Question

An article reports that 1 in 500 people carry the defective gene that causes inherited colon cancer. In a sample of 2000 individuals, what is the approximate distribution of the number who carry this gene?

We can approximate this distribution by the Poisson distribution with μ = 4 .

(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.)

(b) Use the Poisson approximation to calculate the approximate probability that at least 7 carry the gene. (Round your answer to three decimal places.)

Answer 1

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:

$P(X = x) = \frac{e^{-\mu}*\mu^{x}}{(x)!}$

In which

x is the number of sucesses

e = 2.71828 is the Euler number

$\mu$ is the mean in the given interval

In this problem, we have that:

$\mu = 4$

(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.)

$P(3 \leq X \leq 7) = P(X = 3) + P(X = 4) + P(X = 5) + P(X = 6) + P(X = 7)$

$P(X = x) = \frac{e^{-\mu}*\mu^{x}}{(x)!}$

$P(X = 3) = \frac{e^{-4}*(4)^{3}}{(3)!} = 0.195$

$P(X = 4) = \frac{e^{-4}*(4)^{4}}{(4)!} = 0.195$

$P(X = 5) = \frac{e^{-4}*(4)^{5}}{(5)!} = 0.156$

$P(X = 6) = \frac{e^{-4}*(4)^{6}}{(6)!} = 0.104$

$P(X = 7) = \frac{e^{-4}*(4)^{7}}{(7)!} = 0.06$

$P(3 \leq X \leq 7) = P(X = 3) + P(X = 4) + P(X = 5) + P(X = 6) + P(X = 7) = 0.195 + 0.195 + 0.156 + 0.104 + 0.06 = 0.71$

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

$P(X < 7) + P(X \geq 7) = 1$

We want $P(X \geq 7)$. So

$P(X \geq 7) = 1 - P(X < 7)$

In which

$P(X < 7) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) + P(X = 5) + P(X = 6)$

$P(X = x) = \frac{e^{-\mu}*\mu^{x}}{(x)!}$

$P(X = 0) = \frac{e^{-4}*(4)^{0}}{(0)!} = 0.018$

$P(X = 1) = \frac{e^{-4}*(4)^{1}}{(1)!} = 0.073$

$P(X = 2) = \frac{e^{-4}*(4)^{2}}{(2)!} = 0.147$

$P(X = 3) = \frac{e^{-4}*(4)^{3}}{(3)!} = 0.195$

$P(X = 4) = \frac{e^{-4}*(4)^{4}}{(4)!} = 0.195$

$P(X = 5) = \frac{e^{-4}*(4)^{5}}{(5)!} = 0.156$

$P(X = 6) = \frac{e^{-4}*(4)^{6}}{(6)!} = 0.104$

$P(X < 7) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) + P(X = 5) + P(X = 6) = 0.018 + 0.073 + 0.147 + 0.195 + 0.195 + 0.156 + 0.104 = 0.888$

$P(X \geq 7) = 1 - P(X < 7) = 1 - 0.888 = 0.112$

0.112 = 11.2% probability that at least 7 carry the gene.