1 point) As reported in “Runner’s World” magazine, the times of the finishers in the New York City 10 km run are normally distributed with a

Question

1 point) As reported in “Runner’s World” magazine, the times of the finishers in the New York City 10 km run are normally distributed with a mean of 61 minutes and a standard deviation of 9 minutes. Let X denote finishing time for the finishers. a) The distribution of the variable X has mean 61 and standard deviation 9 . b) The distribution of the standardized variable Z has mean and standard deviation

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Emery 1 week 2021-09-15T08:19:19+00:00 1 Answer 0

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    2021-09-15T08:20:19+00:00

    Answer:

    (a) E (X) = 61 and SD (X) = 9

    (b) E (Z) = 0 and SD (Z) = 1

    Step-by-step explanation:

    The time of the finishers in the New York City 10 km run are normally distributed with a mean,μ = 61 minutes and a standard deviation, σ = 9 minutes.

    (a)

    The random variable X is defined as the finishing time for the finishers.

    Then the expected value of X is:

    E (X) = 61 minutes

    The variance of the random variable X is:

    V (X) = (9 minutes)²

    Then the standard deviation of the random variable X is:

    SD (X) = 9 minutes

    (b)

    The random variable Z is the standardized form of the random variable X.

    It is defined as:Z=\frac{X-\mu}{\sigma}

    Compute the expected value of Z as follows:

    E(Z)=E[\frac{X-\mu}{\sigma}]\\=\frac{E(X)-\mu}{\sigma}\\=\frac{61-61}{9}\\=0

    The mean of Z is 0.

    Compute the variance of Z as follows:

    V(Z)=V[\frac{X-\mu}{\sigma}]\\=\frac{V(X)+V(\mu)}{\sigma^{2}}\\=\frac{V(X)}{\sigma^{2}}\\=\frac{9^{2}}{9^{2}}\\=1

    The variance of Z is 1.

    So the standard deviation is 1.

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