Question

A rare form of malignant tumor occurs in 11 children in a million, so its probability is 0.000011. Four cases of this tumor occurred in a certain town, which had 19,458 children.

a) Provides the mean number of cases in groups of 19,458 children
b) Provides the probability that the number of tumor cases is 0 or 1
c) Provides the probability of more than one case
d) Does not answer the questions

Answer 1

To solve this problem, we can use the binomial distribution formula. The mean number of cases can be calculated as n * p, and the probability of different cases can be found using the cumulative distribution function.

Step-by-step explanation:

To solve this problem, we can use the binomial distribution formula. Let's denote the number of cases of the tumor as X. The probability distribution for X follows a binomial distribution with parameters n = 19458 (the number of children in the town) and p = 0.000011 (the probability of the tumor in a child).

a) The mean number of cases in groups of 19458 children can be calculated as n * p, which is 19458 * 0.000011 = 0.214.

b) To find the probability of 0 or 1 case, we can use the cumulative distribution function (CDF) of the binomial distribution. P(X <= 1) = P(X = 0) + P(X = 1) = (1 - p)^n + n * p * (1 - p)^(n - 1).

c) To find the probability of more than one case, we can use the complement of the probability of 0 or 1 case. P(X > 1) = 1 - P(X <= 1).