Question

A certain rare form of cancer occurs in 37 children in a​ million, so its probability is 0.000037. In the city of Normalville there are​ 74,090 children. A Poisson distribution will be used to approximate the probability that the number of cases of the disease in Normalville children is more than 2. Find the mean of the appropriate Poisson distribution​ (the mean number of cases in groups of​ 74,090 children).

Answer 1

Answer:

Mean = 2.7

In a group of 74090 we would expect about 3 (rounding to nearest whole number) children with the rare form of cancer.

Step-by-step explanation:

We are given that the rate of cancer in children is 37 children in 1 million. So the probability of cancer in a child is P(C) = 0.000037

Poisson distribution is used to approximate the number of cases of diseases and we have to find what will be the mean number of cases for 74,090.

In simple words we have to find the expected number of children with cancer in a group of 74,090 children.

The mean value of expected value can be obtained by multiplying the probability with the sample size. So, in this case multiplying probability of child having a cancer with total group size will give us the expected or mean number of children in the group with cancer.

Mean = E(x) = P(C) * Group size

Mean = 0.000037 x 74090

Mean = 2.7

This means in a group of 74090 we would expect about 3 (rounding to nearest whole number) children with the rare form of cancer.