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The Sums of the Deviations of a Set of N Values 𝑥1, 𝑥2, … . 𝑥11 Measured from 15 and −3 Are − 90 and 54 Respectively. Find the ValùE of N and Mean.F - Mathematics

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Question

The sums of the deviations of a set of n values 𝑥1, 𝑥2, … . 𝑥11 measured from 15 and −3 are − 90 and 54 respectively. Find the valùe of n and mean.

Solution

(i)  Given `sum _ (i =1)^n (x_i + 5) = - 90`

⇒`( x+_1 -15) + ( x_2 -15) + ....... + ( x_n -15) = -90`

⇒`( x_1 + x_2 + .......... + x_n ) - (15 +15 + ...... +15) = -90`

⇒ `sumx -15n `= - 90       ........(1)

And `sum _(i=1)^n( x_i + 3) = 54`

⇒ `( x_1 - 3) + ( x_2 - 3) + ....... + ( x_n + 3) = 54`.

⇒ `( x_1 + x_2 + x_3 + .......... + x_n ) + (3 + 3 + 3 + ...... + 37) = 54`

⇒ `sumx + 3n = 54 `      ....(2)

By subtracting equation (1) from equation (2)

`sumx - 30 - sumx + 15n = 54 + 90`

⇒ 18n = 144

⇒`n = 144/ 18 = 8`

Put value of n in equation (1)

`sumx - 15 xx 8 = - 90`

⇒ `sumx - 120 =- 90`

⇒ `sumx = - 90 + 120 = 30`

∴`Mean = (sumx) / n = 30 / 8 = 15 /4` 
 

 

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Chapter 24: Measures of Central Tendency - Exercise 24.1 [Page 10]

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RD Sharma Mathematics [English] Class 9
Chapter 24 Measures of Central Tendency
Exercise 24.1 | Q 22 | Page 10

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