According to a study done by a university student, the probability a randomly se

Question

According to a study done by a university student, the probability a randomly selected individual will not cover his or her mouth when sneezing is 0.267. Suppose you sit on a bench in a mall and observe people’s habits as they sneeze.
(a) What is the probability that among 18 randomly observed individuals exactly 7 do not cover their mouth when sneezing? (b) What is the probability that among 18 randomly observed individuals fewer than 4 do not cover their mouth when sneezing? (c) Would you be surprised if, after observing 18 individuals, fewer than half covered their mouth when sneezing? Why? According to a study done by a university student, the probability a randomly selected individual will not cover his or her mouth when sneezing is 0.267. Suppose you sit on a bench in a mall and observe people’s habits as they sneeze.
(a) What is the probability that among 18 randomly observed individuals exactly 7 do not cover their mouth when sneezing? (b) What is the probability that among 18 randomly observed individuals fewer than 4 do not cover their mouth when sneezing? (c) Would you be surprised if, after observing 18 individuals, fewer than half covered their mouth when sneezing? Why?
(a) What is the probability that among 18 randomly observed individuals exactly 7 do not cover their mouth when sneezing? (b) What is the probability that among 18 randomly observed individuals fewer than 4 do not cover their mouth when sneezing? (c) Would you be surprised if, after observing 18 individuals, fewer than half covered their mouth when sneezing? Why?

Explanation / Answer

P(not covering mouth when sneezing) = 0.267

P(covering mouth when sneezing) = 1 - 0.267 = 0.733

P(X = x) = nCx * p^x * (1 - p)^(n - x)

a) P(X = 7) = 18C7 * (0.267)^7 * (0.733)^11 = 0.1010

b) P(X < 4) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3)

                    = 18C0 * (0.267)^0 * (0.733)^18 + 18C1 * (0.267)^1 * (0.733)^17 + 18C2 * (0.267)^2 * (0.733)^16 + 18C3 * (0.267)^3 * (0.733)^15 = 0.2511

c) P(X < 9) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) + P(X = 5) + P(X = 6) + P(X = 7) + P(X = 8)

                  = 18C0 * (0.733)^0 * (0.267)^18 + 18C1 * (0.733)^1 * (0.267)^17 + 18C2 * (0.733)^2 * (0.267)^16 + 18C03* (0.733)^3 * (0.267)^15 + 18C4 * (0.733)^4 * (0.267)^14 + 18C5 * (0.733)^5 * (0.267)^13 + 18C6 * (0.733)^6 * (0.267)^12 + 18C7 * (0.733)^7 * (0.267)^1 + 18C18 * (0.733)^18 * (0.267)^0 = 0.0089

We will be surprised because the probability of fewer than half covered their mouth when sneezing is very less.

As the probability is less than 0.05(0.0089 < 0.05), so it will be unusual.

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