Question

In a 90-story office building, the amount of time spent waiting on an elevator after pushing the call button is uniformly distributed between 15 seconds and 120 seconds.

Let X = the amount of time spent waiting on the elevator after pushing the call button.

Find the following probabilities. Round your answers to 4 decimal places, if needed.

What is the probability that a randomly selected person will wait more than a minute for the elevator to arrive?  

P(x < 55) =  

Find the probability that a randomly-selected person will wait between 75 and 100 seconds for an elevator to arrive.  

Find the 80th percentile.   seconds

P(x < 30 | x < 90) =  

Answer 1

Mean $\mu=\dfrac{120+15}{2}=\dfrac{135}{2}=67.5$

Standard deviation $\sigma=\dfrac{120-15}{90}=\dfrac{105}{90}=1.16$

 Probability that a randomly selected person will wait more than a minute for the elevator to arrive

$P(X>60)=P(z>\dfrac{60-67.5}{1.16})=p(z>-6.46)=0.999$

$P(X<55)=P(z<\dfrac{55-67.5}{1.16})=P(z>-10.77)=0.99999=1$

probability that a randomly-selected person will wait between 75 and 100 seconds for an elevator to arrive.  

$P(75<x<100)=P(\dfrac{75-67.5}{1.16}<z<\dfrac{100-67.5}{1.16})=P(6.46652<z<28.01)=0.1125$

$80^{th} \text{percentile}=P 80 ​ ​ = ​ μ+z p ​ ×σ= 67.5+0.842×1.16= 68.476 ​$

$P(x < 30 | x < 90) = p(z<\dfrac{30-67.5}{1.16}|z<\dfrac{90-67.5}{1.16})$

$=P(z<-32.12|z<19.04)=\dfrac{0. 0001}{1}=0.001$