The united states department of agriculture (usda) found that the proportion of young adults ages 20–39 who regularly skip eatin

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The united states department of agriculture (usda) found that the proportion of young adults ages 20–39 who regularly skip eatin

g breakfast is 0.238. suppose that lance, a nutritionist, surveys the dietary habits of a random sample of size n=500 of young adults ages 20–39 in the united states.

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1 answer:

Katen [3.5K] · Answer 1 · 2026-03-06T01:00:54+03:00

Answer:

Probability = 0.3557

Explanation:

Data provided

Young adults age range = 20 to 39

Proportion skipping breakfast p = 0.238

Sample size n = 500

Objective

We aim to determine the probability that more than 122 of Lance's sampled individuals do not eat breakfast regularly.

Solution

Utilizing the Normal Approximation for the Binomial Distribution,

let's define our random variable as x:

x ~ Bin (n,p).............1

Using Normal Approximation gives us:

x ~ Normal Approx (np, npq).................2

This simplifies to:

x ~ (500, 0.238)

where we know q will be:

q = 1 - p

q = 1 - 0.238

q = 0.762.............3

Hence, we find:

x ~ Normal Approx (119, 90.678)

Now we want to compute P(X > 122).

First, we convert this to Z:

z = $\frac{x-\mu}{\sigma}$ ................4

Where the mean is $\mu$ = np

and standard deviation is $\sigma = \sqrt{npq}$

For P(X > 122, we evaluate:

$P(\frac{X-\mu}{\sigma}>\frac{122-119}{\sqrt{90.678}})$ ............5

This results in P(Z > 0.37).

Thus,

Probability = 1 - P(Z < 0.37)

Next, we'll refer to the z table:

Probability = 1-0.6443

Probability = 0.3557