ANSWER
0.2546
EXPLANATION
The movie theatre's daily revenue, X, is normally distributed with a mean of $3152 and a standard deviation of $1281.
We have to find the probability that the theatre generates more than $4000 in revenue on a randomly selected day,
[tex]P(X>4000)[/tex]To find this probability, we have to standardize X using the formula,
[tex]Z=\frac{X-\mu}{\sigma}[/tex]So the probability is,
[tex]P\left(\frac{X-\mu}{\sigma}\gt\frac{4000-3152}{1281}\right)=P(Z\gt0.66)[/tex]Now, we have to look up this z-value in a z-score table. These tables usually show the area to the left of the z-score - this means that they show the probability for a z less than the z-score, so we have to find the complement,
[tex]P(Z\gt0.66)=1-P(Z\lt0.66)[/tex]In a z-score table,
So the probability is,
[tex]P(X>4000)=1-P(Z\lt0.66)=1-0.7454=0.2546[/tex]Hence, the probability that the theatre generates more than $4000 in revenue on a randomly selected day is 0.2546.
