Answer to Question #333830 in Statistics and Probability for Darwin

Question

2. A line for tickets to a local concert had an average waiting time of 20 min. and a σ = 4 min. a. What percentage of the people in line will wait for more than 28 minutes? b. If 2000 ticket buyers were in line, how many of them would expect to wait for less than 16 minutes? c. How many minutes of waiting time would include 95% of those who would fall in line?

Answer 1

Let $X=$ waiting time: $X\sim N(\mu, \sigma^2).$

Given $\mu=20\ min, \sigma=4\ min.$

a.

$$P(X>28)=1-P(X\le 28)$$

$$=1-P(Z\le \dfrac{28-20}{4})=1-P(Z\le2)$$

$$\approx0.022750$$

2.275%

b.

$$P(X<16)=P(Z< \dfrac{16-20}{4})=P(Z<-1)$$

$$\approx0.158655$$

$$0.158655(2000)=317$$

317 ticket buyers.

c.

$$P(X<x)=P(Z< \dfrac{x-20}{4})=0.95$$

$$\dfrac{x-20}{4}=1.6449$$

$$x=26.58$$

26.58 minutes