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6 August, 02:11

Suppose you take a sample of 200 Lehman students. The average height is 65 inches (5'4"") with a standard deviation of 5 inches. Using the Standard Normal Distribution: a) What would be the height in inches when the probability is 4.95% (0.0495) and when the probability is 95.05% (0.9505) ? b) What would be the height if the probability is 99.01% (0.9901) ?

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  1. 6 August, 02:16
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    Step-by-step explanation:

    Using the Standard Normal Distribution, the formula for normal distribution is expressed as

    z = (x - u) / s

    Where

    x = height of students

    u = mean score

    s = standard deviation

    From the information given,

    n = 200

    u = 65 inches

    Standard deviation = 5 inches

    a) when the probability is 4.95%. p = 0.0495

    From the normal distribution table,

    z = - 1.65

    Therefore

    - 1.65 = (x - 65) / 5

    x - 65 = 5 * - 1.65 = - 8.25

    x = - 8.25 + 65 = 56.75 inches

    The height is 56.75 inches

    when the probability is 95.05%. p = 0.9505

    From the normal distribution table,

    z = 1.65

    Therefore

    1.65 = (x - 65) / 5

    x - 65 = 5 * 1.65 = 8.25

    x = 8.25 + 65 = 73.25 inches

    The height is 73.25 inches

    b) when the probability is 99.01%.

    p = 0.9901

    From the normal distribution table,

    z = 2.33

    Therefore

    2.33 = (x - 65) / 5

    x - 65 = 5 * 2.33 = 11.65

    x = 11.65 + 65 = 76.65 inches

    The height is 76.65 inches
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