Let a discrete frequency distribution have variate values X_i 1, 2, 3, 4 with corresponding frequencies k, k, k^2, k^2 , where k is a positive integer. If the sum of the variance and the square of the mean of this distribution is 12 , then the value of 10 , where is the mean of the distribution, is equal to

Step-by-step solution

We know that for a frequency distribution, the sum of the variance and the square of the mean is given by ^2 + ^2 = f_i X_i^2 f_i . First, we calculate the sum of frequencies: f_i = k + k + k^2 + k^2 = 2k^2 + 2k Next, we calculate the sum of the squares of the variates multiplied by their frequencies: f_i X_i^2 = 1^2(k) + 2^2(k) + 3^2(k^2) + 4^2(k^2) = k + 4k + 9k^2 + 16k^2 = 25k^2 + 5k Given that ^2 + ^2 = 12 , we have: 25k^2 + 5k 2k^2 + 2k = 12 25k^2 + 5k = 24k^2 + 24k k^2 - 19k = 0 Since k is a positive integer,