Project 23: Sampling Distribution of x-bar

Intro Problem: Suppose that the mean household income of a certain region is 45,000 with standard deviation 30,000.   If we were to take a random sample of 100 households from this region, what is the probability that the mean household income of the 100 households is greater than 47,000?

Since n is large (n = 100 which is at least 30), the sampling distribution of sample means is approximately normal.  The distribution has a mean of 45,000 and standard deviation 3,000 (30,000/sqrt(100)).

Project 23: Variables 'mu', 'sigma', 'n', and 'xbar'.  There are also working methods called normalcdf and invNorm. 

mu represents the population mean (45,000 in the first example)

sigma represents the population standard deviation (30,000 in the first example)

n is the sample size (100 in the first example)

xbar is the sample mean (47,000 in the first example)

Review projects 2-5 if you need a reminder of how normalcdf and invNorm work.

Task: Appropriately assign the value of 'z' and 'prob' where z represents the number of standard deviations that xbar is from the mean and prob is the probability that a random sample would result in a mean larger than the one obtained.

Example: If mu = 45,000, sigma = 30,000, n = 100, xbar = 47,000 then z should be .67 and prob should be .2414.

**If your code works for 5 test cases, you can enter your e-mail address

Universal Computational Math Methods:

pow(5,2) returns 25.0

abs(-3.0) returns 3

sqrt(49.0) returns 7.0