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

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

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

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

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

The probability that 2 out of 8 customers are very satisfied is__________assuming that 85 % of customers are very satisfied. Therefore, it is___________that randomly selecting 8

customers would result in 2 responding that they are very satisfied.

Answer 1

here this is binomial distribution with paramter p=0.85 and n=8

a) probability that exactly seven customers are very satisfied=P(X=7)==0.3847

b)

  probability that more than seven customers are very satisfied=P(X>7)=P(X=8)

= =0.2725

c)

probability that less than six customers are very satisfied =P(X<6) =1-(P(X>=6)

=1-(P(X=6)+P(X=7)+P(X=8)) =1-(+0.3847+0.2725) =0.1052

d)

The probability that 2 out of 8 customers are very satisfied is 0.0002 assuming that 85 % of customers are very satisfied. Therefore, it is unusual that randomly selecting 8 customers would result in 2 responding that they are very satisfied.