A hotel claims that 90% of its customers very satisfied with its service. Comple

Question

A hotel claims that 90% of its customers very satisfied with its service. Complete parts through d below based on a random sample of eight customers a. What is the probability that exactly seven customers are very satisfied? (Round to four decimal places as needed.) b. What is the probability that more than seven customers are very satisfied? (Round to four decimal places at needed.) c. What is the probability that less than six customers are very satisfied? (Round to four decimal places as needed.) d. Suppose that of eight customers selected three responded that they are very satisfied. What conclusions can be drawn about the hotel's clam? The probability that 3 out of 8 customers are very satisfied is assuming that 90% of customers are very satisfied. Based on this sample it is that 90% of customers are very satisfied (Round) to four decimal places as needed)

Explanation / Answer

Ans: binomial distribution

a)P(exactly seven customers are very satisfied)=8C7 *0.97*0.11=8*0.478*0.1=0.3824

b)P(more than 7 customers are very satisfied)=0.3824+P(8 customers satisfied)=0.3824+0.98

=0.3824+0.4305=0.8129

c)P(less than 6 customers satisfied)=1-(8C6*0.96*0.12+0.3824+0.4305)

=1-(0.1488+0.3824+0.4305)

=1-0.9617

=0.0383

d)P(exactly 3 are satisfied)=8C3*0.93*0.15=0.00041

As it is less than 0.05,it is unusual that exactly 3 out of 8 custmers will be satisfied.

For this sample,it is not true that 90 % of customes will be very satisfied.

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.