The mean and variance of 6 observations 3, 9, 11, 17, x, y (where x < y ) are 10 and 25 respectively. If a number is chosen at random from the set of all positive divisors of xy , then the probability that the chosen number is a multiple of x is :

Step-by-step solution

Given the mean of the observations is 10 : 3 + 9 + 11 + 17 + x + y 6 = 10 40 + x + y = 60 x + y = 20 Given the variance is 25 : 3^2 + 9^2 + 11^2 + 17^2 + x^2 + y^2 6 - (10)^2 = 25 9 + 81 + 121 + 289 + x^2 + y^2 6 = 125 500 + x^2 + y^2 = 750 x^2 + y^2 = 250 Using (x+y)^2 = x^2 + y^2 + 2xy : (20)^2 = 250 + 2xy 400 = 250 + 2xy xy = 75 Solving x+y=20 and xy=75 with x The set of all positive divisors of xy = 75 is S = 1, 3, 5, 15, 25, 75 . Total number of outcomes = 6 . We need the probability that the chosen number is