Let Y denote the time it takes a student to complete a statistics assignment. As

Question

Let Y denote the time it takes a student to complete a statistics assignment. Assume this distribution has variance 625 mins^2. Assuming the distribution is exponential. (a) Calculate the value of the parameter A as well as the mean completion time. (b) Calculate the median and 99^th percentile of homework completion time. (c) If you knew that no student (in a class of size 50) takes more than 2 hours to complete the assignment, would you consider the exponential model a good fit? Properly motivate your answer.

Explanation / Answer

(a) Variance = 625 min2 so sstandard deviation = 25 min

Here the distribution is exponetial so in exponential distribution variance = 1/2

so = 1/25 min-1 and mean completion time = 1/ = 25 min

(b) Median of exponential distribution = 1/ ln(2) = (25) * ln2 = 17.33 min

99% percentile means P( t < t0) = 0.99

so cdf of exponential distribution = 1 - e-t

so that would be equal to 0.99

so 1 - e-t = 0.99

0.01 = e-t

t = ln (100)

t = (1/) ln (100)  = 115.12 minutes

(c) Lets calculate how many people take more than 2 hours out of 50 students

so P(t > 120) = 1- ( 1- e-t ) =e-t = e-(1/25) * 120 = e-4.8 =0.00822

so in the class of 50 students probability of students who will take more than 2 hours = 50 * 0.00822 = 0.411 < 1

so the given model is a good fit for the given condition that no student takes more than 2 hours to complete the assignment.

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