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