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A Food Marketing Institute found that 42% of households spend more than $125 a week on groceries. Assume the population proportion is 0.42 and a simple random sample of 126 households is selected from the population. What is the probability that the sample proportion of households spending more than $125 a week is between 0.2 and 0.39?Answer =

Question

A Food Marketing Institute found that 42% of households spend more than $125 a week on groceries. Assume the population proportion is 0.42 and a simple random sample of 126 households is selected from the population. What is the probability that the sample proportion of households spending more than $125 a week is between 0.2 and 0.39?

Answer =

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Step 1

It was found that 42% of the households spend more than $125 on groceries.

That is, the population proportion, p=0.42.

A sample of 126 households is selected from the population. That is, the sample size n = 126.

Let p-hat denote the sample proportion of households spending more than $125 on groceries.

In other words, the mean of the sampling distribution of p-hat is equal to the population proportion p and the standard deviation of the sampling distribution of p-hat is equal to sqrt[(p(1-p)/n].

Thus, the mean and standard deviation of the sample proportion of households spending more than $125 on groceries is,

p 0.42
P(1-p)
n
0.42(1-0.42)
11
126
= 0.04
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p 0.42 P(1-p) n 0.42(1-0.42) 11 126 = 0.04

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Step 2

Computing the probability that the sample proportion of households spending more than $125 a week on groceri...

0.39 0.42
P(0.2 p<0.39) = P/ 0.2-0.42
<Z<
0.04
0.04
P(-5.5<Z<-0.75)
-P(Z<-0.75)-P(Z-5.5)
Using the formulas,
0.2266 0.0000-NORM.S.DIST (-0.75,1)
NORM.S.DIST(-5.5,1)
=0.2266
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0.39 0.42 P(0.2 p<0.39) = P/ 0.2-0.42 <Z< 0.04 0.04 P(-5.5<Z<-0.75) -P(Z<-0.75)-P(Z-5.5) Using the formulas, 0.2266 0.0000-NORM.S.DIST (-0.75,1) NORM.S.DIST(-5.5,1) =0.2266

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