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

Suppose, random variable $\tiny X_i\left ( i=1,2,...,35 \right )$ denotes service time (in minutes) taken to deal with i-th customer.

$\tiny \therefore X_i\sim N\left(\mu=40,\sigma^2=49\right)$

We define,

$\tiny Y=\sum_{i=1}^{35} X_i$

$\tiny \therefore E\left (Y \right )=E\left (\sum_{i=1}^{35} X_i \right )=\sum_{i=1}^{35} E\left (X_i \right )=\sum_{i=1}^{35}40=1400$

$\tiny \therefore Var\left (Y \right )=Var\left (\sum_{i=1}^{35} X_i \right )=\sum_{i=1}^{35} Var\left (X_i \right )=\sum_{i=1}^{35}49=1715$

$\tiny \therefore Y\sim N\left(\mu_y=1400,\sigma_y^2=1715\right)$

Probability that allocated customer support time (i.e. 30 hours) is exceeded is given by

$\tiny P\left ( Y>30*60 \right )=P\left ( Y>1800 \right )=P\left ( \frac{Y-\mu_y}{\sigma_y}>\frac{1800-1400}{\sqrt {1715}} \right )$

$\tiny =P\left ( Z>9.658906 \right )\approx 0$   [Using R-code '1-pnorm(9.658906)']

Hence, probability that the allocated customer support time is exceeded is approximately 0.

We know,

$\tiny P\left ( Z>3.719016 \right )=0.0001$

$\tiny \Rightarrow P\left ( \frac{Y-\mu_y}{\sigma_y}>3.719016 \right )=0.0001$

$\tiny \Rightarrow P\left ( \frac{Y-1400}{\sqrt {1715}}>3.719016 \right )=0.0001$

$\tiny \Rightarrow P\left ( Y>1400+3.719016*\sqrt {1715} \right )=0.0001$

$\tiny \therefore P\left ( Y>1554.014 \right )=0.0001$

We have,

$\tiny \frac{1554.014}{60}=25.90023\approx 26$

Hence, 26 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 0.0001.