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?

Answer 1

Solution:

Given:

Mean = 2090

SD = 150

Part a

P(2000<X<2200) = P(X<2200) – P(X<2000)

Z = (X – mean) / SD

Z = (2200 - 2090)/150

Z = 0.733333

P(Z<0.733333) = 0.768322

(by using z-table)

Z = (2000 - 2090)/150

Z = -0.6

P(Z<-0.6) = 0.274253

(by using z-table)

P(2000<X<2200) = P(X<2200) – P(X<2000)

P(2000<X<2200) = 0.768322 - 0.274253

P(2000<X<2200) = 0.494069

Required probability = 0.494069

Part b

Here, we have to find P(X>2260) = 1 – P(X<2260)

Z = (X – mean) / SD

Z = (2260 - 2090)/150

Z =1.133333

P(Z<1.133333) = 0.871463

(by using z-table)

P(X>2260) = 1 – P(X<2260)

P(X>2260) = 1 – 0.871463

P(X>2260) = 0.128537

Required probability = 0.128537

Required percentage = 12.85%

Part c

X = mean + Z*SD

Z for 90% area = 1.281552

(by using z-table)

X = 2090 + 1.281552*150

X = 2282.233

Required amount = $2,282.23