Let μ be the mean and σ be the standard deviation of the distribution X i 0 1 2 3 4 5 f i k + 2 2 k k 2 - 1 k 2 - 1 k 2 + 1 k - 3 where Σ f i = 62 . If x denotes the greatest integer ≤ x , then μ 2 + σ 2 is equal to

Step-by-step solution

Given, X i 0 1 2 3 4 5 f i k + 2 2 k k 2 - 1 k 2 - 1 k 2 + 1 k - 3 Now we know that, Mean = ∑ X i f i ∑ f i ⇒ μ = 2 k +   2 k 2   – 2   +   3 k 2 – 3   +   4 k 2 + 4   + 5 k   –   15 62 ⇒ μ = 9 k 2 + 7 k - 16 62   . . . . . . . 1 And ∑ f i = 62 ⇒ 3 k 2 + 4 k - 2 = 62 ⇒ 3 k 2 + 4 k - 64 = 0 ⇒ ( 3 k + 16 ) ( k - 4 ) = 0 ⇒ k = 4 Now putting the value of k in above equation 1 we get, &#956