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A population experiences an average of 12 deaths from lung cancer per year. Calculate the probability that 10 or fewer people die, and the probability that at least 3 people die.

Mathematics

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Step-by-step solution

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This problem involves the Poisson distribution, which is used to model the number of events occurring in a fixed interval of time or space, given a known average rate of occurrence (λ\lambda) and that these events occur independently.

Here, the random variable XX represents the number of deaths from lung cancer in a year.
The average number of deaths is given as λ=12\lambda = 12.
The probability mass function for a Poisson distribution is:

P(X=k)=e−λλkk!P(X=k) = \frac{e^{-\lambda} \lambda^k}{k!}

where kk is the number of events (deaths) and ee is Euler's number (e≈2.71828e \approx 2.71828).

Part i

What is the probability that 10 or fewer people will die of the disease?

This asks for P(X≤10)P(X \le 10). This is the cumulative probability from k=0k=0 to k=10k=10.

P(X≤10)=∑k=010P(X=k)=∑k=010e−1212kk!P(X \le 10) = \sum_{k=0}^{10} P(X=k) = \sum_{k=0}^{10} \frac{e^{-12} 12^k}{k!}

Using a Poisson cumulative distribution function calculator for λ=12\lambda=12 and k=10k=10:

P(X≤10)≈0.34723P(X \le 10) \approx 0.34723

The probability that 10 or fewer people will die is approximately 0.34723.

Part ii· 5 steps

What is the probability that there will be at least 3 deaths from cancer?

This asks for P(X≥3)P(X \ge 3). It is easier to calculate this as 1−P(X<3)1 - P(X < 3).

P(X≥3)=1−P(X<3)=1−[P(X=0)+P(X=1)+P(X=2)]P(X \ge 3) = 1 - P(X < 3) = 1 - [P(X=0) + P(X=1) + P(X=2)]

  1. Calculate P(X=0)P(X=0).

    P(X=0)=e−121200!=e−12×1=e−12≈0.00000614421P(X=0) = \frac{e^{-12} 12^0}{0!} = e^{-12} \times 1 = e^{-12} \approx 0.00000614421

  2. Calculate P(X=1)P(X=1).

    P(X=1)=e−121211!=12e−12≈12×0.00000614421≈0.00007373052P(X=1) = \frac{e^{-12} 12^1}{1!} = 12e^{-12} \approx 12 \times 0.00000614421 \approx 0.00007373052

  3. Calculate P(X=2)P(X=2).

    P(X=2)=e−121222!=1442e−12=72e−12≈72×0.00000614421≈0.00044238312P(X=2) = \frac{e^{-12} 12^2}{2!} = \frac{144}{2}e^{-12} = 72e^{-12} \approx 72 \times 0.00000614421 \approx 0.00044238312

  4. Sum these probabilities to find P(X<3)P(X < 3).

    P(X<3)≈0.00000614421+0.00007373052+0.00044238312≈0.00052225785P(X < 3) \approx 0.00000614421 + 0.00007373052 + 0.00044238312 \approx 0.00052225785

  5. Calculate P(X≥3)P(X \ge 3).

    P(X≥3)=1−P(X<3)≈1−0.00052225785≈0.99947774215P(X \ge 3) = 1 - P(X < 3) \approx 1 - 0.00052225785 \approx 0.99947774215

    The probability that there will be at least 3 deaths is approximately 0.99948.

    (i) P(X≤10)≈0.34723(ii) P(X≥3)≈0.99948\boxed{ \begin{array}{l} \text{(i) } P(X \le 10) \approx 0.34723 \\ \text{(ii) } P(X \ge 3) \approx 0.99948 \end{array} }
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