Data Handling - Exercise 24(C)

Find the mode of each of the following data:

(i) 9, 7, 8, 16, 12, 9, 8, 5, 9, 11

(ii) 29, 34, 38, 25, 25, 19, 25, 16, 37, 21, 25

Answer

(i) Arranging the given data in ascending order, we get :

5, 7, 8, 8, 9, 9, 9, 11, 12, 16

Since 9 occurs most frequently in the given data, the mode is 9.

Hence, the mode is 9.

(ii) Arranging the given data in ascending order, we get :

16, 19, 21, 25, 25, 25, 25, 29, 34, 37, 38

Since 25 occurs most frequently in the given data, the mode is 25.

Hence, the mode is 25.

A shoe store sells pairs of shoes of different sizes. The number of pairs of various sizes sold on a particular day are given below :

What is the modal shoe-size?

Answer

From the given frequency distribution, it is clear that the maximum number of pairs sold is 8 which corresponds to shoe size 5.

Hence, the modal shoe-size is 5.

The following table shows the heights of 50 students of a class :

Find the mean and the median.

Using empirical formula, calculate its mode.

Answer

Arranging the terms in ascending order and preparing the cumulative frequency table :

Mean = $\dfrac{∑f_ix_i}{∑f_i}$

= $\dfrac{7511}{50}$

= 150.22

We have, N = 50, which is even.

Median = $\dfrac{1}{2}\Big[\Big(\dfrac{N}{2}\Big)^{th} \text{ term} + \Big(\dfrac{N}{2} + 1\Big)^{th} \text{ term}\Big]$

$= \dfrac{1}{2}\Big[\Big(\dfrac{50}{2}\Big)^{th} \text{ term} + \Big(\dfrac{50}{2} + 1\Big)^{th} \text{ term}\Big] \\[1em] = \dfrac{1}{2}\Big[(25)^{th} \text{ term} + (26)^{th} \text{ term}\Big] \\[1em] = \dfrac{1}{2}\Big[150 + 151\Big] \\[1em] = \dfrac{1}{2}\Big[301\Big] \\[1em] = 150.5$

Now, Mode = 3(Median) - 2(Mean)

= 3 × 150.5 - 2 × 150.22

= 451.5 - 300.44

= 151.06

Hence, Mean = 150.22, Median = 150.5 and Mode = 151.06.

The marks (out of 10) obtained by 35 students of a class are given in the following table :

Find the mean and the median.

Using empirical formula, calculate its mode.

Answer

Arranging the terms in ascending order and preparing the cumulative frequency table:

Mean = $\dfrac{∑f_ix_i}{∑f_i}$

= $\dfrac{223}{35}$

= 6.37 (approx.)

We have, N = 35, which is odd.

Median = $\Big[\dfrac{N+1}{2}\Big]^{th}$ term

= $\Big[\dfrac{35+1}{2}\Big]^{th}$ term

= $\Big[\dfrac{36}{2}\Big]^{th}$ term

= $18^{th}$ term

= 6

Now, Mode = 3(Median) - 2(Mean)

= 3 × 6 - 2 × 6.37

= 18 - 12.74

= 5.26

Hence, Mean = 6.37, Median = 6 and Mode = 5.26.

The following table gives the weights (in grams) of 30 boxes of fruits :

Find the mean and the median.

Using empirical formula, calculate its mode.

Answer

Arranging the terms in ascending order and preparing the cumulative frequency table :

Mean = $\dfrac{∑f_ix_i}{∑f_i}$

= $\dfrac{12825}{30}$

= 427.5

We have, N = 30, which is even.

Median = $\dfrac{1}{2}\Big[\Big(\dfrac{N}{2}\Big)^{th} \text{ term} + \Big(\dfrac{N}{2} + 1\Big)^{th} \text{ term}\Big]$

$= \dfrac{1}{2}\Big[\Big(\dfrac{30}{2}\Big)^{th} \text{ term} + \Big(\dfrac{30}{2} + 1\Big)^{th} \text{ term}\Big] \\[1em] = \dfrac{1}{2}\Big[(15)^{th} \text{ term} + (16)^{th} \text{ term}\Big] \\[1em] = \dfrac{1}{2}\Big[425 + 425\Big] \\[1em] = \dfrac{1}{2}\Big[850\Big] \\[1em] = 425$

Now, Mode = 3(Median) - 2(Mean)

= 3 × 425 - 2 × 427.5

= 1275 - 855

= 420

Hence, Mean = 427.5 g, Median = 425 g and Mode = 420 g.