Question

Your team has allocated 30 hours per month for dealing with customer support. The time to service a customer is a normally distributed RV with a mean of 40 minutes and a variance of 49 min?. If during a given month there are 35 customers, what is the probability that the allocated customer support time is exceeded, (i.e. the time used by the customers is greater than the time allocated)? How many hours per month would have to be allocated to insure that for 35 customers the likelihood of exceeding the allocated time is less than or equal to 10-4? (40 Marks)

Answer 1

solution;

given that

the time service a customer is a normally distributed RV with a mean of 40min

   variance is 49m^2

during this month customers three or 35

the exceed time is less than or equal to 10^-4

Soak The expected amounts of time used for dealing with 35 customer is HO 35 * 2 1400 min. The variance of the time used for dealing with 35 customcy is 35* 49 3 = 1715 mina The time allocated for dealing with customers :30 hrs is 30XGO = 1800 minutes. the Probability that the allocated dealing is 6x ceeded is Customers with 85 q (x > 1800 ( X 1800 - 1400 51715 : 0 (Z > 9.659

Hence, there is almost o probability of the dealling time to exceed the alloted value But, if we want the probability of the time to Gxceed the allocated value to be at most then the Z-score dealing o-oool 3-49 value will be for the allocated x be let so the new allocated time, 1 then X - 1400 51715 a 3.49 x = 1400 + 1715 * 3.49 X a 1544.53 x = 25.74hrs Hence, the Qrobability will be at most 10th if the allocated time is at least por month 25.74 hrs