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The time it takes for a statistics professor to mark a single midterm test is normally distributed with a mean of 4.8 minutes and a standard deviation of 2.3 minutes. There are 55 students in the professor's class. What is the probability that he needs more than 5 hours to mark all of the midterm tests?

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thaovtp1407

Answer:

Step-by-step explanation:

SD of mean = 2.3/[tex]\sqrt{55}[/tex] = 0.418

5 hours = 300 minutes

300 minutes/55 students = 5.45 min/student

The problem asks what is the probability he needs more than an average of 5 min/student

P(XBAR < 5.0) = P((XBAR - 4.8) / 0.418) < (5.0 - 4.8) /0.418) = P(Z < 0.478) = 0.8893        

P(XBAR > 5.0) = 1 - 0.8893 = 0.1107

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