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

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.

1)is not similar since corresponding sides are not in the same ratio

2)are similar since corresponding sides are in the same ratio and corresponding angles are equal

hope it will

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