Question

V- (6pts) The time needed to complete this ECO382 midterm exam is normally distributed with (uo?) a. If 6.68% of the class can expect to complete the exam within at most 1 hour and 45 minutes, and if 2.28% of the class need more than 2 hours and 20 minutes to complete the same exam, then what is the average time needed to complete this exam? What is the standard deviation? (2pts) b. What is the probability that the time spent by a student to complete the exam is between 130 minutes and 150 minutes? (1pt) Assume that the exam's time allocation is 2hours and 15 minutes. What's the probability that a student will be unable to complete the exam within the allocated time? (1pt) d. What is the probability that the average time spent by four randomly selected students in the class to complete the examination is within the allocated time? (1pt) e. What time is exceeded by the late 4.95% of the students? (1pt) c.

Answer 1

ca). Given that PCxE 1.45he ) =0.0668 P ( x < 105 min) = 0.0668 P(Z < 105-u) 20.0668 04 InverseNorm (0.0668) -1.5 105 - M -1.5 u-1.5o = 105 = 0.0228 = 0.0228 = 0.0228 PCX9.20 P(x> 140 min) P(Z 2140-4) 1- P ( 22 14-) P/22 140-M) 0.0228 1 -0.0228 -0.9772 Inverse Normal (0.9772) = 2.0 140 - u z 2.0 = 140 M + 20 equating 0 1 ② equations: we get us 120 16 P(130 < x < 150 P 130 -120 10 L z c. 150-120 10 PISZ 23) p(233) - P(2312 70.9987 - 0.8413 0.1574 ) cs Scanned with CamScanner

o 2 hr 15 minutes > 135 minutes. PL x 2 135). 1- p (x2135) o.1-1/2 < 135/90) 1 - P(Z <15) 21-0.9332 0.0668 (d) = 4 :P(X< 135) P(Z <_135-120 10854 - P(Z <3) 0.9987 (e) plx>A) - 0.0495 1- PC XZA) 0.0495 PIXCA) 31-010495 / ZLA-120 = 0.9505 0.9505 Inverse Noterual distribution (0.9505) =1.65 A720 e 1.65 10 А 1.65 X10 + 120 A 3 136.5 Thus, by 4.951. students , 136.5 minutes are exceeded. CS Scanned with CamScanner

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