what is the Z value

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 178 households is selected from the population. What is the probability that the sample proportion of households spending more than $125 a week is between 0.32 and 0.49?

Note: You should carefully round any z-values you calculate to 4 decimal places to match wamap's approach and calculations.

Answer 1

Solution

The total number is 0.32 and percentage more than is 32%.

If 175 of 0.32 is 56.

Three for for it to be more it must be between 0.32 and 0.49 respectfully.

Another proof: 1 is equivalent to 0.32.

The proportion of 175 inline with 0.32 is 547.

The ratio of 175/547 is 0.32 so if it must be more than it should be between 0.32 and 0.49 respectfully.

Please upvote my answer . thanks