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 n = 500 of young adults ages 20–39 in the United States.

Apply the central limit theorem for the binomial distribution to find the probability that the number of individuals in Lance's sample who regularly skip breakfast is greater than 122.

Solution:

This is a case where we will use Normal Approximation to Binomial Distribution.

Let X be random variable. So we know:

X ~ Bin (n,p)

Then, Normal Approximation would be:

X ~ Normal Approx (np, npq)

Given in the problem, sample n = 500 and probability of skipping, p = 0.238, so:

X ~ (500, 0.238)

Note: q = 1 - p = 1 - 0.238 = 0.762

Thus,

X ~ Normal Approx (119, 90.678)

Now,

We need P(X > 122).

We convert to Z by using formula:

[tex]z=\frac{x-\mu}{\sigma}[/tex]

Where [tex]\mu[/tex] is the mean, or "np" and [tex]\sigma[/tex] is the standard deviation, which is [tex]\sqrt{npq}[/tex]

THus, we have:

[tex]P(X>122)\\=P(\frac{X-\mu}{\sigma}>\frac{122-119}{\sqrt{90.678}})\\=P(Z>0.37)[/tex]

Which can be said as:

1 - P(Z<0.37)

Using Z-Table (normal table), we have:

[tex]1-0.6443=0.3557[/tex]

THis is our answer.

joaobezerra joaobezerra · Answer 2 · 2022-01-13T14:37:20+01:00

Using the normal approximation to the binomial, it is found that there is a 0.3557 = 35.57% probability that the number of individuals in Lance's sample who regularly skip breakfast is greater than 122.

Normal Probability Distribution

In a normal distribution with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the z-score of a measure X is given by:

[tex]Z = \frac{X - \mu}{\sigma}[/tex]

It measures how many standard deviations the measure is from the mean.

After finding the z-score, we look at the z-score table and find the p-value associated with this z-score, which is the percentile of X.

The binomial distribution is the probability of x successes on n trials, with p probability of a success on each trial. It can be approximated to the normal distribution with [tex]\mu = np, \sigma = \sqrt{np(1-p)}[/tex].

In this problem:

The proportion of young adults ages 20–39 who regularly skip eating breakfast is 0.238, hence [tex]p = 0.238[/tex]

A sample of 500 is taken, hence [tex]n = 500[/tex]

The mean and the standard deviation are given by:

[tex]\mu = np = 500(0.238) = 119[/tex]

[tex]\sigma = \sqrt{np(1 - p)} = \sqrt{500(0.238)(0.762)} = 9.5225[/tex]

The probability that the number of individuals in Lance's sample who regularly skip breakfast is greater than 122, using continuity correction, is [tex]P(X > 122 + 0.5) = P(X > 122.5)[/tex], which is 1 subtracted by the p-value of Z when X = 122.5.

[tex]Z = \frac{X - \mu}{\sigma}[/tex]

[tex]Z = \frac{122.5 - 119}{9.5225}[/tex]

[tex]Z = 0.37[/tex]

[tex]Z = 0.37[/tex] has a p-value of 0.6443.

1 - 0.6443 = 0.3557.

0.3557 = 35.57% probability that the number of individuals in Lance's sample who regularly skip breakfast is greater than 122.

To learn more about the normal approximation to the binomial, you can take a look at https://brainly.com/question/14424710