Question

The average demand on a factory store for a certain electricmotoris8 per week. When the storeman places an order for these motors, delivery takes one week. If the demand for motors has a Poisson distribution, how low can the storeman allow his stock to fall before ordering a new supply if he wants to be at least 95%sureofmeetingallrequirementswhilewaitingforhisnewsupplytoarrive? [6marks] b) A bank has 175 000 credit card holders. During one month the average amount spent by each card holder totalled $192,50 with a standard deviation of $60,20. Assuming a normal distribution, determine the number of card holders who spent more than $250. 

Answer 1

a) Mean λ=8

$P(X=k) = \frac{λ^k \times e^{-λ}}{k!} \\ P(X≤n)≥0.95$

Find for value of “n”

$\frac{8^0 \times e^{-8}}{0!} + \frac{8^1 \times e^{-8}}{1!}+\frac{8^2 \times e^{-8}}{2!}+...+\frac{8^n \times e^{-8}}{n!}≥0.95 \\ e^{-8}(\frac{8^0}{0!} + \frac{8^1}{1!}+...+\frac{8^n}{n!} ) ≥0.95 \\ \frac{8^0}{0!} + \frac{8^1}{1!}+...+\frac{8^n}{n!} ≥2831.91009$

From the above series, minimum n=13 satisfied to meet 95% sure of meeting all requirements while waiting for his new supply to arrive. Therefore, storeman can allow minimum n=13.

b) Mean

$\mu =192.50$

Standard deviation

$\sigma =60.20 \\ P(X>250)= 1- P(X≤250) \\ = 1-P(Z≤ \frac{250-192.50}{60.20}) \\ = 1 -P(Z≤ 0.9551) \\ = 1-0.8302 \\ = 0.1698$

Therefore, Required number of card holders who spent more than $250 $= 175000 \times 0.1698= 29715$

Answer= 29715