Question 21. The mean and standard deviation of 15 observations were found to be 12 and 3 respectively. On rechecking it was found that an observation was read as 10 in place of 12 . If \mu and \sigma^{2} denote the mean and variance of the correct observations respectively, then 15\left(\mu+\mu^{2}+\sigma^{2}\right) is equal to
Answer (2521)
Explanation:
Let the incorrect mean be \mu^{\prime} and standard deviation be \sigma^{\prime}
We have
\mu^{\prime}=\frac{\Sigma x_{i}}{15}=12 \Rightarrow \Sigma x_{i}=180
As per given information correct \Sigma x_{i}=180-10+12
\Rightarrow \mu( correct mean )=\frac{182}{15}
Also
\sigma^{\prime}=\sqrt{\frac{\Sigma x_{i}^{2}}{15}-144}=3 \Rightarrow \Sigma x_{i}^{2}=2295Correct \Sigma x_{i}^{2}=2295-100+144=2339
\sigma^{2}( correct variance )=\frac{2339}{15}-\frac{182 \times 182}{15 \times 15}
Required value
=15\left(\mu+\mu^{2}+\sigma^{2}\right) =15\left(\frac{182}{15}+\frac{182 \times 182}{15 \times 15}+\frac{2339}{15}-\frac{182 \times 182}{15 \times 15}\right) =15\left(\frac{182}{15}+\frac{2339}{15}\right) =2521