Solve.
1 answer:
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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
,
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