Question

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?

Answer #1

X ~ Binomial (n,p)

Where n = 5 , p = 0.20

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

a)

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

= 1 - ^{5}C_{0} 0.20^{0}
0.80^{5}

= **0.6723**

b)

P( X = 3) = ^{5}C_{3} 0.20^{3}
0.80^{2}

= **0.0512**

c)

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

= ^{5}C_{4} 0.20^{4} 0.80^{1}
+^{5}C_{5} 0.20^{5} 0.80^{0}

= **0.0067**

d)

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

= ^{5}C_{0} 0.20^{0} 0.80^{5}
+^{5}C_{1} 0.20^{1} 0.80^{4}
+^{5}C_{2} 0.20^{2} 0.80^{3}

= **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**

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