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21 June, 12:51

The lengths of pregnancies are normally distributed with a mean of 268 days and a standard deviation of 15 days.

a. Find the probability of a pregnancy lasting 309 days or longer.

b. If the length of pregnancy is in the lowest 4 %, then the baby is premature. Find the length that separates premature babies from those who are not premature.

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  1. 21 June, 14:25
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    probability of a pregnancy lasting 309 days is 0.0032

    the length of pregnancy is in the lowest 4 % is 241.74 days

    Step-by-step explanation:

    Given data

    mean = 268

    standard deviation = 15

    to find out

    probability of a pregnancy lasting 309 days or longer and the length of pregnancy is in the lowest 4 %

    solution

    we know mean (M) is 268 and standard deviation (SD) is 15

    so probability will be in 1st part where pregnancy lasting 309 days

    P (X > 309) = P (X - mean > 309 - mean)

    we know

    Z = (309 - mean) / standard deviation,

    it will be Z = 309-268/15 = 2.73

    we can say these both are equal

    P (X > 309) = P (Z > 2.73)

    now we use the standard normal z-table i. e.

    P (Z > 2.73) = 0.0032

    so P (X > 309) is 0.0032

    probability of a pregnancy lasting 309 days is 0.0032

    and in 2nd part z value with 4%. i. e 0.04

    z = - 1.7507 from the standard table

    so days = z * standard deviation + mean

    days = - 1.7507 * 15 + 268

    days = 241.74

    the length of pregnancy is in the lowest 4 % is 241.74 days
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