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7 January, 13:30

Assume that the heights of women are normally distributed with a mean of 63.6 inches and a standard deviation of 2.5 inches. The cheerleaders for a local professional basketball team must be between 65.5 and 68.0 inches. If a woman is randomly selected, what is the probability that her height is between 65.5 and 68.0 inches

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  1. 7 January, 13:41
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    Answer: The probability that the height of a selected woman's height is between 65.5 and 68.0 is 0.21

    Step-by-step explanation:

    Assume that the heights of women are normally distributed, we would apply the formula for normal distribution which is expressed as

    z = (x - µ) / σ

    Where

    x = heights women.

    µ = mean height

    σ = standard deviation

    From the information given,

    µ = 63.6 inches

    σ = 2.5 inches

    The probability that the height of a selected woman's height is between 65.5 and 68.0 inches is expressed as

    P (65.5 ≤ x ≤ 68)

    For x = 65.5,

    z = (65.5 - 63.6) / 2.5 = 0.76

    Looking at the normal distribution table, the probability corresponding to the z score is 0.7764

    For x = 68,

    z = (68 - 63.6) / 2 = 2.2

    Looking at the normal distribution table, the probability corresponding to the z score is 0.986

    Therefore,

    P (65.5 ≤ x ≤ 68) = 0.986 - 0.7764 = 0.21
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