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Question
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 ______.
Options
8.25
6.5
3.87
2.87
Solution
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 8.25.
Explanation:
First 10 positive integers are 1, 2, 3, 4, 5, 6, 7, 8, 9, 10
On multiplying each number by – 1, we get
– 1, – 2, – 3, – 4, – 5, – 6, – 7, – 8, – 9, – 10
On adding 1 to each of the number, we get
0, – 1, – 2, – 3, – 4, – 5, – 6, – 7, – 8, – 9
∴ `sumx_i` = 0 – 1 – 2 – 3 – 4 – 5 – 6 – 7 – 8 – 9
= – 45
And `sumx_i^2 = 0^2 + (-1)^2 + (-2)^2 + (-3)^2 + (-4)^2 + ... + (-9)^2`
= `(9 xx 10 xx 19)/6`
= 285 .....`[because sumn^2 = (n(n + 1)(2n + 1))/6]`
∴ S.D. = `sqrt((sumx_i^2)/N - ((sumx_i^2)/N)^2`
= `sqrt(285/10 - ((-45)/10)^2`
= `sqrt(285/10 - 2025/100) - sqrt((2850 - 2025)/100`
= `sqrt(8.25)`
∴ Variance = (S.D.)2
= `(sqrt(8.25))^2`
= 8.25
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