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.

You're on a roll — 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 () and that these events occur independently. Here, the random variable X represents the number of deaths from lung cancer in a year. The average number of deaths is given as = 12. The probability mass function for a Poisson distribution is: P(X=k) = e^- ^kk! where k is the number of events (deaths) and e is Euler's number (e ≈ 2.71828). i) What is the probability that 10 or fewer people will die of the disease? This asks for P(X 10). This is the cumulative probability from k=0 to k=10. P(X 10) = _k=0^10 P(X=k) = _k=0^10 e^-12 12^kk! Using a Poisson cumulative distribution function calculator for =12 and k=10: P(X 10) ≈ 0.34723 The probability that 10 or fewer people will die is approximately 0.34723. ii) What is the probability that there will be at least 3 deaths from cancer? This asks for P(X 3). It is easier to calculate this as 1 - P(X < 3). P(X 3) = 1 - P(X < 3) = 1 - [P(X=0) + P(X=1) + P(X=2)] Step 1: Calculate P(X=0). P(X=0) = e^-12 12^00! = e^-12 × 1 = e^-12 ≈ 0.00000614421 Step 2: Calculate P(X=1). P(X=1) = e^-12 12^11! = 12e^-12 ≈ 12 × 0.00000614421 ≈ 0.00007373052 Step 3: Calculate P(X=2). P(X=2) = e^-12 12^22! = (144)/(2)e^-12 = 72e^-12 ≈ 72 × 0.00000614421 ≈ 0.00044238312 Step 4: Sum these probabilities to find P(X < 3). P(X < 3) ≈ 0.00000614421 + 0.00007373052 + 0.00044238312 ≈ 0.00052225785 Step 5: Calculate P(X 3). P(X 3) = 1 - P(X < 3) ≈ 1 - 0.00052225785 ≈ 0.99947774215 The probability that there will be at least 3 deaths is approximately 0.99948. l (i) P(X 10) ≈ 0.34723 \\ (ii) P(X 3) ≈ 0.99948 4 down, 1 left today. What's next?