MATH SOLVE

3 months ago

Q:
# 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.

Accepted Solution

A:

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