Answer:
a) 0.71 = 71% probability that between 3 and 7 (inclusive) carry the gene.
b) 0.112 = 11.2% probability that at least 7 carry the gene.
Step-by-step explanation:
In a Poisson distribution, the probability that X represents the number of successes of a random variable is given by the following formula:
![P(X = x) = \frac{e^{-\mu}*\mu^{x}}{(x)!}](https://tex.z-dn.net/?f=P%28X%20%3D%20x%29%20%3D%20%5Cfrac%7Be%5E%7B-%5Cmu%7D%2A%5Cmu%5E%7Bx%7D%7D%7B%28x%29%21%7D)
In which
x is the number of sucesses
e = 2.71828 is the Euler number
is the mean in the given interval
In this problem, we have that:
![\mu = 4](https://tex.z-dn.net/?f=%5Cmu%20%3D%204)
(a) Use the Poisson approximation to calculate the approximate probability that between 3 and 7 (inclusive) carry the gene. (Round your answer to three decimal places.)
![P(3 \leq X \leq 7) = P(X = 3) + P(X = 4) + P(X = 5) + P(X = 6) + P(X = 7)](https://tex.z-dn.net/?f=P%283%20%5Cleq%20X%20%5Cleq%207%29%20%3D%20P%28X%20%3D%203%29%20%2B%20P%28X%20%3D%204%29%20%2B%20P%28X%20%3D%205%29%20%2B%20P%28X%20%3D%206%29%20%2B%20P%28X%20%3D%207%29)
![P(X = x) = \frac{e^{-\mu}*\mu^{x}}{(x)!}](https://tex.z-dn.net/?f=P%28X%20%3D%20x%29%20%3D%20%5Cfrac%7Be%5E%7B-%5Cmu%7D%2A%5Cmu%5E%7Bx%7D%7D%7B%28x%29%21%7D)
![P(X = 3) = \frac{e^{-4}*(4)^{3}}{(3)!} = 0.195](https://tex.z-dn.net/?f=P%28X%20%3D%203%29%20%3D%20%5Cfrac%7Be%5E%7B-4%7D%2A%284%29%5E%7B3%7D%7D%7B%283%29%21%7D%20%3D%200.195)
![P(X = 4) = \frac{e^{-4}*(4)^{4}}{(4)!} = 0.195](https://tex.z-dn.net/?f=P%28X%20%3D%204%29%20%3D%20%5Cfrac%7Be%5E%7B-4%7D%2A%284%29%5E%7B4%7D%7D%7B%284%29%21%7D%20%3D%200.195)
![P(X = 5) = \frac{e^{-4}*(4)^{5}}{(5)!} = 0.156](https://tex.z-dn.net/?f=P%28X%20%3D%205%29%20%3D%20%5Cfrac%7Be%5E%7B-4%7D%2A%284%29%5E%7B5%7D%7D%7B%285%29%21%7D%20%3D%200.156)
![P(X = 6) = \frac{e^{-4}*(4)^{6}}{(6)!} = 0.104](https://tex.z-dn.net/?f=P%28X%20%3D%206%29%20%3D%20%5Cfrac%7Be%5E%7B-4%7D%2A%284%29%5E%7B6%7D%7D%7B%286%29%21%7D%20%3D%200.104)
![P(X = 7) = \frac{e^{-4}*(4)^{7}}{(7)!} = 0.06](https://tex.z-dn.net/?f=P%28X%20%3D%207%29%20%3D%20%5Cfrac%7Be%5E%7B-4%7D%2A%284%29%5E%7B7%7D%7D%7B%287%29%21%7D%20%3D%200.06)
![P(3 \leq X \leq 7) = P(X = 3) + P(X = 4) + P(X = 5) + P(X = 6) + P(X = 7) = 0.195 + 0.195 + 0.156 + 0.104 + 0.06 = 0.71](https://tex.z-dn.net/?f=P%283%20%5Cleq%20X%20%5Cleq%207%29%20%3D%20P%28X%20%3D%203%29%20%2B%20P%28X%20%3D%204%29%20%2B%20P%28X%20%3D%205%29%20%2B%20P%28X%20%3D%206%29%20%2B%20P%28X%20%3D%207%29%20%3D%200.195%20%2B%200.195%20%2B%200.156%20%2B%200.104%20%2B%200.06%20%3D%200.71)
0.71 = 71% probability that between 3 and 7 (inclusive) carry the gene.
(b) Use the Poisson approximation to calculate the approximate probability that at least 7 carry the gene. (Round your answer to three decimal places.)
Either less than 7 carry the gene, or at least 7 do. The sum of the probabilities of these events is decimal 1. So
![P(X < 7) + P(X \geq 7) = 1](https://tex.z-dn.net/?f=P%28X%20%3C%207%29%20%2B%20P%28X%20%5Cgeq%207%29%20%3D%201)
We want
. So
![P(X \geq 7) = 1 - P(X < 7)](https://tex.z-dn.net/?f=P%28X%20%5Cgeq%207%29%20%3D%201%20-%20P%28X%20%3C%207%29)
In which
![P(X < 7) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) + P(X = 5) + P(X = 6)](https://tex.z-dn.net/?f=P%28X%20%3C%207%29%20%3D%20P%28X%20%3D%200%29%20%2B%20P%28X%20%3D%201%29%20%2B%20P%28X%20%3D%202%29%20%2B%20P%28X%20%3D%203%29%20%2B%20P%28X%20%3D%204%29%20%2B%20P%28X%20%3D%205%29%20%2B%20P%28X%20%3D%206%29)
![P(X = x) = \frac{e^{-\mu}*\mu^{x}}{(x)!}](https://tex.z-dn.net/?f=P%28X%20%3D%20x%29%20%3D%20%5Cfrac%7Be%5E%7B-%5Cmu%7D%2A%5Cmu%5E%7Bx%7D%7D%7B%28x%29%21%7D)
![P(X = 0) = \frac{e^{-4}*(4)^{0}}{(0)!} = 0.018](https://tex.z-dn.net/?f=P%28X%20%3D%200%29%20%3D%20%5Cfrac%7Be%5E%7B-4%7D%2A%284%29%5E%7B0%7D%7D%7B%280%29%21%7D%20%3D%200.018)
![P(X = 1) = \frac{e^{-4}*(4)^{1}}{(1)!} = 0.073](https://tex.z-dn.net/?f=P%28X%20%3D%201%29%20%3D%20%5Cfrac%7Be%5E%7B-4%7D%2A%284%29%5E%7B1%7D%7D%7B%281%29%21%7D%20%3D%200.073)
![P(X = 2) = \frac{e^{-4}*(4)^{2}}{(2)!} = 0.147](https://tex.z-dn.net/?f=P%28X%20%3D%202%29%20%3D%20%5Cfrac%7Be%5E%7B-4%7D%2A%284%29%5E%7B2%7D%7D%7B%282%29%21%7D%20%3D%200.147)
![P(X = 3) = \frac{e^{-4}*(4)^{3}}{(3)!} = 0.195](https://tex.z-dn.net/?f=P%28X%20%3D%203%29%20%3D%20%5Cfrac%7Be%5E%7B-4%7D%2A%284%29%5E%7B3%7D%7D%7B%283%29%21%7D%20%3D%200.195)
![P(X = 4) = \frac{e^{-4}*(4)^{4}}{(4)!} = 0.195](https://tex.z-dn.net/?f=P%28X%20%3D%204%29%20%3D%20%5Cfrac%7Be%5E%7B-4%7D%2A%284%29%5E%7B4%7D%7D%7B%284%29%21%7D%20%3D%200.195)
![P(X = 5) = \frac{e^{-4}*(4)^{5}}{(5)!} = 0.156](https://tex.z-dn.net/?f=P%28X%20%3D%205%29%20%3D%20%5Cfrac%7Be%5E%7B-4%7D%2A%284%29%5E%7B5%7D%7D%7B%285%29%21%7D%20%3D%200.156)
![P(X = 6) = \frac{e^{-4}*(4)^{6}}{(6)!} = 0.104](https://tex.z-dn.net/?f=P%28X%20%3D%206%29%20%3D%20%5Cfrac%7Be%5E%7B-4%7D%2A%284%29%5E%7B6%7D%7D%7B%286%29%21%7D%20%3D%200.104)
![P(X < 7) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) + P(X = 5) + P(X = 6) = 0.018 + 0.073 + 0.147 + 0.195 + 0.195 + 0.156 + 0.104 = 0.888](https://tex.z-dn.net/?f=P%28X%20%3C%207%29%20%3D%20P%28X%20%3D%200%29%20%2B%20P%28X%20%3D%201%29%20%2B%20P%28X%20%3D%202%29%20%2B%20P%28X%20%3D%203%29%20%2B%20P%28X%20%3D%204%29%20%2B%20P%28X%20%3D%205%29%20%2B%20P%28X%20%3D%206%29%20%3D%200.018%20%2B%200.073%20%2B%200.147%20%2B%200.195%20%2B%200.195%20%2B%200.156%20%2B%200.104%20%3D%200.888)
![P(X \geq 7) = 1 - P(X < 7) = 1 - 0.888 = 0.112](https://tex.z-dn.net/?f=P%28X%20%5Cgeq%207%29%20%3D%201%20-%20P%28X%20%3C%207%29%20%3D%201%20-%200.888%20%3D%200.112)
0.112 = 11.2% probability that at least 7 carry the gene.