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

Complete solution for the problem can be found in the images below with all the necessary steps.
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Service ith Z N(40 7 IS in let X; = time to customer id then x; ~ 49) , 1z1,2,0,35 (X minutes) : Total time to service = x= x + x + -+X35 =) X ~ (40635 49.35 quired Probability = P( 8 x > 30 hrs P(x > 30x60 minutes) = Pl 1800 Re = x > Moj Ka N (152 35, 1943.) 5 (1400, 1715) 7 tet X-1400 ~ N(0, 1) 11 715 Z = X-1400 ~ N(0) J1 715

Now 2 P_X) і Роо ) PT Х – 14оо 2 1 300 - 400 (115 ( 115 ноо )

=P( 2 7 9.6589 1 - 0 (9-6589 3) = cof of N(0,1) dista 1. P(x > 30 hrs) = zalo -22

to to find let so that amount amount of of his be DC. Then we have re Such that. -4 10 P ( x 7 x hars) x > Gon mins PL <lo 4 2 -4 nfc < lo for-140 1715

om Now foro Normal Statistical sof tware functions, we can find z such Z such that -4 Plz 7z > Z) = lo O

This such a Z comes out to be za 3.3527 < P (272 Z Now, P (2) Gon - 140 125

sol Gon-1400 Z. T715 o a 7, Z 1715 + 1400 60 ! r 71 25.647 hrs. This the a which smallest for the likelihood of the the service time you the x lors 35 customers to exceed 0 งา equal to to 4 time less than is is 25.647 was for a month