Calls to a customer service center last on average 2.6 minutes with a standard d
ID: 3330594 • Letter: C
Question
Calls to a customer service center last on average 2.6 minutes with a standard deviation of 1.2 minutes. An operator in the call center is required to answer 89 calls each day. Assume the call times are independent.
a. What is the expected total amount of time in minutes the operator will spend on the calls each day? Give an exact answer.
b. What is the standard deviation of the total amount of time in minutes the operator will spend on the calls each day? Give your answer to four decimal places.
c. What is the approximate probability that the total time spent on the calls will be less than 226 minutes? Give your answer to four decimal places. Use the standard deviation as you entered it above to answer this question.
d. What is the value c such that the approximate probability that the total time spent on the calls each day is less than c is 0.97? Give your answer to four decimal places. Use the standard deviation as you entered it above to answer this question.
Explanation / Answer
Mean time of a call = 2.6 minutes
standard deviation of a call = 1.2 minutes
a. What is the expected total amount of time in minutes the operator will spend on the calls each day?
E(T) = E(89X) where X is the time taken by a call and T = total time taken in calls
E(T) = 2.6 * 89 = 231.4 minutes
b. What is the standard deviation of the total amount of time in minutes the operator will spend on the calls each day?
Answer: Var(T) = 89 Var(X) = 89 * 1.22 = 128.16
STD(T) = sqrt(Var(T) = 11.3208 minutes
c. Pr (T < 226 ; 231.4 minutes ; 11.3208 minutes) = ?
Z = (226 - 231.4)/ 11.3208 = - 0.48
from z table
Pr (T < 226 ; 231.4 minutes ; 11.3208 minutes) = 0.3156
(d) Pr ( T < c ; 231.4 minutes ; 11.3208 minutes) = 0.97
Z - value for p - value = 0.97 is from Z table
Z = 1.88
1.88 = (c - 231.4)/ 11.3208
c = 231.4 + 1.88 * 11.3208 = 252.6831 minutes
Related Questions
drjack9650@gmail.com
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.