Question

Assume that 61% 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

X ~ Bin(n,p)

Where n = 5 , p = 0.61

Binomial probability distribution,

P(X) = ⁿC_x * p^x * ( 1 - p)^n-x

a)

P(X >= 1) = 1 - P(X = 0)

= 1 - ⁵C₀ * 0.61⁰ * ( 1 - 0.61)⁵

= 0.9910

b)

P(X = 3) = ⁵C₃ * 0.61³ * ( 1 - 0.61)²

= 0.3452

c)

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

= ⁵C₄ * 0.61⁴ * ( 1 - 0.61)¹ + ⁵C₅ * 0.61⁵ * ( 1 - 0.61)⁰

= 0.3545

d)

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

= ⁵C₀ * 0.61⁰ * ( 1 - 0.61)⁵ + ⁵C₁ * 0.61¹ * ( 1 - 0.61)⁴ + ⁵C₂ * 0.61² * ( 1 - 0.61)³

= 0.3003

e)

E(X) = n p = 5 * 0.61 = 3.05

f)

Standard deviation = sqrt [ n p ( 1 - p) ]

= sqrt [ 5 * 0.61 ( 1 - 0.61) ]

= 1.0906