Question

1) The following histogram shows the distribution of the amount of money spent on a haircut by a Math 119 class: o.1 If the mean was $33 with a standard deviation $38, can the CLT for sample means be used if we obtained 20 random Math 119 students? How about if we obtained 50? a) b) Suppose we obtained 100 random Math 119 students. Use the CLT for sample means to find the probability that the sample mean of the 100 students will be greater than $25. Use the same mean and standard deviation from part a. Round all results to the nearest hundredth.

Answer 1

The math 119 class mean (mu) = 33 Standard derivation (sigma) = 38 a) CLT (central limit theorem) for sample mean (x^Bar) it sample size (n) = 20 Sampling distribution of sample mean mu_x^Bar = mu = 33 sampling distribution so sigma_x^Bar = sigma/squareroot n = 38/squareroot 20 = If we obtained sample mean form n = 50 then sampling distribution mean mu_x^Bar = mu = 33 Standard deviation sigma_x^Bar = sigma/squareroot n = 38/squareroot 20 = b) n = 100 mu_x^Bar = mu = 33 sigma_x^Bar = sigma/squareroot n = 38/squareroot 100 = 3.8 p(x^Bar > 25) = p(z > - 2.11) = 0.9826. z = 25 - 33/3.8 = -2.11