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 min2. 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

The expected amount of time used for dealing with 35 customers is

= 35 * 40

= 1400 minutes

The variance of the time used for dealing with 35 customers is

= 35 * 49 = 1715 min^2

The time allocated for dealing with customers is = 30 hrs = 1800 minutes

The probability that the allocated dealing time with customers is exceeded is

P(X 1800

$= P( \frac{X - \mu}{\sigma} > \frac{1800 - 1400}{\sqrt{1715}})$

$\approx 0$

Hence, there is almost 0 probability of the dealing time to exceed the allotted value.

But if we want the probability of dealing time to exceed the allocated value to be at most 0.0001, then the z-score for the allocated value will be = 3.49

So, let X be the new allocated time , then

$z = \frac{X - 1400}{\sqrt{1715}} = 3.49$

$X = 1400 + \sqrt{1715} * 3.49$

Hence, the probability will be at most $10^{-4}$ if the allocated time is at least 25.74 hrs per month.

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