Use the reciprocal property to solve for x:
1 answer:
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The lengths of pregnancies are normally distributed with a mean of 266 days and a standard deviation of 15 days.
That is,
Consider X be the length of the pregnancy
Mean and standard deviation of the length of the pregnancy.
Mean
Standard deviation \sigma =15
For part (a) , to find the probability of a pregnancy lasting 308 days or longer:
That is, to find
Using normal distribution,
Thus
So
Thus the probability of a pregnancy lasting 308 days or longer is given by 0.00256.
This the answer for part(a): 0.00256
For part(b), to find the length that separates premature babies from those who are not premature.
Given that the length of pregnancy is in the lowest 3%.
The z-value for the lowest of 3% is -1.8808
Then
This implies
Thus the babies who are born on or before 238 days are considered to be premature.
