Probability of score in z distribution

Reference no: EM13106515

Please provide a very detailed explanation.

The length of time to do a complete full service oil change at Speedy-Lube is normally distributed with a mean of 15.8 minutes and a standard deviation of 2.2 minutes

X P(X < x) P(X > x) Mean Std dev

11 .0146 .9854 15.8 2.2

15 .3581 .6419 15.8 2.2

21 .9910 .0090 15.8 2.2

24 .9999 .0001 15.8 2.2

p(lower) p(upper)

(A) Analyze the output above to determine what percentage of complete full service oil changes will fall between 21 and 24 minutes.

(B) What percentage of complete full service oil changes will take more than 21? If 1000 cars had a complete full service oil change, how many would you expect to be finished in more than 21 minutes?

Reference no: EM13106515

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