Assume that the random variable X is normally​ distributed, with mean mu equals 80μ=80 and standard deviation sigma equals 20.σ=20. Compute

Question

Assume that the random variable X is normally​ distributed, with mean mu equals 80μ=80 and standard deviation sigma equals 20.σ=20. Compute the probability ​P(Xgreater than>9696​).

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Eloise 7 months 2021-10-08T00:29:51+00:00 1 Answer 0 views 0

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    2021-10-08T00:31:35+00:00

    Answer:

    [tex]P(X \geq 86) = 1 – 0.7881 = 0.2119[/tex]

    Step-by-step explanation:

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

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

    [tex]Z = \frac{X – \mu}{\sigma}[/tex]

    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.

    In this problem, we have that:

    [tex]\mu = 80, \sigma = 20[/tex]

    Compute the probability ​P(Xgreater than>96​).

    This is 1 subtracted by the pvalue of Z when X = 96. So

    [tex]Z = \frac{X – \mu}{\sigma}[/tex]

    [tex]Z = \frac{96 – 80}{20}[/tex]

    [tex]Z = 0.8[/tex]

    [tex]Z = 0.8[/tex] has a pvalue of 0.7881

    [tex]P(X \geq 86) = 1 – 0.7881 = 0.2119[/tex]

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