Measures of Central Tendency — Exercise 21.3

Find the mode of the following sets of numbers :

(i) 5, 7, 6, 8, 9, 0, 6, 8, 1, 8

(ii) 9, 0, 2, 8, 5, 3, 5, 4, 1, 5, 2, 7

Answer

(i) In the given data : 5, 7, 6, 8, 9, 0, 6, 8, 1, 8

8 is repeated more number of times than any other number,

∴ mode = 8.

(ii) In the given data : 9, 0, 2, 8, 5, 3, 5, 4, 1, 5, 2, 7

5 is repeated more number of times than any other number,

∴ mode = 5.

Find the mean, median and mode of the following distribution :

8, 10, 7, 6, 10, 11, 6, 13, 10.

Answer

Arithmetic mean (A.M.) = $\dfrac{\text{Sum of terms}}{\text{No.of terms}} = \dfrac{∑x_i}{n}$

Sum of terms = 8 + 10 + 7 + 6 + 10 + 11 + 6 + 13 + 10 = 81.

$\therefore \text{A.M.} = \dfrac{81}{9} = 9.$

∴ Mean = 9.

On arranging the marks in ascending order, we get

6, 6, 7, 8, 10, 10, 10, 11, 13.

Here, n (no. of observations) = 9, which is odd.

$\therefore \text{Median} = \dfrac{n + 1}{2} \text{th observation} \\[1em] = \dfrac{9 + 1}{2} \\[1em] = \dfrac{10}{2} \\[1em] = 5 \text{th observation}$

5th observation = 10.

∴ Median = 10.

In the given data : 8, 10, 7, 6, 10, 11, 6, 13, 10

10 is repeated more number of times than any other number,

∴ Mode = 10.

Hence, mean = 9, median = 10 and mode = 10.

Calculate the mean, median and the mode of the following numbers :

3, 1, 5, 6, 3, 4, 5, 3, 7, 2.

Answer

Arithmetic mean (A.M.) = $\dfrac{\text{Sum of terms}}{\text{No.of terms}} = \dfrac{∑x_i}{n}$

Sum of terms = 3 + 1 + 5 + 6 + 3 + 4 + 5 + 3 + 7 + 2 = 39

$\therefore \text{A.M.} = \dfrac{39}{10} = 3.9$

∴ Mean = 3.9

On arranging the numbers in ascending order we get,

1, 2, 3, 3, 3, 4, 5, 5, 6, 7.

Here, n (no. of observations) = 10, which is even.

$\therefore \text{Median} = \dfrac{\dfrac{n}{2} \text{th observation} + \big(\dfrac{n}{2} + 1\big)\text{ th observation}}{2} \\[1em] = \dfrac{\dfrac{10}{2} \text{th observation} + \big(\dfrac{10}{2} + 1\big)\text{ th observation}}{2} \\[1em] = \dfrac{\text{5th observation + 6th observation}}{2} \\[1em] = \dfrac{3 + 4}{2} \\[1em] = \dfrac{7}{2} \\[1em] = 3.5$

∴ Median = 3.5

In the given data : 3, 1, 5, 6, 3, 4, 5, 3, 7, 2.

3 is repeated more number of times than any other number,

∴ Mode = 3.

Hence, mean = 3.9, median = 3.5 and mode = 3.

The marks of 10 students of a class in an examination arranged in ascending order are as follows :

13, 35, 43, 46, x, x + 4, 55, 61, 71, 80.

If the median marks is 48, find the value of x. Hence, find the mode of the given data.

Answer

Here, n (no. of observations) = 10, which is even.

$\therefore \text{Median} = \dfrac{\dfrac{n}{2} \text{th observation} + \big(\dfrac{n}{2} + 1\big)\text{ th observation}}{2} \\[1em] = \dfrac{\dfrac{10}{2} \text{th observation} + \big(\dfrac{10}{2} + 1\big)\text{ th observation}}{2} \\[1em] = \dfrac{\text{5th observation + 6th observation}}{2} \\[1em] = \dfrac{x + (x + 4)}{2} \\[1em] = \dfrac{2x + 4}{2} \\[1em] = x + 2.$

Given, median marks = 48.

∴ x + 2 = 48
⇒ x = 46.

Putting value of x in data we get,

13, 35, 43, 46, 46, 50, 55, 61, 71, 80.

In the given data 46 is repeated more number of times than any other number.

Hence, the value of x = 46 and mode = 46.

Find the mode and median of the following frequency distribution :

Answer

The variates are already in ascending order. We construct the cumulative frequency table as under :

Total number of observations = 29, which is odd.

$\therefore \text{Median} = \dfrac{n + 1}{2} \text{th observation} \\[1em] = \dfrac{29 + 1}{2} \\[1em] = \dfrac{30}{2} \\[1em] = 15 \text{th observation}$

All observations from 13th to 17th are equal, each = 13, so median = 13.

As the variate 14 has maximum frequency 9, so mode = 14.

Hence, median = 13 and mode = 14.

In a class of 40 students marks obtained by the students in a class test (out of 10) are given below :

Calculate the following for the given distribution :

(i) median

(ii) mode.

Answer

(i) The variates are already in ascending order. We construct the cumulative frequency table as under :

Total number of observations = 40, which is even.

$\therefore \text{Median} = \dfrac{\dfrac{n}{2} \text{th observation} + \big(\dfrac{n}{2} + 1\big)\text{th observation}}{2} \\[1em] = \dfrac{\dfrac{40}{2} \text{th observation} + \big(\dfrac{40}{2} + 1\big)\text{th observation}}{2} \\[1em] = \dfrac{\text{ 20th observation + 21st observation}}{2} \\[1em]$

All observations from 16th to 25th are equal, each = 6.

Hence, median

$= \dfrac{6 + 6}{2} \\[1em] = \dfrac{12}{2} \\[1em] = 6.$

Hence, median = 6.

(ii) As the variate 6 has maximum frequency 10, so mode = 6.

Hence, mode = 6.

The marks obtained by 30 students in a class assessment of 5 marks is given below :

Calculate the mean, median and mode of the above distribution.

Answer

The variates (marks) are already in ascending order. We construct the cumulative frequency table as under :

Mean = $\dfrac{∑f_ix_i}{∑f_i} = \dfrac{90}{30}$ = 3.

Total number of observations = 30, which is even.

$\therefore \text{Median} = \dfrac{\dfrac{n}{2} \text{th observation} + \big(\dfrac{n}{2} + 1\big)\text{th observation}}{2} \\[1em] = \dfrac{\dfrac{30}{2} \text{th observation} + \big(\dfrac{30}{2} + 1\big)\text{ th observation}}{2} \\[1em] = \dfrac{\text{ 15th observation + 16th observation}}{2} \\[1em]$

All observations from 11th to 20th are equal, each = 3.

Hence, median

$= \dfrac{3 + 3}{2} \\[1em] = \dfrac{6}{2} \\[1em] = 3.$

As the variate 3 has maximum frequency 10, so mode = 3.

Hence, mean = 3, median = 3 and mode = 3.

The distribution table given below shows the marks obtained by 25 students in an aptitude test. Find the mean, median and mode of the distribution.

Answer

The variates (marks) are already in ascending order. We construct the cumulative frequency table as under :

Mean = $\dfrac{∑f_ix_i}{∑f_i} = \dfrac{171}{25}$ = 6.84.

Total number of observations = 25, which is odd.

$\therefore \text{Median} = \dfrac{n + 1}{2} \text{th observation} \\[1em] = \dfrac{25 + 1}{2} \\[1em] = \dfrac{26}{2} \\[1em] = 13 \text{th observation}$

All observations from 13th to 18th are equal, each = 7, so median = 7.

As the variate 6 has maximum frequency 9, so mode = 6.

Hence, mean = 6.84, median = 7 and mode = 6.

The following table gives the weekly wages (in ₹) of workers in a factory:

Calculate:

(i) the mean.

(ii) the modal class.

(iii) the number of workers getting weekly wages below ₹800

(iv) the number of workers getting ₹650 or more but less than ₹850 as weekly wages.

Answer

(i) We construct the following table :

By formula,

$\text{Mean }= \dfrac{∑f_iu_i}{∑f_i}\\[1em] = \dfrac{55200}{80}\\[1em] = 690.$

Hence, mean = ₹ 690.

(ii) The class 550 - 600 has maximum frequency 20.

Hence, modal class = 550-600.

(iii) From table,

The number of workers getting weekly wages below ₹800 = 60.

(iv) From table,

The number of workers getting ₹650 or more but less than ₹850 as weekly wages = 72 - 35 = 37.