Question

Assume that 11% 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

a)

Here, n = 5, p = 0.11, (1 - p) = 0.89 and x = 1
As per binomial distribution formula P(X = x) = nCx * p^x * (1 - p)^(n - x)

We need to calculate P(X >= 1).
P(X >= 1) = (5C1 * 0.11^1 * 0.89^4) + (5C2 * 0.11^2 * 0.89^3) + (5C3 * 0.11^3 * 0.89^2) + (5C4 * 0.11^4 * 0.89^1) + (5C5 * 0.11^5 * 0.89^0)
P(X >= 1) = 0.3451 + 0.0853 + 0.0105 + 0.0007 + 0
P(X >= 1) = 0.4416

b)
Here, n = 5, p = 0.11, (1 - p) = 0.89 and x = 3
As per binomial distribution formula P(X = x) = nCx * p^x * (1 - p)^(n - x)

We need to calculate P(X = 3)
P(X = 3) = 5C3 * 0.11^3 * 0.89^2
P(X = 3) = 0.0105

c)
Here, n = 5, p = 0.11, (1 - p) = 0.89 and x = 4
As per binomial distribution formula P(X = x) = nCx * p^x * (1 - p)^(n - x)

We need to calculate P(X >= 4).
P(X >= 4) = (5C4 * 0.11^4 * 0.89^1) + (5C5 * 0.11^5 * 0.89^0)
P(X >= 4) = 0.0007 + 0
P(X >= 4) = 0.0007

d)

Here, n = 5, p = 0.11, (1 - p) = 0.89 and x = 2
As per binomial distribution formula P(X = x) = nCx * p^x * (1 - p)^(n - x)

We need to calculate P(X < 2).
P(X < 2) = (5C0 * 0.11^0 * 0.89^5) + (5C1 * 0.11^1 * 0.89^4)
P(X < 2) = 0.5584 + 0.3451
P(X <2) = 0.9035

e)

np = 5 * 0.11 = 0.55

f)
std.dev = sqrt(npq)
= sqrt(5 * 0.11 * 0.89)
= 0.6996