SOLUTION: A Food Marketing Institute found that 25% of households spend more than $125 a week on groceries. Assume the population proportion is 0.25 and a simple random sample of 378 househo

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Question 1203993: A Food Marketing Institute found that 25% of households spend more than $125 a week on groceries. Assume the population proportion is 0.25 and a simple random sample of 378 households is selected from the population. What is the probability that the sample proportion of households spending more than $125 a week is less than 0.24?
There is a blank probability that the sample proportion of households spending more than $125 a week is less than 0.24. Round the answer to 4 decimal places.
Answer by Theo(13342) About Me  (Show Source):
You can put this solution on YOUR website!
population p = .25
population q = 1 - p = .75
sample size is 378.
standard error = sqrt(.25 * .75 / 378) = .022272.
z-score = (.24 - .25) / .022272 = -.448994.
z is the z-score
x is the sample proportion
m is the population proportion
s is the standad error

the probability of getting a z-score less than -.448994 is equal to .326718.
that's the probaability of getting a sample mean less than .24.

here's what it looks like on a z-score calculator.