The lengths of pregnancies are normally distributed with a mean of 208 days and a standard deviation of 15 days. A wife claimed
to have given birth 308 days after a brief visit from her husband who was serving in the Navy. What is the probability of a pregnancy lasting 308 days or longer
1 answer:
Answer:
0%
Step-by-step explanation:
We have the following information:
mean (m) = 208 days
standard deviation (s) = 15 days
We have that the z value would be:
z = (x - m) / sd
replacing we have:
z = (308 - 208) / 15
z = 6.66
However,
P (x> 308) = 1 - (x <308)
P (x> 308) = 1 - (z <6.66)
if we look for this value of z in the table we assume that it is 1, since the table reaches 3 and the value is 1. Therefore:
P (x> 308) = 1 - 1
P (x> 308) = 0
Which means that the probability is 0%, therefore it is a very unusual pregnancy.
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