Answer :
Answer:
a) 0.6062
b) 0.9505
c) 0.679
Step-by-step explanation:
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 will be
(a) less than 10 minutes?
(b) longer than 5 minutes?
(c) between 8 and 15 minutes?
a) The Z score (z) is given by the equation:
[tex]z=\frac{x-\mu}{\sigma}[/tex],
Where:
μ is the mean = 9.3 minutes,
σ is the standard deviation = 2.6 minutes and x is the raw score
[tex]z=\frac{x-\mu}{\sigma}=\frac{10-9.3}{2.6}=0.27[/tex]
From the z tables, P(X < 10) = P(z < 0.27) = 0.6062 = 60.62%
b) The Z score (z) is given by the equation:
[tex]z=\frac{x-\mu}{\sigma}[/tex],
[tex]z=\frac{x-\mu}{\sigma}=\frac{5-9.3}{2.6}=-1.65[/tex]
From the z tables, P(X > 5) = P(z > -1.65) = 1 - P(z < -1.65) = 1 - 0.0495 = 0.9505 = 95.05%
c) For 8 minutes
[tex]z=\frac{x-\mu}{\sigma}=\frac{8-9.3}{2.6}=-0.5[/tex]
For 15 minutes
[tex]z=\frac{x-\mu}{\sigma}=\frac{15-9.3}{2.6}=2.19[/tex]
From the z tables, P(8< X < 15) = P(-0.5 < z < 2.19) = P(z < 2.19) - P(z< -0.5) = 0.9875 - 0.3085 = 0.679 = 67.9%