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

Step-by-step explanation:

We know that for binomial distribution :-

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

Given : p=0.80  ;   n=20

Then, $\mu=20(0.8)=16\ \ ;\ \sigma=\sqrt{20(0.8)(0.2)}=1.79$

Let X be the binomial variable.

z-score : $z=\dfrac{x-\mu}{\sigma}$

For x=9

$z=\dfrac{9-16}{1.79}=-3.91$

For x=17

$z=\dfrac{17-16}{1.79}=0.56$

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