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An expert witness in a paternity suit testifies that the length (in days) of distributed with parameters μ=270 and σ^2=100. The defendant in the suit is able to prove that he was out of the country during a period that began 290 days before the birth of the child and ended 240 days before birth.
If the defendant was in fact, the father of the child, what is the probability that the mother could have had the very long or very short pregnancy indicated by the testimony?

  1.   0.241
  2.   0.0241

how to solve such questions??

1 Answer

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variance is given as 100.

Hence, standard deviation is 10. Mean = 270.

For the women to have a long pregenancy more than 300 days or less than 240 days.Here depending upon the defintion of long and short pregenany there answer may vary. As we know mean is 270 and standard deviation is 10. this range specified $\mu + 3\sigma and \mu -3\sigma$v i.e that the suspect must have  been out of country in that range. So answer should 0.996. It must be true that he should've been out of country at that range. May be if they had asked negation of that question then it will be 0.004. I don't know how its 0.241.

For her to have a long pregenancy the suspect must've been within the country300 days before pregenancy. And for her to have a short pregnancy he must've been within the country 240 days before pregnancy.

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