The lengths of pregnancies are normally distributed with a mean of days and a standard deviation of days. a. Find the probability of a pregn

Question

The lengths of pregnancies are normally distributed with a mean of days and a standard deviation of days. a. Find the probability of a pregnancy lasting days or longer. b. If the length of pregnancy is in the lowest ​%, then the baby is premature. Find the length that separates premature babies from those who are not premature.

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Thành Đạt 6 months 2021-07-23T11:11:17+00:00 1 Answers 5 views 0

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    2021-07-23T11:12:31+00:00

    Answer:

    a) The probability of a pregnancy lasting X days or longer is given by 1 subtracted by the p-value of Z = \frac{X - \mu}{\sigma}, in which \mu is the mean and \sigma is the standard deviation.

    b) We have to find X when Z has a p-value of \frac{a}{100}, and X is given by: X = \mu - Z\sigma, in which \mu is the mean and \sigma is the standard deviation.

    Step-by-step explanation:

    Normal Probability Distribution

    Problems of normal distributions can be solved using the z-score formula.

    In a set with mean \mu and standard deviation \sigma, the z-score 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 p-value, we get the probability that the value of the measure is greater than X.

    In this question:

    Mean \mu, standard deviation \sigma

    a. Find the probability of a pregnancy lasting X days or longer.

    The probability of a pregnancy lasting X days or longer is given by 1 subtracted by the p-value of Z = \frac{X - \mu}{\sigma}, in which \mu is the mean and \sigma is the standard deviation.

    b. If the length of pregnancy is in the lowest a​%, then the baby is premature. Find the length that separates premature babies from those who are not premature.

    We have to find X when Z has a p-value of \frac{a}{100}, and X is given by: X = \mu - Z\sigma, in which \mu is the mean and \sigma is the standard deviation.

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