Question 715927: The help desk for a software company logs the lengths of times of each call that technicians have with customers. Historical records based on the past 3 years indicate that call length (L) is a normally distributed variable with an average (μ) of 9.00 minutes and a standard deviation (σ) of 0.95 min. For each of the problems below, find the appropriate probability.
A) P(L ≤ 10.22)
B) P(L > 10.74)
C) P(7.67 ≤ L < 11.02)
D) The call times that enclose the central 81% of the distribution are:
Note: Round your answers to two decimal places for this problem.
I really need step by step help. Im not coming up with the correct answers..
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Answer by stanbon(75887)   (Show Source):
You can put this solution on YOUR website! call length (L) is a normally distributed variable with an average (μ) of 9.00 minutes and a standard deviation (σ) of 0.95 min. For each of the problems below, find the appropriate probability.
A) P(L ≤ 10.22)
z(10.22) = (10.22-9)/0.95 = 1.2842
Ans: P(L<= 10.22) = P(z<=1.2842) = normalcdf(-100,1.2842) = 0.9005
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B) P(L > 10.74)
Find the z-value of 10.74.
Find the P(z is above that z-value)
Ans: 0.0335
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C) P(7.67 ≤ L < 11.02)
Find the z-value of those numbers.
Find the P(z is between those z-values)
Ans: 0.9025
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D) The call times that enclose the central 81% of the distribution are:
Note: Round your answers to two decimal places for this problem.
Draw the picture:
You have a left tail of 0.5000- 0.4150 = 0.085
invNorm(0.085) = -1.3722 for the left tail and +1.3722 for the right tail:
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Find the corresponding x-value using x = z*s+u
x = -1.3722*0.95+9 = 7.70
x = +1.3622*0.95+9 = 10.30
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Cheers,
Stan H.
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