Asked by Aryaa
Let X denote the time in hours needed to locate and correct a problem in the software that governs the timing of traffic lights in the downtown area of a large city. Assume that X is normally distributed with mean 10 hours and variance 9.
(a) Find the probability that the next problem will require at most 15 hours to find and correct.
(b) The fastest 5% of repairs take at most how many hours to complete?
How do we do part b only using standard normal table?
(a) Find the probability that the next problem will require at most 15 hours to find and correct.
(b) The fastest 5% of repairs take at most how many hours to complete?
How do we do part b only using standard normal table?
Answers
Answered by
R_Scott
standard deviation is the √ of the variance
the fastest 5% are more than 2 s.d. below the mean
the fastest 5% are more than 2 s.d. below the mean
Answered by
Aryaa
How do we get that : 2 s.d below mean?
Answered by
R_Scott
sorry ... my mistake
... approx. 95% of the population lies within 2 s.d. of the mean
... but we want the bottom 5%
s.d. = √(variance) = √9 = 3
looking at a z-score table
... the bottom 5% is approx. 1.64 s.d. below the mean
1.64 * 3 hr =?
... approx. 95% of the population lies within 2 s.d. of the mean
... but we want the bottom 5%
s.d. = √(variance) = √9 = 3
looking at a z-score table
... the bottom 5% is approx. 1.64 s.d. below the mean
1.64 * 3 hr =?
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