Measures of Central Tendency — Exercise 24(B)

The mean of given observations is :

1. 41

2. 27

3. 71

4. 91

Answer

By formula,

Mean = $\dfrac{Σfx}{Σf} = \dfrac{205}{5}$ = 41.

Hence, Option 1 is the correct option.

For data given in the adjoining table, the mean is :

1. 17

2. 27

3. 25

4. 60

Answer

In the table,

x is the mean value or class mark,

i = class-size = 10,

By formula,

Mean = A + $\dfrac{Σfu}{Σf} \times i$

Substituting values we get :

Mean = 15 + $\dfrac{5}{25} \times 10 = 15 + \dfrac{50}{25}$ = 15 + 2 = 17.

Hence, Option 1 is the correct option.

The mean of observations, given in the adjoining table, is 24, the value of a is :

1. 20

2. 30

3. 10

4. 25

Answer

By formula,

Mean = $\dfrac{Σfx}{Σf}$

Substituting values we get :

$\Rightarrow 24 = \dfrac{1000 + 20a}{40 + a}$

⇒ 24(40 + a) = 1000 + 20a

⇒ 960 + 24a = 1000 + 20a

⇒ 24a - 20a = 1000 - 960

⇒ 4a = 40

⇒ a = 10.

Hence, Option 3 is the correct option.

If the mean of the data given in adjoining table is 20, the relation between x₁ and x₂ is :

1. x₁ + x₂ = 30

2. x₁ - x₂ = 15

3. x₁ - x₂ = 0

4. x₁ + x₂ = 20

Answer

By formula,

Mean = $\dfrac{Σfx}{Σf}$

Substituting values we get :

$\Rightarrow 20 = \dfrac{400 + 10x_1 + 30x_2}{20 + x_1 + x_2} \\[1em] \Rightarrow 20(20 + x_1 + x_2) = 400 + 10x_1 + 30x_2 \\[1em] \Rightarrow 400 + 20x_1 + 20x_2 = 400 + 10x_1 + 30x_2 \\[1em] \Rightarrow 20x_1 - 10x_1 + 20x_2 - 30x_2 = 400 - 400 \\[1em] \Rightarrow 10x_1 - 10x_2 = 0 \\[1em] \Rightarrow 10(x_1 - x_2) = 0 \\[1em] \Rightarrow x_1 - x_2 = 0.$

Hence, Option 3 is the correct option.

The mean of data, represented by given diagram is :

1. 53

2. 47

3. 42

4. 51

Answer

Table from the given graph is :

By formula,

Mean = $\dfrac{Σfx}{Σf} = \dfrac{2350}{50}$ = 47.

Hence, Option 2 is the correct option.

The following table gives the ages of 50 students of a class. Find the arithmetic mean of their ages.

Answer

By formula,

Mean = $\dfrac{Σfx}{Σf} = \dfrac{1074}{50}$ = 21.48

Hence, mean = 21.48

The following are the marks obtained by 70 boys in a class test.

Calculate the mean by :

(i) Short-cut method

(ii) Step-deviation method

Answer

(i) Let assumed mean (A) be 65.

n = Σf = 70.

By formula,

Mean = A + $\dfrac{Σfd}{n} = 65 + \dfrac{-270}{70}$

= 65 - 3.86 = 61.14

Hence, mean = 61.14

(ii) We get mean values from part (i) and assumed mean (A) = 65. Let i = 10.

n = Σf = 70.

By formula,

Mean = A + $\dfrac{Σft}{n} \times i$

= 65 + $\dfrac{-27}{70} \times 10$

= 65 - $\dfrac{27}{7}$

= 65 - 3.86

= 61.14

Hence, mean = 61.14

Find mean by 'step-deviation method' :

Answer

Let assumed mean (A) be 87.5 and i = 7.

n = Σf = 160.

By formula,

Mean = A + $\dfrac{Σft}{n} \times i$

= 87.5 + $\dfrac{29}{160} \times 7$

= 87.5 + $\dfrac{203}{160}$

= 87.5 + 1.3

= 88.8

Hence, mean = 88.8

The mean of following frequency distribution is $21\dfrac{1}{7}$. Find the value of 'f'.

Answer

By formula,

Mean = $\dfrac{Σfx}{Σf}$

$\Rightarrow 21\dfrac{1}{7} = \dfrac{1235 + 35f}{63 + f} \\[1em] \Rightarrow \dfrac{148}{7} = \dfrac{1235 + 35f}{63 + f} \\[1em] \Rightarrow 148(63 + f) = 7(1235 + 35f) \\[1em] \Rightarrow 9324 + 148f = 8645 + 245f \\[1em] \Rightarrow 245f - 148f = 9324 - 8645 \\[1em] \Rightarrow 97f = 679 \\[1em] \Rightarrow f = \dfrac{679}{97} = 7.$

Hence, f = 7.

Using the information given in the adjoining histogram; calculate the mean.

Answer

The table for the adjoining histogram is :

By formula,

Mean = $\dfrac{Σfx}{Σf} = \dfrac{2850}{75}$ = 38.

Hence, mean = 38.

If the mean of the following observations is 54, find the value of p.

Answer

By formula,

Class mark = $\dfrac{\text{Upper limit + Lower limit}}{2}$

By formula,

Mean = $\dfrac{Σfx}{Σf}$

⇒ 54 = $\dfrac{2370 + 30p}{39 + p}$

⇒ 54(39 + p) = 2370 + 30p

⇒ 2106 + 54p = 2370 + 30p

⇒ 54p - 30p = 2370 - 2106

⇒ 24p = 264

⇒ p = $\dfrac{264}{24}$

⇒ p = 11.

Hence, p = 11.

The mean of the following distribution is 62.8 and the sum of all the frequencies is 50. Find the missing frequencies f₁ and f₂.

Answer

By formula,

Class mark = $\dfrac{\text{Upper limit + Lower limit}}{2}$

Given,

Sum of frequencies = 50

⇒ f₁ + f₂ + 30 = 50

⇒ f₁ + f₂ = 20

⇒ f₁ = 20 - f₂ ........(1)

By formula,

Mean = $\dfrac{Σfx}{Σf}$

⇒ 62.8 = $\dfrac{2060 + 30f_1 + 70f_2}{50}$

⇒ 2060 + 30f₁ + 70f₂ = 3140

⇒ 30f₁ + 70f₂ = 1080

Substituting value of f₁ in above equation from (1), we get :

⇒ 30(20 - f₂) + 70f₂ = 1080

⇒ 600 - 30f₂ + 70f₂ = 1080

⇒ 40f₂ = 480

⇒ f₂ = $\dfrac{480}{40}$ = 12.

⇒ f₁ = 20 - f₂ = 20 - 12 = 8.

Hence, f₁ = 8 and f₂ = 12.

Calculate the mean of the distribution, given below, using the short cut method :

Answer

The above distribution is discontinuous, converting into continuous distribution, we get :

Adjustment factor = (Lower limit of one class - Upper limit of previous class) / 2

= $\dfrac{21 - 20}{2} = \dfrac{1}{2}$

= 0.5

Subtract the adjustment factor (0.5) from all the lower limits and add the adjustment factor (0.5) to all the upper limits.

Let assumed mean (A) be 45.5

n = Σf = 50

Mean = A + $\dfrac{Σfd}{n}$

= $45.5 + \dfrac{70}{50}$

= 45.5 + 1.4

= 46.9

Hence, mean = 46.9