Answer:
0.9792
Step-by-step explanation:
Data provided in the question:
Average gross sales = $1,240
Standard deviation = $180
sample size = 40
Now,
standard deviation of sample average
= [tex]\frac{\textup{Standard deviation}}{\sqrt{n}}[/tex]
= [tex]\frac{180}{\sqrt{40}}[/tex]
= 28.46
Now,
z value for 1200 = [tex]\frac{(1200-1240)}{28.46}[/tex] = -1.4,
and,
p value for (z = -1.4) = 0.0808
therefore,
P(average < $1200) = 0.0808
Thus,
probability that the average over the next 40 weekdays will exceed $1,200
= 1 - 0.808
= 0.9792