Question

8.4.20 Given the discrete uniform population shown to the right find the probability that a random sample of size 96, selected with replacement, will yield a sample mean greater than 9.2 but less than 9.9. Assume the means are measured to the nearest tenth. x 1, 9, 17 f(x)= 3 0, elsewhere Click here to view page 1 of the standard normal distribution table Click here to view page 2 of the standard normal distribution table The probability is (Round to four decimal places as needed) lick Check Answer

Answer 1

Let X be the discrete uniform random variable with the probability distribution junction fo First, we determine the moon: My = 1. f(1) + 9. f (9) + 17. f(17). = 1x 1 + 9x5 + 17xų = q and the variance of Xu of = f(1). (1 - Mx + f(9). (1-4x) + f (17). (17 = 4x)? - Ir (1-9)?+ (9-9)+ (17-9127 - 128 Now, we have given the sample size n=96, using sample site and value obtained above, we can compute the mean and standard deviation of somple mean: 3 us - l op = = 9 = 1128/2 = 0.6667

and we know, Z = X uy - X-ly ~ N(0, 1) Thus, we get P(9.2 < X < 9. 9) = P(9.2-9 <X-9 29.99 10.6667 0.6667 0.6667 = P(0.29999<2<1.3499) =P(Z <1,3499)- P(Z <0. 29.999) 2 0.9125–0.6179 - 0 = 0.2936 In , we obtained values using standard normal distrubution table? P(9.2 < X < 9.9) = 0.2436