Answer:
a) P ( X < 50 ) = 0.56540
b) P ( x < 50 ) = 0.1193
Explanation:
Given:-
- The average time between consecutive phone inquiries = 60 seconds
Find:-
a What is the probability that the time between two consecutive phone inquiries is less than 50 seconds?
b Consider a random selection of 100 consecutive phone inquiries. What is the probability that the average time between consecutive inquiries in this random sample is less than 50 seconds?
Solution:-
- A random variable (X) follows an exponential distribution with rate scale parameter λ = 1 / average = 1/60.
X ~ Exp ( 1 / 60 )
a)
- The CDF of exponential distribution is given by:
P ( X < x ) = 1 - e^(-λx)
P ( X < 50 ) = 1 - e^(-50/60)
= 0.56540
b)
- A sample of n = 100 consecutive phone inquiries was taken to determine probability that the average time between consecutive inquiries in this random sample is less than 50 seconds.
- We will approximate the sample taken to be normally distributed. The mean (u) and standard deviation (s) of the normally distributed sample would be:
u = 1 / λ = 60 seconds
s= 1 / √n*λ = 60 / √50 = 8.48528
Then,
x ~ N ( 60 , 8.48528^2 )
- We will standard our test value:
P ( x < 50 ) = P ( z < (50 - 60) / 8.48528 )
P ( z < -1.17851 ) = 0.1193
- The result is P ( x < 50 ) = 0.1193