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Vadim26 [7]
3 years ago
6

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
Mathematics
1 answer:
Slav-nsk [51]3 years ago
5 0

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

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

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