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 . Suppose you sit on a bench in a mall and observe​ people's habits as they sneeze. ​(a) What is the probability that among randomly observed individuals exactly do not cover their mouth when​ sneezing? ​(b) What is the probability that among randomly observed individuals fewer than do not cover their mouth when​ sneezing? ​(c) Would you be surprised​ if, after observing ​individuals, fewer than half covered their mouth when​ sneezing? Why? ​(a) The probability that exactly individuals do not cover their mouth is nothing. ​(Round to four decimal places as​ needed.) ​(b) The probability that fewer than individuals do not cover their mouth is nothing. ​(Round to four decimal places as​ needed.) ​(​c) Fewer than half of individuals covering their mouth ▼ would would not be surprising because the probability of observing fewer than half covering their mouth when sneezing is nothing​, which ▼ is is not an unusual event. ​(Round to four decimal places as​ needed.)

Answer 1

Complete Question

The complete question is shown on the first uploaded image

Answer:

a

 $P(X = 8) = 0.0037$

b

 $P(X < 5) = 0.805$

c

 $P(X > 6) = 0.0206$

I would be surprised because the value is very small , less the 0.05

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 $p = 0.267$

Generally data collected from this study follows  binomial  distribution because the number of trials is  finite , there are only two outcomes, (covering  , and  not covering mouth when sneezing ) , the trial are independent

Hence for a randomly selected variable  X we have that  

   $X \ \ \~ \ \ { B ( p , n )}$

The probability distribution function for binomial  distribution is  

    $P(X = x ) = ^nC_x * p^x * (1 -p) ^{n-x}$

Considering question a

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

     $P(X = 8) = ^{12} C_8 * (0.267)^8 * (1- 0.267)^{12-8}$

Here C denotes  combination

So

     $P(X = 8) = 495 * 0.000025828 * 0.28867947$

    $P(X = 8) = 0.0037$

Considering question b

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

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

=>   $P(X < 5 ) =[ ^{12} C_0 * (0.267)^0 * (1- 0.267)^{12-0} + \cdots + ^{12} C_4 * (0.267)^4 * (1- 0.267)^{12-4} ]$

=> $P(X < 5 ) = 0.02406 + 0.10516 + 0.21067 + 0.25580 + 0.20964$

=>  $P(X < 5) = 0.805$

Considering question c

Generally the probability that fewer than half(6) covered their mouth when​ sneezing(i.e the probability the greater than half do not cover their mouth when sneezing) is mathematically represented as

      $P(X > 6) = 1 - p(X \le 6)$

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

=>    $P(X > 6)=1 - [^{12} C_0 * (0.267)^0 * (1- 0.267)^{12-0}+ \cdots + ^{12} C_4 * (0.267)^6 * (1- 0.267)^{12-6} ]$

=>    $P(X > 6)= 1 - [0.02406 + \cdots + 0.0519 ]$  

=>    $P(X > 6) = 0.0206$

I would be surprised because the value is very small , less the 0.05