Answer:
The probability that the mean daily revenue for the next 30 days will exceed $7500 is 0.0855
Step-by-step explanation:
It has been given that
[tex]\mu=7200\\\\\sigma=1200,n=30,x=7500[/tex]
Now, the formula for z-score of a normal distribution is given by
[tex]z=\frac{x-\mu}{\frac{\sigma}{\sqrt{n}}}[/tex]
Substituting the known values, we get
[tex]z=\frac{7500-7200}{\frac{1200}{\sqrt{30}}}[/tex]
Simplifying, we get
[tex]z=\frac{300}{219.1}\\\\z=1.369238[/tex]
Now, we have to find that the daily revenue for next 30 days will exceed $7500.
Thus, we have to find
P(z>1.369238)
From the normal distribution table, we have
P(z>1.369238)= 0.0855
Therefore, required probability is 0.0855
