Question

A Food Marketing Institute found that 30% of households spend more than $125 a week on groceries. Assume the population proportion is 0.3 and a simple random sample of 245 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 0.32

Answer 1

Answer:

75.17% probability that the sample proportion of households spending more than $125 a week is less than 0.32

Step-by-step explanation:

Problems of normally distributed samples are solved using the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$, the zscore of a measure X is given by:

$Z = \frac{X - \mu}{\sigma}$

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

For the sampling distribution of a proportion p with size n, we have that $\mu = p, \sigma = \sqrt{\frac{p(1-p)}{n}}$

In this problem, we have that:

$p = 0.3, n= = 245$

So

$\mu = 0.3, \sigma = \sqrt{\frac{0.3*0.7}{245}} = 0.0293$

What is the probability that the sample proportion of households spending more than $125 a week is less than 0.32

This is the pvalue of Z when X = 0.32. So

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{0.32 - 0.3}{0.0293}$

$Z = 0.68$

$Z = 0.68$ has a pvalue of 0.7517

75.17% probability that the sample proportion of households spending more than $125 a week is less than 0.32