Question on: JAMB Mathematics - 1998
Find the variance of the numbers k, k+1, k+2,
A
\(\frac{2}{3}\)
B
1
C
k + 1
D
(k + 1)2
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Correct Option: A
mean (x) = \(\frac{\sum x}{N}\)
= k + k + 1 + k + 3
= \(\frac{3k + 3}{3}\)
= k + 1
\(\begin{array}{c|c} X & (X -X) & (X -X)^2\ \hline k & -1 & 1 \ k + 1 & 0 & 0\ k + 2 & 1 \ \hline & & 2\end{array}\)
Variance (52) = \(\frac{\sum (x - x)^2}{N}\)
= \(\frac{2}{3}\)
= k + k + 1 + k + 3
= \(\frac{3k + 3}{3}\)
= k + 1
\(\begin{array}{c|c} X & (X -X) & (X -X)^2\ \hline k & -1 & 1 \ k + 1 & 0 & 0\ k + 2 & 1 \ \hline & & 2\end{array}\)
Variance (52) = \(\frac{\sum (x - x)^2}{N}\)
= \(\frac{2}{3}\)
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