  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?

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, help_outlineImage Transcriptionclosep 0.42 P(1-p) n 0.42(1-0.42) 11 126 = 0.04 fullscreen
Step 2

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

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