Question

Assume that 20% of people are left-handed. If we select 5 people at random, find the...

Assume that 20% 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?

X ~ Binomial (n,p)

Where n = 5 , p = 0.20

P(X) =  nCx px (1-p)n-x

a)

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

= 1 -  5C0 0.200 0.805

= 0.6723

b)

P( X = 3) = 5C3 0.203 0.802

= 0.0512

c)

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

= 5C4 0.204 0.801 +5C5 0.205 0.800

= 0.0067

d)

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

= 5C0 0.200 0.805 +5C1 0.201 0.804 +5C2 0.202 0.803

= 0.9421

e)

E(X) = np

= 5 * 0.20

= 1

f)

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

= sqrt( 5 * 0.20 * 0.80)

= 0.8944