Question

Question 2) (40 Marks) 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)

need urgent, no need to write it neat and clean, just send solutions in 10 to 15 minutes..

b0bGr..

Answer 1

The time to service a customer
X
has normal distribution.
X; ~ N(40,72)
. Then,

35 ΣΧ, ο Ν(40 x 35, 35 x 72) = N(1400, 41.41262) =1

The required probability is

35 35 Ρ(ΣΧ > 30 x 60) = P(ΣΧ > 1800) =1 =1 35 ΡΙΣX > 30 x 60 - ΡΙΖ 1800 – 1400 41.4126 =1 35 Ρ (ΣΧ. > 30 x 60 X > 30 x 60 ) = P(Z > 9.6588961) =1 35 Ρ (ΣΧ. > 30 x 60 ) = Φ(-9.6588961) 35 Ρ (ΣΧ > 30 x 60 ) = 0.0000 =1

We need to find such that

35 Φ Ρ (ΣΧ > T) < 0.0001 =1 Τ– 1400 P(Z> <0.0001 41.4126 Τ– 1400 <0.0001 41.4126 Τ – 1400 «Φ-1 (0.0001). 41.4126 Τ– 1400 -3.719016 41.4126 Τ– 1400 > 3.719016 41.4126 T> 1554.0141 T> 1554.0141/60 = 25.9 hours

At least per month has to be allocated.

25.9 hours