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1 votes
1 votes
consider the first 10 positive integers . If we multiply each number by -1 and then add 1 to each number , the variance of the numbers so obtained is

1) 6.5

2) 8.25

3) 3.87

4) 2.87

2 Answers

Best answer
2 votes
2 votes
Let $\sigma$ be variance and $\mu$ be mean

$\sigma = \sum_{i=1}^{10} \frac{(x_i - \mu)^2}{n} \\= \sum_{i=1}^{10} \frac{{x_i}^2}{n} - \mu^2 \\= \frac{0^2 + (-1)^2 + \dots + (-9)^2}{10} - (-4.5)^2 \\= \frac{ 9 . 10 . 19} {6 . 10} - 20.25\\ \left(\because \text{Sum of squares of first $n$ natural numbers $=  \frac{n.(n+1).(2n+1)}{6}$} \right)  \\= 28.5 - 20.25 = 8.25$
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0 votes
0 votes

the sequence ranges from -9 to 0.

mean=-4.5

var=sum(-i+4.5)^2)/10, i=0 to 9

=19.25+12.25+6.25+2.25+0.25+0.25+2.25+6.25+12.25+19.25=80.5/10

nearest to 2nd option.

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