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It is given that 4 cases of neuroblastoma have occurred in the population of 12429 children in Oak Park, Illinois.

a.

The probability ofoccurrence of neuroblastoma cases is equal to p=0.000011.

The number of children in the group is equal to 12429.

The mean number of cases in Illinois:

\(\begin{aligned}{c}\mu = np\\ = 12429\left( {0.000011} \right)\\ = 0.137\end{aligned}\)

Thus, the mean number of cases of neuroblastoma is equal to 0.137.

b.

Let X be the number of cases in a group of 12429 children.

Here, X follows a Poisson distribution with mean equal to\({\kern 1pt} \mu = 0.137\).

The probability that the number of neuroblastoma cases in a group of 12429 children is equal to 0 or 1 is computed below:

\(P\left( x \right) = \frac{{{\mu ^x}{e^{ - \mu }}}}{{x!}}\)

\(\begin{aligned}{c}P\left( {0\;{\rm{or}}\;1} \right) = P\left( 0 \right) + P\left( 1 \right)\\ = \frac{{{{\left( {0.137} \right)}^0}{{\left( {2.71828} \right)}^{ - 0.137}}}}{{0!}} + \frac{{{{\left( {0.137} \right)}^1}{{\left( {2.71828} \right)}^{ - 0.137}}}}{{1!}}\\ = 0.87197 + 0.11946\\ = 0.991430\end{aligned}\)

\( \approx 0.991\)

Therefore, the probability that the number of neuroblastoma cases in a group of 12429 children is equal to 0 or 1 is equal to 0.991.

c.

The probability of more than 1 case is computed below:

\(\begin{aligned}{c}P\left( {X > 1} \right) = 1 - P\left( {X \le 1} \right)\\ = 1 - \left( {P\left( {X = 0} \right) + P\left( {X = 1} \right)} \right)\\ = 1 - 0.991\\ = 0.009\end{aligned}\)

Therefore, the probability that the number of neuroblastoma cases in a group of 12429 children is more than 1 is equal to 0.009.

d.

It can be seen that the probability of more than 1 case is extremely small.

Thus, the probability of 4 cases will be even smaller.

Therefore, it can be concluded that the occurrence of 4 cases cannot be considered a random event because the probability of the event happening is negligible.