Question

A movie theatre's daily revenue follows a normal distribution, with an average daily revenue of $3,152. The standard deviation for the distribution is $1281. What is the probability that the theatre generates more than $4,000 in revenue on a randomly selected day?

Answer 1

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,

$P(X>4000)$

To find this probability, we have to standardize X using the formula,

$Z=\frac{X-\mu}{\sigma}$

So the probability is,

$P\left(\frac{X-\mu}{\sigma}\gt\frac{4000-3152}{1281}\right)=P(Z\gt0.66)$

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,

$P(Z\gt0.66)=1-P(Z\lt0.66)$

In a z-score table,

So the probability is,

$P(X>4000)=1-P(Z\lt0.66)=1-0.7454=0.2546$

Hence, the probability that the theatre generates more than $4000 in revenue on a randomly selected day is 0.2546.