The United States Department of Agriculture (USDA) found that the proportion of young adults ages 20–39 who regularly skip eating breakfast

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

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Alaia 2 weeks 2021-09-08T22:11:49+00:00 1 Answer 0

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    2021-09-08T22:12:52+00:00

    Answer:

    0.3557

    Step-by-step explanation:

    Complete 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 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:

    z=\frac{x-\mu}{\sigma}

    Where \mu is the mean, or “np” and \sigma is the standard deviation, which is \sqrt{npq}

    THus, we have:

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

    Which can be said as:

    1 – P(Z<0.37)

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

    1-0.6443=0.3557

    THis is our answer.

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