Question

20% of individuals participating in a study possess Gene A. On a particular day, what is the probability that, out of 10 individuals participating, 2 of them do not possess the Gene A?

What is the probability that the number of individuals participating with Gene A will be within 0.3 of the true population mean?

How many individuals must participate to have the probability of at least one Gene A individual over 0.80?

Answer 1

a) Probability that out of 10 people participating, 2 do not possess Gene A is computed here as:

= Probability that 8 of the 10 individuals possess gene A

$P(X = 8) = \binom{10}{8}0.2^8(1- 0.2)^2 = 0.000074$

Therefore 0.000074 is the required probability here.

b) The true population mean here is computed here as:
Mean = np = 0.2*10 = 2

Therefore, the probability here is computed as:
P(2 - 0.3 < X < 2 + 0.3) = P(X = 2) = (10c2)*0.2²*(1 - 0.2)⁸ = 0.3020

Therefore 0.3020 is the required probability here.

c) For a sample size of n, the probability of having at least one individual with gene A would be computed as:
= 1 - Probability of noone with gene A

= 1 - (1 - 0.2)ⁿ

= 1 - 0.8ⁿ

Therefore, we have here:
1 - 0.8ⁿ >= 0.8

0.8ⁿ <= 0.2

Taking natural log both sides, we have here:
nLn(0.8) = Ln(0.2)

n = Ln(0.2) / Ln(0.8) = 7.2

Therefore 8 is the minimum sample size required here.