Suppose the ages of multiple birth mothers (4 or more births) are normally distributed with a mean age of 35.5 years and a sta

Question

Suppose the ages of multiple birth mothers (4 or more births) are normally distributed with a mean age of 35.5 years and a standard deviation of 7. 5 years What percent of these mothers are between the ages of 32 to 40?………………………….. What percent of these mothers are less than 30 years old?………………………………….. What percent of these mothers are more than 38 years old?

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Brielle 18 hours 2021-10-13T03:09:02+00:00 1 Answer 0

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    2021-10-13T03:10:15+00:00

    Answer:

    40.65% of these mothers are between the ages of 32 to 40

    23.27% of these mothers are less than 30 years old

    37.07% of these mothers are more than 38 years old

    Step-by-step explanation:

    Problems of normally distributed samples can be 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 = 35.5, \sigma = 7.5

    What percent of these mothers are between the ages of 32 to 40?

    This is the pvalue of Z when X = 40 subtracted by the pvalue of Z when X = 32.

    X = 40

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

    Z = \frac{40 - 35.5}{7.5}

    Z = 0.6

    Z = 0.6 has a pvalue of 0.7257

    X = 32

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

    Z = \frac{32 - 35.5}{7.5}

    Z = -0.47

    Z = -0.47 has a pvalue of 0.3192

    0.7257 – 0.3192 = 0.4065

    40.65% of these mothers are between the ages of 32 to 40

    What percent of these mothers are less than 30 years old?

    This is the pvalue of Z when X = 30.

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

    Z = \frac{30 - 35.5}{7.5}

    Z = -0.73

    Z = -0.73 has a pvalue of 0.2327

    23.27% of these mothers are less than 30 years old

    What percent of these mothers are more than 38 years old?

    This is 1 subtracted by the pvalue of Z when X = 38.

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

    Z = \frac{38 - 35.5}{7.5}

    Z = 0.33

    Z = 0.33 has a pvalue of 0.6293

    1 – 0.6293 = 0.3707

    37.07% of these mothers are more than 38 years old

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