An article written for a magazine claims that 78% of the magazine’s subscribers report eating healthily the previous day. Suppose we select

Question

An article written for a magazine claims that 78% of the magazine’s subscribers report eating healthily the previous day. Suppose we select a simple random sample of 675 of the magazine’s approximately 50,000 subscribers to check the accuracy of this claim. Assuming the article’s 78% claim is correct, what is the approximate probability that less than 80% of the sample would report eating healthily the previous day?

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Mary 2 weeks 2021-09-14T06:54:16+00:00 1 Answer 0

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    2021-09-14T06:55:30+00:00

    Answer:

    89.44% probability that less than 80% of the sample would report eating healthily the previous day

    Step-by-step explanation:

    We use the binomial approximation to the normal to solve this question.

    Binomial probability distribution

    Probability of exactly x sucesses on n repeated trials, with p probability.

    Can be approximated to a normal distribution, using the expected value and the standard deviation.

    The expected value of the binomial distribution is:

    E(X) = np

    The standard deviation of the binomial distribution is:

    \sqrt{V(X)} = \sqrt{np(1-p)}

    Normal probability distribution

    Problems of normally distributed samples can be solved using the z-score formula.

    In a set with mean \mu and standard deviation \sigma, the zscore of a measure X is given by:

    Z = \frac{X - \mu}{\sigma}

    The Z-score 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. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

    When we are approximating a binomial distribution to a normal one, we have that \mu = E(X), \sigma = \sqrt{V(X)}.

    In this problem, we have that:

    p = 0.78, n = 675

    So

    \mu = E(X) = np = 675*0.78 = 526.5

    \sigma = \sqrt{V(X)} = \sqrt{np(1-p)} = \sqrt{675*0.78*0.22} = 10.76

    What is the approximate probability that less than 80% of the sample would report eating healthily the previous day?

    This is the pvalue of Z when X = 0.8*675 = 540. So

    Z = \frac{X - \mu}{\sigma}

    Z = \frac{540 - 526.5}{10.76}

    Z = 1.25

    Z = 1.25 has a pvalue of 0.8944

    89.44% probability that less than 80% of the sample would report eating healthily the previous day

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