Answer to Question #235759 in Statistics and Probability for Ben

Question

a) The average demand on a factory store for a certain electric motor is 8 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% sure of meeting all requirements while waiting for his new supply to arrive?

[6 marks]

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 hold-

ers who spent more than $250.

[4 marks]

[Total: 10 marks]

Answer 1

QUESTION

a) The average demand on a factory store for a certain electric motor is 8 per week. When the store man places an order for these motors, delivery takes one week. If the demand for motors has a Poisson distribution, how low can the store man allow his stock to fall before ordering a new supply if he wants to be at least 95% sure of meeting all requirements while waiting for his new supply to arrive?

[6 marks]

SOLUTION

The average demand for the electric motor is 8 vehicles weekly.

That mean, x = 8

The demand for the electric motors is then distributed in a poison distribution manner,

Where: $P\left( x = k\right)$ =$\frac{(e^{-\lambda}*\lambda^k)}{(k!)}$

P(x ≤ n) ≥ 0.95

Let us go on to get the value of n from the equation above that represent the least stock at all time before arrival of new stock.

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

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

$\Big (\frac{8^0}{0!}$ + $\frac{8^1}{1!}$ + $\frac{8^2}{2!}$ + ... + $\frac{8^n}{n!} \Big )$ ≥ 283191009

The series display the least value of n = 13, and this meet the 95% sales demand requirement before additional supply reaches the store in about a week. Thus, the store man need to have a minimum of 13 electric motors in the store at all time.

ANSWER: 13 Electric motors

QUESTION

b) A bank has 175 000 credit card holders. During one month the average amount spent by each card holder totaled $192,50 with a standard deviation of $60,20. Assuming a normal distribution, determine the number of card holders who spent more than $250.

[4 marks]

SOLUTION:

If the Mean, $\mu$ = 192.50, and the Standard deviation $\sigma$ = 60.20, therefore:

$P\left(X \gt 250\right)$ = $1- P\left( X ≤ 250\right)$

Let us evaluate the value of P = probability of the number of card holders who spent more than $250.

= 1 − P$\Big ($Z ≤ $\frac{250 - 192.50}{60.20} \Big)$

= 1 − P$\left (Z ≤ 0.9551\right)$

= 1 − 0.8302

P = 0.1698

Use the probability to find the required number of card holders who spent more than $250

= 175000 * 0.1698 = 29715

ANSWER = 29715