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 12 randomly observed individuals exactly 8 do not cover their mouth when​ sneezing?
B) What is the probability that among 12 randomly observed individuals fewer than 5 do not cover their mouth when​ sneezing?​(
C) Would you be surprised​ if, after observing 12 individuals, fewer than half covered their mouth when​ sneezing? Why?

Answer 1

Answer:

A

 $P(X = 8 ) = 0.0037$

B

    $P(X < 5) = 0.805$

C

I will not be surprised because the probability that fewer than half covered their mouth when​ sneezing is less than 0.5

Step-by-step explanation:

From the question we are told that

   The probability a randomly selected individual will not cover his or her mouth when sneezing is $p = 0.267$

     

    The probability a randomly selected individual will cover his or her mouth when sneezing is

                   $q = 1 -p$

                   $q = 1 -0.267$

                    $q = 0.733$

Generally the probability that among 12 randomly observed individuals exactly 8 do not cover their mouth when​ sneezing is mathematically represented as

        $P(X = 8 ) = \left 12 } \atop {}} \right. C_8 * p^8 * q^{12-8}$

        $P(X = 8 ) = 495 * (0.267)^8 * (0.733)^{12-8}$

        $P(X = 8 ) = 0.0037$

Generally the probability that among 12 randomly observed individuals fewer than 5 do not cover their mouth when​ sneezing is mathematically represented as

       $P(X < 5 ) = P[ P(X = 0) + \cdots + P(X = 4)]$

=>        $P(X < 5) = \left 12 } \atop {}} \right. C_0 * (0.267)^0 * (0.733)^{(12- 0) }+\cdots + \left 12 } \atop {}} \right. C_4 * (0.267)^0 * (0.733)^{(12- 4) }$

=>     $P(X < 5) = 0.805$

Give that half of  12 is  6 then

The probability that fewer than half covered their mouth when​ sneezing is mathematically represented as

      $P(X > 6) = 1 - P( X \le 6 )$

  => $P(X > 6) = 1 - [P(X = 0 ) +\cdots+P(X = 6) ]$

 => $P(X > 6) = 1 - [\left 12 } \atop {}} \right. C_0 *(0.267)^0 * (0.733)^{12 - 0 } +\cdots + \left 12 } \atop {}} \right. C_6 * (0.267)^6 * (0.733)^{12-6}]$

=>    $P(X > 6) = 0.0206$