Suppose that in a population of adults between the ages of 18 and 49, glucose follows a normal distribution with a mean of 93.5 and standard

Question

Suppose that in a population of adults between the ages of 18 and 49, glucose follows a normal distribution with a mean of 93.5 and standard deviation of 19.8. What is the probability that glucose exceeds 120 in this population

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Autumn 1 month 2021-10-17T16:15:42+00:00 2 Answers 0 views 0

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    0
    2021-10-17T16:17:07+00:00

    Answer:

    0.0904 or 9.04%

    Step-by-step explanation:

    Mean glucose (μ) = 93.5

    Standard deviation (σ) = 19.8

    In a normal distribution, the z-score for any glucose value, X, is given by:

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

    For X = 120, the z-score is:

    Z= \frac{120-93.5}{19.8}\\ Z=1.3384

    A z-score of 1.3384 corresponds to the 90.96th percentile of a normal distribution. Therefore, the probability that glucose exceeds 120 in this population is:

    P(X>120) = 1-0.9096=0.0904 = 9.04\%

    0
    2021-10-17T16:17:17+00:00

    Answer:

    Probability that glucose exceeds 120 in this population is 0.09012.

    Step-by-step explanation:

    We are given that in a population of adults between the ages of 18 and 49, glucose follows a normal distribution with a mean of 93.5 and standard deviation of 19.8, i.e.; \mu = 93.5  and  \sigma = 19.8 .

    Let X = amount of glucose i.e. X ~ N(\mu = 93.5 , \sigma^{2} = 19.8^{2})

    Now, the Z score probability is given by;

               Z = \frac{X-\mu}{\sigma} ~ N(0,1)

    So,Probability that glucose exceeds 120 in this population =P(X>120)

    P(X > 120) = P( \frac{X-\mu}{\sigma} > \frac{120-93.5}{19.8} ) = P(Z > 1.34) = 1 – P(Z <= 1.34)

                                                       = 1 – 0.90988 = 0.09012 .

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