3. You own a hotel where customers make reservations but sometimes do not show u

Question

3. You own a hotel where customers make reservations but sometimes do not show up. On weekdays percent of all customers making a reservation show up. Suppose that 20 individuals make reservations and that these individuals make independent decisions to show up 70 up for their reservation? What is the standard deviation of people showing up for their rseration on a wedkday? What is the probability that less than 4 people fail to show up for their reservation at this hotel when 20 individuals make a reservation? (3 points) On weekends, a customer who makes a reservation has a 0.93 probability of showing up. Suppose 15 individuals make reservations. What is the probability that more than one individual will not show up for their reservation? (2 points)

Explanation / Answer

Ans:

Probability of showing up,p=0.7

Probability of not showing up,1-p=1-0.7=0.3

n=20

Expected number of people who show up=np=20*0.7=14

standard deviation of people who show up=sqrt(np(1-p))=sqrt(20*0.7*0.3)=sqrt(4.2)=2.05

Probability of less than 4 people fail to show up=P(x<4)=P(x<=3)

=20C0*0.30*0.720+20C1*0.31*0.719+20C2*0.32*0.718+20C3*0.33*0.717

=0.0008+0.0068+0.0278+0.0716

=0.1071

P(x<4)=P(x<=3)=BINOMDIST(3,20,0.3,TRUE)=0.1071

Now,

Probability of showing up,p=0.93

Probability of not showing up,1-p=1-0.93=0.07

Probability that more than one will not show up=1-P(x<=1)=1-15C0*0.070*0.9315-15C1*0.071*0.9314

=1-0.3367-0.3801

=0.2832

Related Questions

Navigate

• Browse (All)
• Browse 3
• Subjects

---

---

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