Question 24. If the mean and variance of the data 65,68,58,44, 48, 45, 60, \alpha,\beta,60 where \alpha>\beta, are 56 and 66.2 respectively, then the value of \alpha^{2}+\beta^{2} is
Answer: 6344
Explanation:
Mean of the data is \bar{x}=56.
Variance of the data is \sigma^{2}=66.2.
Number of observations is 10.
Using the formula of variance,
\sigma^{2}=\frac{\sum x^{2}}{10}-\bar{x}^{2}.
So, \frac{\alpha^{2}+\beta^{2}+25678}{10}-(56)^{2}=66.2.
Solving, \alpha^{2}+\beta^{2}=6344.
Hence, the required value is 6344.