Type numbers in the boxes. Part 1: 4 points Part 2: 3 points Part 3: 3 points Pa
ID: 3075116 • Letter: T
Question
Type numbers in the boxes. Part 1: 4 points Part 2: 3 points Part 3: 3 points Part 4: 4 points Part 5: 4 points Part 6: 4 points 22 points Assume that 89% 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?
Explanation / Answer
p = 0.89
n = 5
This is a binomial distribution
P(X = x) = 5Cx * 0.89x * (1 - 0.89)5-x
a) P(X > 1) = 1 - P(X = 0)
= 1 - 5C0 * 0.890 * 0.115
= 1 - 0
= 1
b) P(X = 3) = 5C3 * 0.893 * 0.112 = 0.0853
c)P(X > 4) = P(X = 4) + P(X = 5)
= 5C4 * 0.894 * 0.111 + 5C5 * 0.895 * 0.110
= 0.9035
d) P(X < 2) = 1 - P(X > 2)
= 1 - [P(X = 3) + P(X > 4)]
= 1 - [0.0853 + 0.9035]
= 1 - 0.9888
= 0.0112
e) Expected number = n * p = 5 * 0.89 = 4.45
f) Standard deviation = sqrt(n * p * (1 - p)) = sqrt(5 * 0.89 * (1 - 0.89)) = 0.70
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