The average diameter of sand dollars on a certain island is 3.00 centimeters with a standard deviation of 1.00 centimeters. If 9 sand dollars are chosen at random for a collection, find the probability that the average diameter of those sand dollars is more than 2.60 centimeters. Assume that the variable is normally distributed.

Lets X be the average diameter of 9 sand dollars. We obtain X from taking the average of 9 identically distributed and independant normal random variables. Since X is the average of Nomral variables, then X is Normal. The mean of the average is equal to the mean of one random sand dollar, so the mean of X is 3, while the standard deviation can be obtained by dividing the standard deviation of one sand dollar by the square root of the length of the sample. Thus, the standard deviation of X is 1/√9 = 1/3.

In order to make computations, we standarize X. We take Y = (X-μ)/σ = (X-3)/(1/3) = (X-3)*3. Y has distribution Normal, with mean 0 and standard deviation 1. The cummulative function of y, Φ, is tabulated. You can find the values of Φ in the attached file.

We want to compute P(X > 2.60), using the standarization, we have

P(X > 2.60) = P((X-3)*3 > (2.60-3)*3) = P(Y > -1.2) = 1 - P(Y ≤ -1.2) = 1-Φ(-1.2)

The value of Φ(-1.2) doesnt appear on the table. Since the dinsity funciton of a standard normal distribution is symmetric throught the origin, then we have that Φ(-1.2) = 1-Φ(1.2), hence

P(X > 2.60) = 1-Φ(-1.2) = 1-(1-Φ(1.2)) = Φ(1.2) = 0.8849

As a result, the probability for a random sample of lenght 9 to have average diameter higher than 2.60 is 0.8849.

I hope that works for you!