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

6

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?

?(?a)

The probability that exactly

6

individuals do not cover their mouth is

0.0220.

?(Round to four decimal places as? needed.)

?(?b)

The probability that fewer than

55

individuals do not cover their mouth is

nothing .

?(Round to four decimal places as? needed.)

?(?c)

This

would

would not

would

be surprising because the probability of observing fewer than half covering their mouth when sneezing is

nothing ?,

which

is

is

is not

an unusual event.

?(Round to four decimal places as? needed.)

Answer 1

X ~ Binomial (n,p)

Binomial probability distribuion is

P(X) = ⁿC_x p^x ( 1 - p)^n-x

a)

n = 12 , p = 0.267 , x = 6

P( X = 6) = ¹²C₆ 0.267⁶ 0.733⁶

= 0.0519

b)

P( X < 5) = P( X <= 4)

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

= ¹²C₀ 0.267⁰ 0.733¹² +¹²C₁ 0.267¹ 0.733¹¹ +¹²C₂ 0.267² 0.733¹⁰ +¹²C₃ 0.267³ 0.733⁹ +¹²C₄ 0.267⁴ 0.733⁸

= 0.8053

c)

P( individual covered their mouth when sneezing) = 1 - 0.267 = 0.733

P( X < 6) = P( X <= 5)

= P( X = 0) +P( X = 1) +P( X = 2) +P( X = 3) +P( X = 4) +P( X = 5)

= ¹²C₀ 0.733⁰ 0.267¹² +¹²C₀ 0.733¹ 0.267¹¹ +¹²C₂ 0.733² 0.267¹⁰ +¹²C₃ 0.733³ 0.267⁹ +

¹²C₄ 0.733⁴ 0.267⁸ + ¹²C₅ 0.733⁵ 0.267⁵

= 0.0206

We would surprised if less than half of the 12 covered their mouth when sneezing since probability is

less than 0.05.

Since probability is less than 0.05, it is unusual event.