Question

Assume that 41% 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

Probability that a random person is left-handed: P(X) = 0.41
a.
P(X ≥ 1) = 1 - P(X=0)
P(X = 0) = ⁵C₀ (0.41)⁰ ( 1 - 0.41)⁵ = 0.0715
P(X ≥ 1) = 1 - 0.0715 = 0.9285
b.
P(X = 3) = ⁵C₃(0.41)³( 1 - 0.41)² = 0.24

c.
P(X ≥ 4) = P(X = 4) + P(X = 5)
P(X = 4) = ⁵C₄(0.41)⁴( 1 - 0.41)¹ = 0.0834
P(X = 5) = ⁵C₅(0.41)⁵( 1 - 0.41)⁰ = 0.0116
P(X ≥ 4) = 0.0834 + 0.0116 = 0.095

d.
P(X ≤ 2) = P(X = 0) + P(X = 1) + P(X = 2)
P(X = 0) = ⁵C₀(0.41)⁰( 1 - 0.41)⁵ = 0.0715
P(X = 1) = ⁵C₁(0.41)¹( 1 - 0.41)⁴ = 0.2484
P(X = 2) = ⁵C₂(0.41)²( 1 - 0.41)³ = 0.3452
P(X ≤ 2) = 0.0715 + 0.2484 + 0.3452 = 0.6651

e.
E[X] = ∑ xP(X = x)
E[X] = (0 x P(X = 0)) + (1 x P(X = 1)) + (2 x P(X = 2)) + (3 x P(X = 3)) + (4 x P(X = 4)) + (5 x P(X = 5))
E[X] = (0 x 0.0715) + (1 x 0.2484) + (2 x 0.3452) + (3 x 0.24) + (4 x 0.0834) + (5 x 0.0116) = 2.0504
f.
SD[X] = E[X²] - E[X]²
E[X²] = (0² x P(X = 0)) + (1² x P(X = 1)) + (2² x P(X = 2)) + (3² x P(X = 3)) + (4² x P(X = 4)) + (5² x P(X = 5))

E[X²] = (0² x 0.0715) + (1² x 0.2484) + (2² x 0.3452) + (3² x 0.24) + (4² x 0.0834) + (5² x 0.0116) = 5.4136
E[X] ²= 4.2041
E[X²] = 5.4136 - 4.2041 = 1.2095