Question

1. About 5% of the population has a particular genetic mutation. 179 people are randomly selected.

Find the mean for the number of people with the genetic mutation in such groups of 179.

Find the standard deviation for the number of people with the genetic mutation in such groups of 179. (Round to the nearest 2 decimal places.)

2. The Wilson family had 8 children. Assuming that the probability of a child being a girl is 0.5, find the probability that the Wilson family had

at least 7 girls?

at most 7 girls?

Round your answers to 3 decimal places.

Answer 1

1)

mean = np
= 179 * 0.05
= 8.95

std.deviation = sqrt(npq)
=sqrt(179 * 0.05 * 0.95)
= 2.92

2)

Here, n = 8, p = 0.5, (1 - p) = 0.5 and x = 7
As per binomial distribution formula P(X = x) = nCx * p^x * (1 - p)^(n - x)

We need to calculate P(X >= 7).
P(X >= 7) = (8C7 * 0.5^7 * 0.5^1) + (8C8 * 0.5^8 * 0.5^0)
P(X >= 7) = 0.031 + 0.004
P(X >= 7) = 0.035

Here, n = 8, p = 0.5, (1 - p) = 0.5 and x = 7
As per binomial distribution formula P(X = x) = nCx * p^x * (1 - p)^(n - x)

We need to calculate P(X <= 7).
P(X <= 7) = (8C0 * 0.5^0 * 0.5^8) + (8C1 * 0.5^1 * 0.5^7) + (8C2 * 0.5^2 * 0.5^6) + (8C3 * 0.5^3 * 0.5^5) + (8C4 * 0.5^4 * 0.5^4) + (8C5 * 0.5^5 * 0.5^3) + (8C6 * 0.5^6 * 0.5^2) + (8C7 * 0.5^7 * 0.5^1)
P(X <= 7) = 0.004 + 0.031 + 0.109 + 0.219 + 0.273 + 0.219 + 0.109 + 0.031
P(X <= 7) = 0.995