The monthly expenditures on food by single adults living in one neighborhood of Los Angeles are normally distributed with a mean of $410 and a standard deviation of $75. Determine the percentage of samples of size 9 that have mean expenditures within $20 of the population mean expenditure of $410.

0.5763

Step-by-step explanation:

We need to estimate the standard error of the mean.

Standard error of the mean = standard deviation of the original distribution/√sample size

Standard error of the mean = 75/√9 = 25

Now we can use this Standard error to estimate z as follows:

Z = (x - mean) / standard deviation

We want the mean expenditures within $20, so x - mean = 20 and - 20

Z = (20) / 25

Z = 0.8

Z = (-20) / 25

Z = - 0.8

Using a Z table we can find probability

P (-0.8