Question

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.)

Answer 1

x	P(X=x)	xP(x)	x2P(x)
4	0.17000	0.68000	2.72000
5	0.38000	1.90000	9.50000
6	0.31000	1.86000	11.16000
7	0.10500	0.73500	5.14500
8	0.03500	0.28000	2.24000
total		5.4550	30.7650
	E(x) =μ=	ΣxP(x) =	5.4550
	E(x2) =	Σx2P(x) =	30.7650
	Var(x)=σ2 =	E(x2)-(E(x))2=	1.0080
	std deviation=	σ= √σ2 =	1.0040
	Coefficient of variation=(σ/µ)*100 =		18.40

a)

average lead time is =5.455

b) coefficient of variation is =18.40 %