A report indicated that 37% of adults had received a bogus email intended to steal personal information. Suppose a random sample of 700 adul

Question

A report indicated that 37% of adults had received a bogus email intended to steal personal information. Suppose a random sample of 700 adults is obtained. In a random sample of 700 adults, what is the probability that no more than 34% had received such an email?

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Savannah 1 week 2021-11-17T14:18:26+00:00 1 Answer 0 views 0

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    2021-11-17T14:19:28+00:00

    Answer:

    5.05% probability that no more than 34% had received such an email.

    Step-by-step explanation:

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

    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:

    n = 700, p = 0.37

    \mu = E(X) = np = 700*0.37 = 259

    \sigma = \sqrt{V(X)} = \sqrt{np(1-p)} = \sqrt{700*0.37*0.63} = 12.77

    In a random sample of 700 adults, what is the probability that no more than 34% had received such an email?

    34% is 0.34*700 = 238

    So this probability is the pvalue of Z when X = 238.

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

    Z = \frac{238 - 259}{12.77}

    Z = -1.64

    Z = -1.64 has a pvalue of 0.0505

    5.05% probability that no more than 34% had received such an email.

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