Question

9. A local fast-food restaurant finds that its daily profit has mean µ =50 dollars and standard deviation σ =100 dollars. If an auditor randomly selects 30 days, what is the probability that the average profit of the 30 days is a loss (i.e., average profit is less than 0)? (4pts)

Answer 1

Solution :

Given that,

mean = = 50

standard deviation = = 100

n = 30

= 50

=  / n = 100 / 30=18.2574

P( < 30) = P[( - ) / < (30-50) / 18.2574]

= P(z <-1.10 )

Using z table  

= 0.1357   

probability= 0.1357