Question

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(20) The scores on a university entrance examination are normally distributed with a mean of 400 and a standard deviation of 100. If a student is selected at random, find the probability that the student scored a) above 580 b) between 300 and 550. c) what is the lowest value of the top 5%

Answer 1

The following information has been provided: A = 400, J = 100 We need to compute Pr(x > 580). The corresponding z-value needed to be computed is z_ X - 0 580 - 400 100 - = 1.8 Therefore, we get that 580 - 400 Pr(x > 580) = Pr ( 2 > "100" ) = Pr(Z > 1.8) = 1 – 0.9641 = 0.0359 The following is obtained graphically: Normal Distribution: Pr(x>580)=0.0359 0.0040 0.0035 0.0030 0.0025 0.0020 0.0015 0.0010 0.0005 0.0000 0 50 100 150 200 250 300 350 400 450 500 550 600 650 700 750 800

The following information has been provided: u = 400, 0 = 100 We need to compute Pr(300 < X < 550). The corresponding z-values needed to be computed are: z - X1 - M 300 - 400 -= o -1 100 Z = X2 - H_550 - 400 = 1.5 o 100 Therefore, we get (300 - 400 Pr(300 < X < 550) = Pr ( 100 <Z 550 - 400 "") = Pr(-15Z <1.5) 100 = Pr(Z < 1.5) - Pr(Z < -1) = 0.9332 -0.1587 = 0.7745 The following is obtained graphically: Normal Distribution: Pr(300<x<550)=0.7745 0.0040 0.0035 0.0030 0.0025 0.0020 0.0015 0.0010 0.0005 0.0000 0 50 100 150 200 250 300 350 400 450 500 550 600 650 700 750 800

c) PLZyx)=0.05 P(2**) -0.05 p(222-400) = 1-0.05 PLZ 22-100 = 0.95 y- 400 = 1.645 100 x=400+ 1.645*100 = 400 + 1645 x 564.5

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