Test Index

Measures of Central Tendency and Dispersion

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Question : 125 of 211

Marks: +1, -0

The variance of 1,2,4,5 and

x

2,

x

is

3
6
1
2

Solution:

Concept:

Use the variance formula:

\text{Variance} = \frac{1}{n}\sum x_i^2 - \left(\frac{\sum x_i}{n}\right)^2

.

Explanation:

Given data:

1, 2, 4, 5, x

;

n = 5

; variance

= 2

.

Sum of numbers:

1+2+4+5+x = 12 + x

.

Sum of squares:

1^2 + 2^2 + 4^2 + 5^2 + x^2 = 1+4+16+25+x^2 = 46 + x^2

.

Mean

\mu = \frac{12+x}{5}

.

Variance formula:

\frac{1}{5}(46+x^2) - \left(\frac{12+x}{5}\right)^2 = 2

.

Multiply by

25

:

5(46+x^2) - (12+x)^2 = 50

.

Expand:

230 + 5x^2 - (144 + 24x + x^2) = 50

.

Simplify:

230 + 5x^2 - 144 - 24x - x^2 = 50

→

86 + 4x^2 - 24x = 50

.

4x^2 - 24x + 36 = 0

→ divide by

4

:

x^2 - 6x + 9 = 0

.

Factor:

(x - 3)^2 = 0

→

x = 3

.

Answer:

x = 3

, which is option A.

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