X ~ Exp(2)
P (X < 1) = 1 - e-(1/2)(1) = 0.39 (C)
It is percentile of sampling distribution, thus qnorm
Poisson distribution indicates mean = variance, thus sd = sqrt(2/36) = sqrt(2)/6 (D)
X ~ Exp(8)
P (X <= 7) = 1 - e-(1/8)(7) = 0.58 (E)
Wilcox A.
asked 03/28/201 - The the time between incoming customer service calls to a computer-repair hotline follows an exponential distribution with an expectation of 2 minutes between calls. What is the probability that the time between calls will be less than 1 minute for a randomly selected period?
Select one:
a. .44
b. .31
c. .39
d. .61
e. .0
2 - Traffic at the airport changed recently, and you collected 36 weeks of data in an eort to estimate the new rate of tire replacements. You found that the average tire replacements per week was 2 and the numbers followed a Poisson distribution, but you realize that your sample might not be 100% accurate. What is the 99th percentile of sampling distribution of the mean of tires replaced per week (based on the observed mean of 2 and the sample size of 36)?
Select one:
a. qnorm(.99, mean=2, sd=sqrt(1/2))
b. pnorm(.99, mean=2, sd=2)
c. pnorm(.99, mean=2, sd=1/sqrt(2))
d. qnorm(.99, mean=2, sd=sqrt(2)/6)
e. qnorm(.99, mean=2, sd=1/
3 - At a small store, a customer enters the front door on average every 8 minutes. A prior study indicated that the time between customers entering the front door during weekdays follows an exponential distribution. What is the probability that the time between customers entering the store on a weekday will be less than or equal to 7?
Select one:
a. .62
b. .43
c. 1/8
d. 7/8
e. .58
X ~ Exp(2)
P (X < 1) = 1 - e-(1/2)(1) = 0.39 (C)
It is percentile of sampling distribution, thus qnorm
Poisson distribution indicates mean = variance, thus sd = sqrt(2/36) = sqrt(2)/6 (D)
X ~ Exp(8)
P (X <= 7) = 1 - e-(1/8)(7) = 0.58 (E)
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