The lengths of pregnancies are normally distributed with a mean of 266 days and

Question

The lengths of pregnancies are normally distributed with a mean of 266 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 2%, then the baby is premature. Find the length that separates babies from those who are not premature (round to the nearest integer as needed). The lengths of pregnancies are normally distributed with a mean of 266 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 2%, then the baby is premature. Find the length that separates babies from those who are not premature (round to the nearest integer as needed). A) Find the probability of a pregnancy lasting 309 days or longer. B) If the length of pregnancy is in the lowest 2%, then the baby is premature. Find the length that separates babies from those who are not premature (round to the nearest integer as needed).

Explanation / Answer

the PDF of normal distribution is = 1/ * 2 * e ^ -(x-u)^2/ 2^2
standard normal distribution is a normal distribution with a,
mean of 0,
standard deviation of 1
equation of the normal curve is ( Z )= x - u / sd ~ N(0,1)
mean ( u ) = 266
standard Deviation ( sd )= 15
a.
P(X < 309) = (309-266)/15
= 43/15= 2.8667
= P ( Z <2.8667) From Standard Normal Table
= 0.9979
P(X > = 309) = (1 - P(X < 309)
= 1 - 0.9979 = 0.0021
b.
P ( Z < x ) = 0.02
Value of z to the cumulative probability of 0.02 from normal table is -2.053749
P( x-u/s.d < x - 266/15 ) = 0.02
That is, ( x - 266/15 ) = -2.053749
--> x = -2.053749 * 15 + 266 = 235.1938

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