Unit 3 Chapter 7: Lesson 7.7 Assignment
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Question 11
A Food Marketing Institute found that $29 \%$ of households spend more than $\$ 125$ a week on groceries. Assume the population proportion is 0.29 and a simple random sample of 230 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 0.26 ?
There is a probability that the sample proportion of households spending more than $\$ 125$ a week is less than 0.26 . Round the answer to 4 decimal ploces.
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Step 1 ：We are given that the population proportion (p) is 0.29 and the sample size (n) is 230. We want to find the probability that the sample proportion is less than 0.26.

Step 2 ：We first calculate the mean of the distribution, which is equal to the population proportion (p). So, the mean is \(0.29\).

Step 3 ：Next, we calculate the standard deviation of the distribution. The formula for the standard deviation is \(\sqrt{p*(1-p)/n}\). Substituting the given values, we get the standard deviation as approximately \(0.02992\).

Step 4 ：We then find the z-score for the sample proportion 0.26. The z-score is calculated as \((x - mean) / std_dev\), where x is the sample proportion. Substituting the given values, we get the z-score as approximately \(-1.00267\).

Step 5 ：Finally, we use the standard normal distribution to find the probability that the z-score is less than \(-1.00267\). This gives us the probability as approximately \(0.1580\).

Step 6 ：So, the probability that the sample proportion of households spending more than $125 a week is less than 0.26 is approximately \(\boxed{0.1580}\).