Stress at work: In a poll conducted by the General Social Survey, 81% of respondents said that their jobs were sometimes or always stressful

Question

Stress at work: In a poll conducted by the General Social Survey, 81% of respondents said that their jobs were sometimes or always stressful. Two hundred workers are chosen at random. a. Approximate the probability that 160 or fewer find their jobs stressful. b. Approximate the probability that more than 150 find their jobs stressful. c. Approximate the probability that the number who find their jobs stressful is between 155 and 162, inclusive.

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Mary 1 month 2021-09-15T17:55:56+00:00 1 Answer 0

Answers ( )

  1. Emma
    0
    2021-09-15T17:57:06+00:00

    Answer:

    a) 35.94% probability that 160 or fewer find their jobs stressful.

    b) 98.46% probability that more than 150 find their jobs stressful.

    c) 39.62% probability that the number who find their jobs stressful is between 155 and 162, inclusive.

    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:

    n = 200, p = 0.81. So

    \mu = E(X) = np = 200*0.81 = 162

    \sigma = \sqrt{V(X)} = \sqrt{np(1-p)} = \sqrt{200*0.81*0.19} = 5.55

    a. Approximate the probability that 160 or fewer find their jobs stressful.

    This is the pvalue of Z when X = 160. So

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

    Z = \frac{160 - 162}{5.55}

    Z = -0.36

    Z = -0.36 has a pvalue of 0.3594.

    35.94% probability that 160 or fewer find their jobs stressful.

    b. Approximate the probability that more than 150 find their jobs stressful.

    This is 1 subtracted by the pvalue of Z when X = 150. So

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

    Z = \frac{150 - 162}{5.55}

    Z = -2.16

    Z = -2.16 has a pvalue of 0.0154.

    1 – 0.0154 = 0.9846

    98.46% probability that more than 150 find their jobs stressful.

    c. Approximate the probability that the number who find their jobs stressful is between 155 and 162, inclusive.

    This is the pvalue of Z when X = 162 subtracted by the pvalue of Z when X = 155. So

    X = 162

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

    Z = \frac{162 - 162}{5.55}

    Z = 0

    Z = 0 has a pvalue of 0.5

    X = 155

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

    Z = \frac{155 - 162}{5.55}

    Z = -1.26

    Z = -1.26 has a pvalue of 0.1038

    0.5 – 0.1038 = 0.3962

    39.62% probability that the number who find their jobs stressful is between 155 and 162, inclusive.

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