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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.

Respuesta :

Answer:

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<z<0.8)= 0.5763

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