Consider a bell-shaped symmetric distribution with mean of 16 and standard deviation of 1.5. Approximately what percentage of data lie betwe

Question

Consider a bell-shaped symmetric distribution with mean of 16 and standard deviation of 1.5. Approximately what percentage of data lie between 13 and 19?

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Isabella 2 weeks 2021-09-09T10:48:01+00:00 1 Answer 0

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    2021-09-09T10:49:19+00:00

    Answer: 95.45 %

    Step-by-step explanation:

    Given : The distribution is bell shaped , then the distribution must be normal distribution.

    Mean : \mu=\ 16

    Standard deviation :\sigma= 1.5

    The formula to calculate the z-score :-

    z=\dfrac{x-\mu}{\sigma}

    For x = 13

    z=\dfrac{13-16}{1.5}=-2

    For x = 19

    z=\dfrac{19-16}{1.5}=2

    The p-value = P(-2<z<2)=P(z<2)-P(z<-2)

    0.9772498-0.0227501=0.9544997\approx0.9545

    In percent, 0.9545\times100=95.45\%

    Hence, the percentage of data lie between 13 and 19 = 95.45 %

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