Question

A Food Marketing Institute found that 39% of households spend more than $125 a week on groceries. Assume the population proportion is 0.39 and a simple random sample of 87 households is selected from the population. What is the probability that the sample proportion of households spending more than $125 a week is between 0.29 and 0.41?

Answer 1

Answer:

0.6210

Step-by-step explanation:

Given that a Food Marketing Institute found that 39% of households spend more than $125 a week on groceries

Sample size n =87

Sample proportion will follow a normal distribution with p =0.39

and standard error = $\sqrt{\frac{0.39(1-0.39)}{87} } \\=0.0523$

the probability that the sample proportion of households spending more than $125 a week is between 0.29 and 0.41

=$P(0.29$

There is 0.6210 probability that the sample proportion of households spending more than $125 a week is between 0.29 and 0.41