Question

A Food Marketing Institute found that 32% of households spend more than $125 a week on groceries. Assume the population proportion is 0.32 and a simple random sample of 113 households is selected from the population. What is the probability that the sample proportion of households spending more than $125 a week is between 0.3 and 0.48?

Answer 1

Given that the Food Marketing Institute found that 32% of households spend more than $125 a week on groceries. Assume the population proportion is p = 0.32 and a simple random sample of n = 113 households is selected from the population.

Now to calculate the probability that the  sample proportion of households spending more than $125 a week is between 0.3 and 0.48 we need to use Z score, which is calculated as;

Where the denominator is calculated as;

Now the probability is calculated as;

Now the probability is calculated using the excel formula for normal distribution which is =NORM.S.DIST(3.65, TRUE)- NORM.S.DIST(-0.46, TRUE), thus the probability is computed as:

= 0.6711