Question

1. If a sandwich delivery takes longer than 30 minutes, the sandwich is free. If the average time for delivery is 18 minutes with a standard deviation of 4 minutes and delivery time is normally distributed:

1. What percent of sandwich will be free?

2. Delivery drivers who complete deliveries in fewer than 10 minutes get a bonus. What percent of deliveries grant bonuses to drivers?

3. Delivery drivers who are “buzzer beaters” – meaning they deliver the sandwich with less than two minutes to spare – get a reprimand. What percent of deliveries are within the 30-minute threshold, but still lead to a driver reprimand?

Answer 1

a)

Let X be the continuous random variable i.e. delivery time

Mean, 18 minutes

Standard deviation=4 minutes

It means,we need to find the probability of X being more than 30.

P(X>30)=P(Z>3)=0.5-P(0<Z<3)=0.5-0.4987=0.0013

(You can refer standard normal probability curve for Z=3.0 to get the probability value or can use a calculator/Excel)

We can say that 0.13% of sandwich will be free.

b)

It means,we need to find the probability of X being less than 10.

P(X<10)=P(Z<-2)=0.5-P(-2<Z<0)=0.5-P(0<Z<2)=0.5-0.4772=0.0228

We can say that 2.28% of deliveries will get bonus.

c)

In this we need to find the probability of deliveries made between 28 and 30 minutes

P(28<X<30)=P(2.5<Z<3)=P(0<Z<3)-P(0<Z<2.5)=0.4987-4938=0.0049

It means that around 0.49% of deliveries will lead to driver reprimand though within 30 minutes threshold