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

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 and standard deviation , the zscore of a measure X is given by:

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

In this problem, we have that:

So

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

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

4444444444444 i think