Consider a frequency distribution for the data values 0, 5, and 10 , with corresponding frequencies a, 4, and 6 - a , respectively. If the variance of this distribution is a prime number, then the sum of all possible values of a is:

Step-by-step solution

Let the values be x_i and frequencies be f_i . Total frequency, N = a + 4 + (6 - a) = 10 . Mean, = f_i x_i N = a(0) + 4(5) + (6 - a)(10) 10 = 20 + 60 - 10a 10 = 8 - a Variance, ^2 = f_i x_i^2 N - ^2 f_i x_i^2 = a(0^2) + 4(5^2) + (6 - a)(10^2) = 100 + 600 - 100a = 700 - 100a ^2 = 700 - 100a 10 - (8 - a)^2 ^2 = 70 - 10a - (64 - 16a + a^2) = 6 + 6a - a^2 Since frequencies must be non-negative integers, a 0 and 6 - a 0 , which gives 0 a 6 . We check the value of ^2 for each integer a in this range: For a = 0 , ^2 = 6 F