Question

Sales delay is the elapsed time between the manufacture of a product and its sale. According an article it is quite common for investigators to model sales delay using a lognormal distribution. For a particular product, the cited article proposes this distribution with parameter values μ = 2.04 and σ2 = 0.08 (here the unit for delay is months).

(a) What are the variance and standard deviation of delay time? (Round your answers to two decimal places.) variance months2 standard deviation months

(b) What is the probability that delay time exceeds 12 months? (Round your answer to four decimal places.)

(c) What is the probability that delay time is within one standard deviation of its mean value? (Round your answer to four decimal places.)

(d) What is the median of the delay time distribution? (Round your answer to two decimal places.) months

(e) What is the 99th percentile of the delay time distribution? (Round your answer to two decimal places.) months

(f) Among 10 randomly selected such items, how many would you expect to have a delay time exceeding 9 months? (Round your answer to three decimal places.) items

Answer 1

a)

variance=σ2=	e(2ϴ+ω2)*(eω2-1)=	5.34
standard deviation=		2.31

b)

probability =P(X>12)=1-P(X<12)=1-P(Z<(ln(12)-2.04)/0.282842712474619)=1-P(Z<1.57)=0.0582

c)

probability =P(5.694<X<10.315)=P((ln(10.315)-2.04)/0.283)<Z<P((ln(5.694)-2.04)/0.283)=P(-1.06<Z<1.04)=0.8508-0.1446=0.7062

d)

for 50th percentile critical value of z=			0
therefore 50 th percentile =e(ϴ+ωz)=exp(2.04+0.08*0) =			7.69

e)

for 99th percentile critical value of z=			2.33
therefore 99 th percentile =e(ϴ+ωz)=			14.87

f)

probability =P(X>9)=1-P(X<9)=1-P(Z<(ln(9)-2.04)/0.282842712474619)=1-P(Z<0.56)=0.2877

expected to exceed 9 months =10*0.2877 =2.877