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
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