Introduction to statistics

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Question

The mean and variance of 8 observations are 10 and 13.5 respectively. If 6 of these observations are 5, 7, 10, 12, 14, 15, then the absolute difference of the remaining two observations is

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Solution

Let the remaining two observations be x and y. Then,
Mean =10
$\begin{array}{l} \Rightarrow \frac{1}{8} (5 + 7 + 10 + 12 + 14 + 15 + x + y) = 10 \\ \Rightarrow 63 + x + y = 80 \Rightarrow x + y = 17 \end{array}$
Variance =13.5
$\begin{array}{l} \Rightarrow \frac{1}{8} (25 + 49 + 100 + 144 + 196 + 225 + x^{2} + y^{2}) - 10^{2} = 13.5 \\ \Rightarrow x^{2} + y^{2} + 739 = 908 \\ \Rightarrow x^{2} + y^{2} = 169 \end{array}$
$\begin{array}{ll} ∴ (x - y)^{2} + (x + y)^{2} = 2 (x^{2} + y^{2}) \\ \Rightarrow (x - y)^{2} = 2 \times 169 - 289 = 49 \\ \Rightarrow | x - y | = 7 \end{array}$

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Introduction to statistics

Remember concepts with our Masterclasses.

The mean and variance of 8 observations are 10 and 13.5 respectively. If 6 of these observations are 5, 7, 10, 12, 14, 15, then the absolute difference of the remaining two observations is

Let the remaining two observations be x and y. Then,Mean =10⇒ 18(5+7+10+12+14+15+x+y)=10⇒ 63+x+y=80⇒x+y=17Variance =13.5⇒ 1825+49+100+144+196+225+x2+y2−102=13.5⇒ x2+y2+739=908⇒ x2+y2=169∴ (x−y)2+(x+y)2=2x2+y2⇒ (x−y)2=2×169−289=49⇒ |x−y|=7

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