A Food Marketing Institute found that 35% of households spend more than $125 a week on groceries. Assume the population proportion is 0.35 and a simple random sample of 75 households is selected from the population. What is the probability that the sample proportion of households spending more than $125 a week is between 0.36 and 0.42

The probability that the sample proportion of households spending more than $125 a week is between 0.36 and 0.42 is 0.3115

Step-by-step explanation:

From the question, we can deduce the following;

n = 75

p = 0.35

where q = 1-p = 1-0.35 = 0.65

To compute the probability that the sample proportion of households spending more than $125 a week is between 0.36 and 0.42, we start by calculating the standard deviation of sample proportion.

Mathematically;

Standard deviation of sample proportion = √pq/n

SD = √(0.35)(0.65)/75 = 0.055 which is approximately 0.06

Let’s now compute the z-scores

For 0.36, we have ; (0.36-0.35)/0.06 = 0.01/0.06 = 0.17

For 0.42, we have; (0.42-0.35)/0.06 = 0.07/0.06 = 1.17

So the probability we want to calculate is :

P(0.17<z<1.17) = P(z<1.17) - P(z < 0.17) = 0.8790 - 0.5675 = 0.3115