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The mean of 5 observations is 15 and variance is 9. If two observations having values -5 and 13 are combined with these observations, then what will be the new variance?
Options:
Solution:
1641 Upvotes
Verified Answer
The correct answer is:
\(\frac{2659}{49}\)
Let the observations are \(x_1, x_2, x_3, x_4, x_5\), so the mean \(\bar{x}=\frac{x_1+x_2+x_3+x_4+x_5}{5}=15\) (given)
\(\Rightarrow \quad x_1+x_2+x_3+x_4+x_5=75\) ...(i)
and variance
\(\begin{aligned}
& \sigma^2=\frac{x_1^2+x_2^2+x_3^2+x_4^2+x_5^2}{5}-(\bar{x})^2=9 \quad \text{(given)} \\
& \Rightarrow x_1^2+x_2^2+x_3^2+x_4^2+x_5^2=5[9+225]=1170 \quad \ldots (ii)
\end{aligned}\)
Now, after combined the -5 and 13 with the observations, the new variance
\(\begin{aligned}
& =\frac{x_1^2+x_2^2+x_3^2+x_4^2+x_5^2+25+169}{7} -\left(\frac{75-5+13}{7}\right)^2 \\
& =\frac{1170+25+169}{7}-\left(\frac{83}{7}\right)^2=\frac{9548-6889}{49}=\frac{2659}{49}
\end{aligned}\)
Hence, option (d) is correct.
\(\Rightarrow \quad x_1+x_2+x_3+x_4+x_5=75\) ...(i)
and variance
\(\begin{aligned}
& \sigma^2=\frac{x_1^2+x_2^2+x_3^2+x_4^2+x_5^2}{5}-(\bar{x})^2=9 \quad \text{(given)} \\
& \Rightarrow x_1^2+x_2^2+x_3^2+x_4^2+x_5^2=5[9+225]=1170 \quad \ldots (ii)
\end{aligned}\)
Now, after combined the -5 and 13 with the observations, the new variance
\(\begin{aligned}
& =\frac{x_1^2+x_2^2+x_3^2+x_4^2+x_5^2+25+169}{7} -\left(\frac{75-5+13}{7}\right)^2 \\
& =\frac{1170+25+169}{7}-\left(\frac{83}{7}\right)^2=\frac{9548-6889}{49}=\frac{2659}{49}
\end{aligned}\)
Hence, option (d) is correct.
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