Suppose that the duration of a routine doctor’s visit is known to be normally distributed with a mean of 21 minutes and a standard deviation

Question

Suppose that the duration of a routine doctor’s visit is known to be normally distributed with a mean of 21 minutes and a standard deviation of seven minutes. If one of the visits is randomly chosen, what is the probability that it lasted at least 24 minute?

in progress 0
Josephine 1 week 2021-09-13T23:40:48+00:00 1 Answer 0

Answers ( )

    0
    2021-09-13T23:41:56+00:00

    Answer:

    33.36% probability that it lasted at least 24 minutes

    Step-by-step explanation:

    Problems of normally distributed samples are 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.

    In this problem, we have that:

    \mu = 21, \sigma = 7

    If one of the visits is randomly chosen, what is the probability that it lasted at least 24 minute?

    This is 1 subtracted by the pvalue of Z when X = 24. So

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

    Z = \frac{24 - 21}{7}

    Z = 0.43

    Z = 0.43 has a pvalue of 0.6664

    1 – 0.6664 = 0.3336

    33.36% probability that it lasted at least 24 minutes

Leave an answer

45:7+7-4:2-5:5*4+35:2 =? ( )