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.

Answer:0.5763Step-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<z<0.8)= 0.5763