Question

Assume that 12% of people are left-handed. If we select 5 people at random, find the probability of each outcome described below, rounded to four decimal places:

a. There are some lefties (≥ 1) among the 5 people.

b. There are exactly 3 lefties in the group.

c. There are at least 4 lefties in the group.

d. There are no more than 2 lefties in the group.

e. How many lefties do you expect?

f. With what standard deviation?

Answer 1

n = 5

P = 0.12

It is a binomial distribution

P(X = x) = nCx * p^x * (1 - p)^{n - x}

a) P(X > 1) = 1 - P(X < 1)

= 1 - P(X = 0)

= 1 - (5C0 * (0.12)^0 * (0.88)^5)

= 1 - 0.5277 = 0.4723

b) P(X = 3) = 5C3 * (0.12)^3 * (0.88)^2 = 0.0134

C) P(X > 4) = P(X = 4) + P(X = 5)

= 5C4 * (0.12)^4 * (0.88)^1 + 5C5 * (0.12)^5 * (0.88)^0 = 0.0009

d) P(X < 2) = P(X = 0) + P(X = 1) + P(X = 2)

= 5C0 * (0.12)^0 * (0.88)^5 + 5C1 * (0.12)^1 * (0.88)^4 + 5C2 * (0.12)^2 * (0.88)^3 = 0.9857

e) Expected number of lefties = np = 5 * 0.12 = 0.6

f) Standard deviation = sqrt(np(1 - p))

= sqrt(5 * 0.12 * (1 - 0.12))

= 0.7266