Question

1. A review of the sales records for Kara’s Carwash indicates that daily revenue can be represented by a normal distribution with mean of $2000 and standard deviation of $400.
   1. What is the probability that tomorrow’s profit will be greater than $2500? (Be sure to indicate the relevant Z-score(s) for the problem).

2. What is the probability that profits fall between $1700 and $2300? (Be sure to indicate the relevant Z-score(s) for the problem).

3. Suppose one day’s revenue is independent from every other day.   What is the probability that that daily revenue exceeds $2400 two days in a row?

Answer 1

μ=2000

σ=400

a)

x=2500

z=(x-μ)/σ

=(2500-2000)/400

=500/400

=1.25

P(x>2500) = P(z>1.25)

= 1- P(z<1.25)

= 1- 0.8944

=0.1056

b)

For x=1700

z=(x-μ)/σ

=(1700-2000)/400

=-300/400

=-0.75

For x=2300

z=(x-μ)/σ

=(2300-2000)/400

=300/400

=0.75

P(1700<x<2300) = P(-0.75 <z<0.75)

= P(z<0.75)-P(z<-0.75)

=0.7734 - 0.2266

=0.5468

c)

For x=2400

z=(x-μ)/σ

=(2400-2000)/400

=400/400

=1

P(x>2400) = P(z>1)

= 1-P(z<1)

= 1- 0.8413

=0.1587

P(daily revenue exceeds $2400 two days in a row) = 0.1587*0.1587

=0.0252