Homework SourseThe customer service center in a large new york department s
The customer service center in a large new york department store has determined tha the amount of time spent with a customer about a complaint is normally distributed, with a mean of 9.3 minutes and a standard deviation of 2.5 minutes. What is the probability that for a randomly chosen customer with a complaint, the amount of time spent resolving the complaint wil be
(a) less than 10 minutes?
(b) longer than 5 minutes?
(c) between 8 and 15 minutes?
Solution
a)
We first get the z score for the critical value. As z = (x - u) / s, then as
x = critical value = 10
u = mean = 9.3
s = standard deviation = 2.5
Thus,
z = (x - u) / s = 0.28
Thus, using a table/technology, the left tailed area of this is
P(z < 0.28 ) = 0.610261248 [ANSWER]
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b)
We first get the z score for the critical value. As z = (x - u) / s, then as
x = critical value = 5
u = mean = 9.3
s = standard deviation = 2.5
Thus,
z = (x - u) / s = -1.72
Thus, using a table/technology, the right tailed area of this is
P(z > -1.72 ) = 0.957283779 [ANSWER]
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c)
We first get the z score for the two values. As z = (x - u) / s, then as
x1 = lower bound = 8
x2 = upper bound = 15
u = mean = 9.3
s = standard deviation = 2.5
Thus, the two z scores are
z1 = lower z score = (x1 - u)/s = -0.52
z2 = upper z score = (x2 - u) / s = 2.28
Using table/technology, the left tailed areas between these z scores is
P(z < z1) = 0.301531788
P(z < z2) = 0.988696156
Thus, the area between them, by subtracting these areas, is
P(z1 < z < z2) = 0.687164368 [ANSWER]
