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Statistics Part 1

Section: Mathematics

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Question : 25 of 100

Marks: +1, -0

The mean and variance of 7 observations are 8 and 16 respectively. If two observations are 6 and 8 , then the variance of the remaining 5 observations is

[31 Aug 2021 Shift 2]

$\frac{92}{5}$
$\frac{134}{5}$
$\frac{536}{25}$
$\frac{112}{5}$

Solution:

Let a, b, c, d and e are 5 remaining observations n = 7, Mean = 8, Variance = 16

Sum of observations

=7 \times 8=56

Mean of remaining observations

=\frac{56-8-6}{5}=\frac{42}{5}

and Variance

=\frac{\sum x^{2}}{n}-(\overline{x})^{2}

16=\frac{\sum x^{2}}{7}-64

\Rightarrow \sum x^{2}=560

\Rightarrow a^{2}+b^{2}+c^{2}+d^{2}+e^{2}=560-8^{2}-6^{2}

=460

Variance of 5 remaining observations

=\frac{460}{5}-\left(\frac{42}{5}\right)^{2}=\frac{536}{25}

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