Question

The daily revenue at a university snack bar has been recorded for the past five years. Records indicate that the mean daily revenue is $2700 and the standard deviation is $400. Suppose that 100 days are randomly selected. What is the probability that the average daily revenue of the sample is higher than $2600?

Answer 1

Answer: 0.9938

Step-by-step explanation:

Let x be the random variable that represents the daily revenue at a university snack bar.

As per given , we have

$\mu=2700$ , $\sigma=400$ and n= 100

Using formula $z=\dfrac{x-\mu}{\dfrac{\sigma}{\sqrt{n}}}$ ,

z-score for x= 2600

$z=\dfrac{2600-2700}{\dfrac{400}{\sqrt{100}}}=-2.5$

The probability that the average daily revenue of the sample is higher than $2600 :

$P(x>2600)=P(z>-2.5)=P(z$   [P(Z>-z)=P(Z<z)]

$=0.9937903\approx0.9938$

Therefore, the probability that the average daily revenue of the sample is higher than $2600 = 0.9938