Normal distribution 3

2009-07-18 12:41 am
When subjected to a dose of insecticide, a particular breed of insect will die if the dose exceeds the tolerance of the insect. For a large population of this breed of insect, the tolerance (measured in appropriate units) is normally distributed with mean 30 and variance 9. The population is subjected to a dose of insecticide of strength 27. Assuming that the effect of each dose is independent of the previous dose, show that the probability that an insect selected at random from the survivors would die when subjected to a dose of strength 34 is approximately 0.89 .

回答 (2)

2009-07-18 1:09 am
✔ 最佳答案
The probability of insect killed using a dose of strength 27 is 0.1587
The probability of insect remained alive after this dose is 1 – 0.1587 = 0.8413
The probability of insect killed using a dose of strength of 34 is 0.9088 assuming that previous dose of strength 27 was not applied.
Since already 0.1587 was killed by the dose of strength 27, the insect killed by the dose of strength 34 is 0.9088 – 0.1587 = 0.7501
Insect killed by strength 34 / insect survived strength 27 = 0.7501 / 0.8413 = 0.89
2009-07-18 2:38 am
Let X be the tolerance of an insect. So, X~N(30, 9)

Since it is given that the insect survived with strength 27, so X>27, also the question ask for the probability that it would die under strength 34, so it comes up with 27<X<34

Notice that is P(-1<Z<4/3)=0.9088-0.1587=0.7501

By theory of conditional probability

Since the dose are independent so the required probability is
P(X<34 given X>27)= P(27<X<34)/P(X>27)
=0.7501/(1-0.1587)=0.8916

so approx equal to0.89
參考: 自己


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