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 8 randomly observed individuals exactly 4 do not cover their mouth when​ sneezing? ​

(​b) What is the probability that among 18 randomly observed individuals fewer than 6 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?

Answer 1

Solution :

Given that P(not cover their mouth when​ sneezing) = p = 0.267

=> q = 1 - p = 0.733

=> For binomial distribution , P(x = r) = nCr*p^r*q^(n-r)

(a) n = 8

=> P(x = 4) = 8C4*0.267^4*0.733^4

= 0.1027

(b) n = 18

=> P(x < 6) = P(x = 5) + P(x = 4) + P(x = 3) + P(x = 2) + P(x = 1) + P(x = 0)

= 18C5*0.267^5*0.733^13 + 18C4*0.267^4*0.733^14 + 18C3*0.267^3*0.733^15 + 18C2*0.267^2*0.733^16 + 18C1*0.267^1*0.733^17 + 18C0*0.267^0*0.733^18

= 0.2050 + 0.2010 + 0.1472 + 0.0757 + 0.0245 + 0.0037

= 0.6571

(c) given that n = 18 , P(cover their mouth when​ sneezing) = p = 1 - P(not cover their mouth when​ sneezing)

= 1 - 0.267

= 0.733

=> q = 1 - p = 0.267

=> P(x < 9) = 0.0089

=> Fewer than half of 18 individuals covering their mouth would be surprising because the probability of observing fewer than half covering their mouth when sneezing is 0.0089,which is an unusual event