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?

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

, and n= 100

Using formula ,

z-score for x= 2600

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

  [P(Z>-z)=P(Z<z)]

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

The answer is $5,000.73.
Explanation: If one Canadian dollar bill is $1.00 and a Trinidad’s $1.00 is $2.73 then that means if you wanna convert it you have to double the amount including the €73 and you have 2,500 you need to double the amount then add the remaining €73.