Mean & Median — Exercise 18(A)

The mean of x - 2, x, x + 2, x + 4 is :

1. x + 1

2. x

3. 4x + 4

4. x + 2

Answer

Mean = $\dfrac{\text{Sum of all observation }}{\text{Number of all observation }}$

Mean = $\dfrac{(x - 2) + x + (x +2) + (x + 4)}{4}$

= $\dfrac{x - 2 + x + x +2 + x + 4}{4}$

= $\dfrac{4x + 4}{4}$

= x + 1

Hence, option 1 is the correct option.

Mean of 10 observations (numbers) is 20. If one number is included, the mean becomes 21, the included number is :

1. 431

2. 31

3. 231

4. 131

Answer

Given:

Number of observations = 10

Mean = 20

Let a be the sum of all observations.

Mean = $\dfrac{\text{Sum of all observations }}{\text{Number of all observations }}$

⇒ 20 = $\dfrac{a}{10}$

⇒ 20 x 10 = a

⇒ 200 = a

Let b be the number added to the observation.

New mean = 21

New mean = $\dfrac{\text{Sum of all old observations + b}}{\text{Number of all old observations + 1}}$

⇒ 21 = $\dfrac{200 + \text{b}}{11}$

⇒ 21 x 11 = 200 + b

⇒ 231 = 200 + b

⇒ b = 231 - 200

⇒ b = 31

Hence, option 2 is the correct option.

Mean of 20 observations (numbers) is 30. If one number is excluded, the mean of remaining numbers becomes 28. The excluded number is :

1. 532

2. 1132

3. 68

4. 572

Answer

Given:

Mean of 20 observations = 30

⇒ Sum of all 20 observations = 30 x 20 = 600

On excluding an observation, the mean of the remaining 19 observations = 28

∵ Sum of all remaining 19 observations = 28 x 19 = 532

⇒ Excluded observation = Sum of all 20 observations - Sum of all remaining 19 observations

= 600 - 532

= 68

Hence, option 3 is the correct option.

If each observation of the data is decreased by 15; then the mean :

1. remains same

2. is increased by 15

3. is decreased by 15

4. is multiplied by 15

Answer

According to property 3, if each observation of the data is decreased by a quantity a, the mean is also decreased by the same quantity a.

Therefore, when each observation of the data is decreased by 15, the mean is decreased by 15.

Hence, option 3 is the correct option.

If the mean of x₁ and x₂ is 12.5 and mean of x₁, x₂ and x₃ is 16. The value of x₃ is :

1. 32

2. 3.5

3. 285

4. 23

Answer

Given:

The mean of x₁ and x₂ = 12.5

Mean = $\dfrac{\text{Sum of all observations }}{\text{Number of all observations }}$

⇒ 12.5 = $\dfrac{x_1 + x_2}{2}$

⇒ 12.5 x 2 = ${x_1 + x_2}$

⇒ ${x_1 + x_2}$ = 25 ...............(1)

The mean of x₁, x₂ and x₃ = 16

⇒ 16 = $\dfrac{x_1 + x_2 + x_3}{3}$

⇒ 16 x 3 = ${x_1 + x_2 + x_3}$

⇒ ${x_1 + x_2 + x_3}$ = 48 ...............(2)

Subtracting equation (1) from (2), we get

⇒ ${(x_1 + x_2 + x_3) - (x_1 + x_2)}$ = 48 - 25

⇒ ${x_3}$ = 23

Hence, option 4 is the correct option.

Find the mean of 43, 51, 50, 57 and 54.

Answer

Mean = $\dfrac{\text{Sum of all observations }}{\text{Number of all observations }}$

= $\dfrac{43 + 51 + 50 + 57 + 54}{5}$

= $\dfrac{255}{5}$

= 51

Hence, the mean is 51.

Find the mean of first six natural numbers.

Answer

Mean = $\dfrac{\text{Sum of all observations }}{\text{Number of all observations }}$

= $\dfrac{1 + 2 + 3 + 4 + 5 + 6}{6}$

= $\dfrac{21}{6}$

= 3.5

Hence, the mean of first six natural numbers is 3.5.

Find the mean of first ten odd natural numbers.

Answer

Mean = $\dfrac{\text{Sum of all observations }}{\text{Number of all observations }}$

= $\dfrac{1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 + 17 + 19}{10}$

= $\dfrac{100}{10}$

= 10

Hence, the mean of first ten odd natural numbers is 10.

Find the mean of all factors of 10.

Answer

Factors of 10 = 1 , 2 , 5 , 10

Mean = $\dfrac{\text{Sum of all observations }}{\text{Number of all observations }}$

= $\dfrac{1 + 2 + 5 + 10}{4}$

= $\dfrac{18}{4}$

= 4.5

Hence, the mean of all factors of 10 is 4.5.

Find the mean of x + 3, x + 5, x + 7, x + 9 and x + 11.

Answer

Mean = $\dfrac{\text{Sum of all observations }}{\text{Number of all observations }}$

= $\dfrac{(x + 3) + (x + 5) + (x + 7) + (x + 9) + (x + 11)}{5}$

= $\dfrac{5x + 35}{5}$

= x + 7

Hence, the mean is x + 7.

If different values of variable x are 9.8, 5.4, 3.7, 1.7, 1.8, 2.6, 2.8, 8.6, 10.5 and 11.1; find

(i) the mean $\overline{x}$

(ii) the value of $∑(x - \overline{x})$

Answer

(i) Given:

The observation are 9.8, 5.4, 3.7, 1.7, 1.8, 2.6, 2.8, 8.6, 10.5 and 11.1.

Mean = $\dfrac{\text{Sum of all observations }}{\text{Number of all observations }}$

= $\dfrac{9.8 + 5.4 + 3.7 + 1.7 + 1.8 + 2.6 + 2.8 + 8.6 + 10.5 + 11.1}{10}$

= $\dfrac{58}{10}$

= 5.8

Hence, the mean $\overline{x}$ = 5.8.

(ii)

$∑(x - \overline{x})$ = 4 + (-0.4) + (-2.1) + (-4.1) + (-4) + (-3.2) + (-3) + 2.8 + 4.7 + 5.3

= 4 - 0.4 - 2.1 - 4.1 - 4 - 3.2 - 3 + 2.8 + 4.7 + 5.3

= 0

Hence, $∑(x - \overline{x})$ = 0.

The mean of 15 observations is 32. Find the resulting mean, if each observation is :

(i) increased by 3

(ii) decreased by 7

(iii) multiplied by 2

(iv) divided by 0.5

(v) increased by 60%

(vi) decreased by 20%

Answer

(i) According to property 2, if each observation is increased by quantity a, then the mean is also increased by the same quantity a.

The mean of 15 observations is 32.

If each observation is increased by 3, then the mean increases by 3 ( 32 + 3 = 35).

Hence, the mean of the new observations = 35.

(ii) According to property 3, if each observation is decreased by a quantity a, then the mean is also decreased by the same quantity a.

The mean of 15 observations is 32.

If each observation is decreased by 7, then the mean is decreased by 7 ( 32 - 7 = 25).

Hence, the mean of the new observations = 25.

(iii) According to property 4, if each observation is multiplied by a quantity a, then the mean is also multiplied by the same quantity a.

The mean of 15 observations is 32.

If each observation is multiplied by 2, then the mean is multiplied by 2 ( 32 x 2 = 64).

Hence, the mean of the new observations = 64.

(iv) According to property 5, if each observation is divided by a quantity a, then the mean is also divided by the same quantity a.

The mean of 15 observations is 32.

If each observation of the data is divided by 0.5, then the mean is divided by 0.5 $\Big(\dfrac{32}{0.5} = 64\Big)$.

Hence, the mean of the new observations = 64.

(v) According to property 2, if each observation is increased by a quantity a, then the mean is also increased by the same quantity a.

The mean of 15 observations is 32.

If each observation is increased by 60%, then the mean is increased by 60% ( 32 + $\dfrac{60}{100} \times 32$ = 32 + 19.2 = 51.2).

Hence, the mean of the new observations = 51.2.

(vi) According to property 3, if each observation is decreased by a quantity a, then the mean is also decreased by the same quantity a.

The mean of 15 observations is 32.

If each observation is decreased by 20%; then the mean is decreased by 20% ( 32 - $\dfrac{20}{100} \times 32$ = 32 - 6.4 = 25.6).

Hence, the mean of the new observations = 25.6.

The mean of 5 numbers is 18. If one number is excluded, the mean of remaining numbers becomes 16. Find the excluded number.

Answer

Given:

Number of observations = 5

Mean = 18

⇒ Sum of all 5 observations = 5 x 18 = 90

On excluding an observation, the mean of the remaining 4 observations = 16

∵ Sum of all remaining 4 observations = 4 x 16 = 64

⇒ Excluded observation = Sum of all 5 observations - Sum of all remaining 4 observations

= 90 - 64

= 26

Hence, the excluded number is 26.

If the mean of observations x, x + 2, x + 4, x + 6 and x + 8 is 11, find the value of x.

Answer

Given, observations = x, x + 2, x + 4, x + 6 and x + 8

Mean = 11

Number of observations = 5

By formula,

Mean = $\dfrac{\text{Sum of observations}}{\text{Number of observation}}$

Substituting values we get :

$\Rightarrow 11 = \dfrac{x + x + 2 + x + 4 + x + 6 + x + 8}{5} \\[1em] \Rightarrow 11 = \dfrac{5x + 20}{5} \\[1em] \Rightarrow 55 = 5x + 20 \\[1em] \Rightarrow 5x = 55 - 20 \\[1em] \Rightarrow 5x = 35 \\[1em] \Rightarrow x = \dfrac{35}{5} \\[1em] \Rightarrow x = 7.$

Hence, the value of x = 7.