Question

A hotel claims that 85​% of its customers are very satisfied with its service. Complete parts a through d below based on a random sample of seven customers.

a. What is the probability that exactly six customers are very​ satisfied?

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

b. What is the probability that more than six customers are very​ satisfied?

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

c. What is the probability that less than five customers are very​ satisfied?

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

d. Suppose that of seven customers​ selected, two responded that they are very satisfied. What conclusions can be drawn about the​ sample?The probability that 22 out of 77 customers are very satisfied is

assuming that 85​% of customers are very satisfied.​ Therefore, it is ▼ likely not likely that randomly selecting 77 customers would result in 22 responding that they are very satisfied.

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

Answer 1

Answer:

A hotel claims that 85​% of its customers are very satisfied with its service. Complete parts a through d below based on a random sample of seven customers.

Binomial distribution used.

n=7, p=0.85

P(X=x) = (_nC_x) p^x (1-p)^n-x

Binomial Probabilities
Data
Sample size	7
Probability of an event of interest	0.85
Binomial Probabilities Table
	X	P(X)
	0	0.0000
	1	0.0001
	2	0.0012
	3	0.0109
	4	0.0617
	5	0.2097
	6	0.3960
	7	0.3206

a. What is the probability that exactly six customers are very​ satisfied?

​P= 0.3960 (Round to four decimal places as​ needed.)

b. What is the probability that more than six customers are very​ satisfied?

​P( x >6) = P( x=7) = 0.3206

(Round to four decimal places as​ needed.)

c. What is the probability that less than five customers are very​ satisfied?

P( x <5)= P( x=0)+ P( x=1)+ P( x=2)+ P( x=3)+ P( x=4)

= 0.0000+0.0001+0.0012+0.0109+0.0617

=0.0739

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

d. Suppose that of seven customers​ selected, two responded that they are very satisfied. What conclusions can be drawn about the​ sample?

The probability that 2 out of 7 customers are very satisfied is 0.0012 assuming that 85​% of customers are very satisfied.​ Therefore, it is not likely that randomly selecting 7 customers would result in 2 responding that they are very satisfied.

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