Suppose that 10% of the world’s population is left handed. (a) In a random sampl

Question

Suppose that 10% of the world’s population is left handed. (a) In a random sample of 5 people, compute the probability there is no left handed person (b) (continued from part (a)) compute the probability that there is one left handed person in the sample. (c) In a random sample of 10 people what is the probability of at least one left handed person in the sample. (d) (continued from part (c)) Compute the probability of 4 or fewer left handed people in the sample. (e) (continued still) Compute the probability that between 2 and 5 left handed (inclusive) people are in the sample. (f) In a random sample of 25 people, what is the mean number of left handed people we can expect? How about the standard deviation? (g) (continued from part (f)) Compute the probability the number of left handed people in the sample is within 2 standard deviations of the mean.

Explanation / Answer

As per binomial distribution,
P(X=r) = nCr * p^r * (1-p)^(n-r)

Here p = 0.1

a)
P(X=0) = 5C0 * 0.1^0 * (1-0.1)^(5-0) = 0.5905

b)
P(X=1) = 5C1 * 0.1^1 * (1-0.1)^(5-1) = 0.3281

c)
P(X >= 1) = 1 - P(X <= 1)
P(X <= 1) = P(X = 0) + P(X = 1)
= 10C0 * 0.1^0 * (1-0.1)^(10-0) + 10C1 * 0.1^1 * (1-0.1)^(10-1)
= 0.7361

P(X >= 1) = 1 - P(X <= 1) = 0.2639

d)
P(X <= 4) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4)
= 10C0 * 0.1^0 * (1-0.1)^(10-0) + .... + 10C4 * 0.1^4 * (1-0.1)^(10-4)
= 0.9984

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