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^2. 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? I

Answer 1

We know that sum of iid normally distributed random follows normal distribution with mean 
nu
and variance $n\sigma^{2}$ .

Since standard normal curve is symmetric about zero hence we can write P(Z>z) = 0.5 - P(0<Z<z)

ta Given Allocated the hours/m /month X be jondon varaste Honet is sovvice time to e-simple Customer. giver to NC 40 minutes, 49 nini? 1933S. Let a a randem variable Y io. Y x, the t Ya Eri - X35 ECY) = E(IK) = Elli) = 35x 40 ECY) 1400 minutes 23.33 hours. ECY) = ard. 27 (91 V (Y) = V8xi) E VCXT) +35849 V(Y) = 1715 min² 0.4164 hours We know that for Tid random variables Xi following N(4,5%) the sum of variables follows N(nu, no 2) therefor. YO-N(23.33,07476+) Probability of service is greniter thom altococted trowe, Ply30) = P(Y-23-33 10:4764 To 1764 =PfZ+9166)---D-5- PCO2Z29066) 075 -0.Jo IP(279.66) Eo Hene the required probability ý zero. 30-23:33 N

let су time is to be allocated to insure that Is customers the probability of exceedrag the allocated time is less than or equal to let ie PCY>y) < 0.0ool f(-3 1-23.33 y-23-33 > soooool 10:4764 10.474 P(Z > 9-23.33 < On0001 0.69 0-69 ☆ 0.69 015+Pfo<z« 7-23:13 $ 0.000) P(OLZ C 4-23-13) 7 0.4999 Y-23.33 > 6. 69 y = 23.33 + 3.7X0.69. 3,7 y = 25,88 } Hence the time allocated should be 25.88 hours.