Question

2. Using the 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 here, where the random variable x is the number of days. I answered part a. please complete parts b thru c below.

x P(x)

3 .............. 0.14

4 .............. 0.46

5 .............. 0.29

6 ............... 0.075

7 ................ 0.035

a. What is the average lead time for the​ component?

The average lead time is 4.405 days.

b. What is the coefficient of variation for delivery lead​ time?

The coefficient of variation is ____________ %.

​(Round to two decimal places as needed. Do not​ round.)

c. How might the manufacturing firm use this​ information?

Choose one of the following:

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

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

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

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

Answer 1

X	P(X)	X . P(X)	X². P(X)
3	0.14	0.42	1.26
4	0.46	1.84	7.36
5	0.29	1.45	7.25
6	0.075	0.45	2.7
7	0.035	0.245	1.715
Total	1	4.405	20.285
Mean , E(X)=	Ʃ[x.P(X)]		4.405
Variance, V(X) =	Ʃ(X².P(X)) - E(X)²	= 20.285 - 4.4052	0.880975
Standard deviation, SD(X)	SQRT(V(X))	= SQRT(0.880975)	0.9386

a) Average = 4.405 days

b) Coefficient of variation, CV = SD(X) /E(X)

= 0.9386/4.405

= 0.2131 = 21.31%

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