Question

Let m be the mean and σ be the standard deviation of the distribution

xi	0	1	2	3	4	5
fi	k+2	2k	k2-1	k2-1	k2+1	k-3

where ∑f_i = 62. if [x] denotes the greatest integer ≤ x, then [μ² + σ²] is equal

• A. 6
• B. 7
• C. 8
• D. 9

Answer 1

The Correct Option is C

Given $\sum f_i = 62$. $(k+2) + 2k + (k^2-1) + (k^2-1) + (k^2+1) + (k-3) = 62$ $3k^2 + 4k - 2 = 62$ $3k^2 + 4k - 64 = 0$ Solving for $k$, we get $k = 4$ (choosing the positive integer solution). 
Frequencies are: 6, 8, 15, 15, 17, 1. 
Mean $\mu = \frac{\sum x_i f_i}{\sum f_i} = \frac{0(6) + 1(8) + 2(15) + 3(15) + 4(17) + 5(1)}{62} = \frac{156}{62} \approx 2.516$. 
Variance $\sigma^2 = \frac{\sum f_i (x_i - \mu)^2}{\sum f_i} \approx 1.733$. $\mu^2 + \sigma^2 \approx (2.516)^2 + 1.733 \approx 6.33 + 1.733 \approx 8.063$. $[\mu^2 + \sigma^2] = [8.063] = 8$. 
Answer: 8.