Question

Suppose pizza delivery times are a normally distributed random variable with an expected value of 20 minutes and a standard deviation of 4 minutes.

a) What is the probability a delivery will take between 15 and 25 minutes?

b) If the pizza doesn’t arrive in 30 minutes, it is free. What is the probability the pizza will be free?

c) Find the shortest range of times that includes 40% of all delivery times.

d) In a particular month 1,000 pizzas were delivered. Assuming that the delivery times for each of these pizzas are independent from each other, what is the probability at least ten will be free? (Hint: You have to use the answer from part b.)

Answer 1

a) probability a delivery will take between 15 and 25 minutes=P(15<X<25)

=P((15-20)/4<Z<(25-20)/4)=P(-1.25<Z<1.25)=0.8944-0.1056 =0.7888

b)probability the pizza will be free =P(X>30)=P(Z>(30-20)/4)=P(Z>2.5)=0.0062

c)for 40 th percentile z =-0.25

shortest range of times that includes 40% of all delivery times =mean +z*strd deviation =20-0.25*4=19 minute

d)expected number of pizza that will be free =np=1000*0.0062=6.2

std deviaition =(np(1-p)))^1/2 =2.4822

probability at least ten will be free =P(X>=10)=1-P(X<=9)=1-P(Z<(9.5-6.2)/2.4822)=1-P(Z<1.33)=1-0.9082=0.0918