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 =

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,

Step 2

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

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