Question

A firm has invoices which follows a normal distribution with a mean of $2.090 and a standard deviation of $150.

a. What is the probability that the invoice will be between $2000 and $2200?

b. What percentage of invoices will be over $2260?

c. Above what amount will 90% of the invoices lie? please show details with this part of the question.

Answer 1

Ans:

a)

z(2000)=(2000-2090)/150=-0.6

z(2200)=(2200-2090)/150=0.733

P(-0.60<z<0.733)=P(z<0.7330)-P(z<-0.60)

=0.7683-0.2743=0.4940

b)

z=(2260-2090)/150

z=1.133

P(z>1.133)=0.1285 or 12.85%

c)

P(Z>z)=0.90

P(Z<=z)=1-0.9=0.1

z=normsinv(0.1)=-1.282

x=2090-1.282*150=1897.7