Can anyone give me a detailed solution to this problem because my teacher has already given me a very brief solution, but I still do not understand how to do it. Thanks!
An expert witness in a paternity lawsuit testifies that the length (in days) of a pregnancy, from conception to delivery, is approximately normally distributed, with parameters μ = 270 and σ = 10. The defendant is able to provide evidence that he was out of the country during the period from 290 to 240 days before the birth of the child. What is the probability that the defendant was in the country when the child was conceived?
Solution :
Given that,
mean = = 270
standard deviation = = 10
P (290 < x < 240 )
P ( 290 - 270 / 10) < ( x - / ) < ( 240 - 270 / 10)
P ( 20 / 10 < z < - 30 / 10 )
P ( 2 < z < - 3 )
P ( z < 2 ) - P ( z < -3)
Using z table
= 0.9772 - 0.0013
= 0.9759
Probability = 0.9759
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