Question

according the the american hotel and motel association, women are expected to account for half of all business travelers by the year 2002. the ahma found tht 80% offer hair dryers in the bathrooms. consider a sample of 20 hotels. find the probability that more than 9 but less than 17 of the hotels in the sample

Answer 1

Answer: 0.71221

Explanation:

We know that for binomial distribution :-

`\mu=np\ \ ;\ \sigma=√(np(1-p))`, where p is the proportion of success in each trial and n is the sample size.

Given : p=0.80 ; n=20

Then,
`\mu=20(0.8)=16\ \ ;\ \sigma=√(20(0.8)(0.2))=1.79`

Let X be the binomial variable.

z-score :
`z=(x-\mu)/(\sigma)`

For x=9

`z=(9-16)/(1.79)=-3.91`

For x=17

`z=(17-16)/(1.79)=0.56`

Then, the probability that more than 9 but less than 17 of the hotels in the sample is given by :-

`P(9<X<17)=P(-3.91<z<0.56)=P(z<0.56)-P(z<-3.91)\\\\=0.7122603-0.0000461=0.7122142\approx0.71221`

Hence, the probability that more than 9 but less than 17 of the hotels in the sample = 0.71221