Question

This question has two parts.

Part 1)

Using historical records, a manufacturing firm has developed the following probability distribution for the number of days required to get components from its suppliers. The distribution is shown here, where the random variable x is the number of days.

x	P(x)
4	0.17
5	0.38
6	0.31
7	0.105
8	0.035

a. The average lead time is ______ days. (Type the entire number, do not round.)

b. The coefficient of variation is ______ %. (Round to two decimal places as needed. Do not round. Do not include % in your answer.)

Part 2)

Using historical records, a manufacturing firm has developed the following probability distribution for the number of days required to get components from its suppliers. The distribution is shown in the question above, where the random variable x is the number of days.

How might the manufacturing firm use this information? Choose the most correct answer.

a. The manufacturing firm may be able to project the most likely range for the number of days required to get components from its suppliers.

b. The manufacturing firm may be able to project the minimum number of days required to get components from its suppliers.

c. The manufacturing firm may be able to project the exact number of days required to get components from its suppliers.

d. The manufacturing firm may be able to project the maximum number of days required to get components from its suppliers.

Answer 1

Part 1) The computations here are made using the following table:

x	P(x)	xP(x)	x^2P(x)
4	0.17	0.68	2.72
5	0.38	1.9	9.5
6	0.31	1.86	11.16
7	0.105	0.735	5.145
8	0.035	0.28	2.24
	1	5.455	30.765

a) The average lead time here is computed as:

$\mu = \sum xP(X = x) = 4*0.17 + 5*0.38 + ....$

$\mu = 5.455$

Therefore 5.455 days is the required mean value here.

b) The second moment of X is computed here as:

(Using the last column of the above table )

E(X) = x²P(X = 1) = 42 * 2²P(X = z) = 42 *0.17 +52 +0.38+

Therefore, the standard deviation here is computed as;

0 = E(X2) - H2 = 30.765 – 5.4552 = 1.0040

Therefore the coefficient of variation here is obtained as:

$CV = \frac{\sigma}{\mu} = \frac{1.0040}{5.455} = 0.1840$

Therefore 0.1840 is the required coefficient of variation here.

Part 2) The interpretation of mean in the above part is that we are getting the expected number of lead days time that is an average lead time in number of days. This is the most likely value .

We cannot obtain minimum, exact or max lead time here.

Therefore A) is the correct answer here.