Question

The United States Department of Agriculture (USDA) found that the proportion of young adults ages 20–39 who regularly skip eating breakfast is 0.238 . Suppose that Lance, a nutritionist, surveys the dietary habits of a random sample of size =500 of young adults ages 20–39 in the United States.Use a normal approximation to find the probability that the number of individuals, , in Lance's sample who regularly skip breakfast is greater than 124 .(>124)= (Round to 3 decimal places)

Answer 1

The answer is 0.300

Problem Statement

We are asked to find the probability that the number of individuals in a survey of 500 people would skip breakfast given that the proportion of people who skip breakfast, in general, is 0.238.

Method

- The proportion of people greater than 124 out of 500 is easily gotten to be:

`\begin{gathered} p>(124)/(500) \\ p>0.248 \end{gathered}`

- We now need to know the probability that the proportion of people that skip breakfast would be greater than 0.248.

- To calculate this probability, we need to find the Z-score associated with this value. This is a good way to approximate the probability because the number of people in the survey is well above 30 and we have been told to apply a normal approximation.

- Once we have the Z-score associated with this proportion of 0.248 in relation to the general population proportion statistic of 0.238, we can then convert the Z-score into a probability using a Z-score calculator or a Z-table.

- If the Z-score is "z", then, the probability we are looking for on the Z-score table or calculator is P(x > z).

- Thus, we can solve the question using the following steps:

1. Calculate the Z-score using the formula below:

`\begin{gathered} z=\frac{p-p_0}{\sqrt[]{(p_0(1-p_0))/(n)}} \\ \\ \text{where,} \\ p=\text{sample proportion} \\ p_0=\text{population proportion} \\ n=\text{ Total number of people in the survey} \end{gathered}`

2. Convert the Z-score into probability

Implementation

Step 1: Calculate the Z-score:

`\begin{gathered} p=0.248,p_0=0.238 \\ \\ z=\frac{0.248-0.238}{\sqrt[]{(0.238(1-0.238))/(500)}} \\ \\ z=(0.01)/(0.019045) \\ \\ z=0.5251 \end{gathered}`

2. Convert the Z-score into probability:

Using the Z-score calculator, we have:

Because we are asked to find the probability that the number of people who skipped breakfast is greater than 124, the correct probability here is P(x > Z).

Thus, the probability that the number of individuals that skipped breakfast is greater than 124 is 0.29977 ≅ 0.300 (To 3 decimal places)

Final Answer

The answer is 0.300.