right\}\) × h

Median = 7 + \(\left\{\frac{\frac{100}{2}-36}{40}\right\}\) × 3

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

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

= 7 + \(\frac{21}{20}\) = 7 + 1.05 = 8.05
Hence, the median of letters in the surnames is 8.05.

For Mean:

From above data, Assumed Mean (a) = 8.5
Width of class (h) = 3
∴ \(\bar{u}=\frac{\Sigma f_{i} u_{i}}{\Sigma f_{i}}=\bar{u}=\frac{-6}{100}\) = – 0.06

Using formula, Mean \((\overline{\mathrm{X}})=a+h \bar{u}\)
\(\bar{X}\) = 8.5 + 3 (- 0.06) = 8.5 – 0.18 = 8.32
Hence, mean number of letters in the surnames is 8.32.

For Modal:
In the given data Maximum frequency is 44 and it corresponds to interval 7 — 10
∴ Modal class = 7 – 10
So l = 7; f₁ = 40; f₀ = 30; f₂ = 16 and h = 3
Using formula, Mode = l + \(\left(\frac{f_{1}-f_{0}}{2 f_{1}-f_{0}-f_{2}}\right)\) × h

Mode = 7 + \(\left(\frac{40-30}{2(40)-30-16}\right)\) × 3

= 7 + \(\frac{10}{80-46}\) × 3

= 7 + \(\frac{30}{34}\) = 7 + 0.882352941

= 7 + 0.88 (approx.) = 7.88.
Hence. modal size of the surnames is 7.88 letters.

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

Solution:

Here, \(\Sigma f_{i}\) = n = 30
∴ \(\frac{n}{2}=\frac{30}{2}\) = 15; which lies in the interval 55 – 60.
∴ Median class = 55 – 60
So, l = 55; n = 30; f = 6; cf = 13 and h = 5
Using formula, Median = l + \(\left\{\frac{\frac{n}{2}-c f}{f}\right\}\) × h

Median = 55 + \(\left\{\frac{\frac{30}{2}-13}{6}\right\}\) × 5

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

= 55 + \(\frac{2 \times 5}{6}\)

= 55 + \(\frac{5}{3}\) = 55 + 1.66666
= 55 + 1.67 (approx.) = 56.67
Hence, median weight of the students are 56.67 kg.