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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?

Respuesta :

JeanaShupp

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

[tex]\mu=2700[/tex] , [tex]\sigma=400[/tex] and n= 100

Using formula [tex]z=\dfrac{x-\mu}{\dfrac{\sigma}{\sqrt{n}}}[/tex] ,

z-score for x= 2600

[tex]z=\dfrac{2600-2700}{\dfrac{400}{\sqrt{100}}}=-2.5[/tex]

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

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

[tex]=0.9937903\approx0.9938[/tex]

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

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