Question 47

waecmaths question: 

Calculate the variance 2, 3, 3, 4, 5, 5, 5, 7, 7 and 9

Option A: 

2.2

Option B: 

3.4

Option C: 

4.0

Option D: 

4.2

waecmaths solution: 

$\begin{align}  & mean=\bar{x}=\frac{2+3+3+4+5+5+5+7+7+9}{10}=\frac{50}{10}=5 \\ & variance={{\sigma }^{2}}=\frac{{{(2-5)}^{2}}+{{(3-5)}^{2}}+{{(3-5)}^{2}}+{{(4-5)}^{2}}+{{(5-5)}^{2}}+{{(5-5)}^{2}}+{{(5-5)}^{2}}+{{(7-5)}^{2}}+{{(7-5)}^{2}}+{{(9-5)}^{2}}}{10} \\ & {{\sigma }^{2}}=\frac{9+4+4+1+0+0+0+4+4+16}{10} \\ & {{\sigma }^{2}}=\frac{42}{10}=4.2 \\\end{align}$ 

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