The total amount of time it takes a JC Auto Repair mechanic to complete a full-service, 16-point oil change is distributed as a uniform random variable, ranging from 18 to 32 minutes.
What is the standard deviation of service time? Report your answer to 2 decimal places using conventional rounding rules.
ANSWER:
What is the probability that the amount of time to complete a full service, 16-point oil change will be less than 28 minutes? Report your answer to 4 decimal places using conventional rounding rules.
ANSWER:
What is the probability that the amount of time to complete a full service, 16-point oil change will be between 19 and 23 minutes? Report your answer to 4 decimal places using conventional rounding rules.
ANSWER:
What is the average amount of time to complete a full service, 16-point oil change at Grease Monkey? Report your answer to 2 decimal places using conventional rounding rules.
ANSWER:
a)
Here, the given values of lower limit, a = 18 and upper limit, b =
32
For Uniform distribution,
Standard Deviation = sqrt((b - a)^2/12)
Standard Deviation = sqrt((32 - 18)^2/12 = 4.04
b)
For Uniform Distribution,
P(X <= x) = (x - a)/(b - a)
P(X <= 28) = (28 - 18)/(32 - 18)
P(X <= 28) = 0.7143
c)
For Uniform Distribution,
P(x1 <= X <= x2) = (x2 - x1)/(b - a)
P(19 <= X <= 23) = (23 - 19)/(32 - 18)
P(19 <= X <= 23) = 0.2857
d)
Mean = (a + b)/2
Mean = (18 + 32)/2 = 25
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