Suppose that on the CSSAT (Computer Science Achievement Test) students from UJJ

Question

Suppose that on the CSSAT (Computer Science Achievement Test) students from UJJ achieve scores which are normally distributed with mean 540 and standard deviation 9. Students from UHV achieve scores which are normally distributed with mean 560 and standard deviation 13. Suppose X1,X2,...,X4 are the scores of 4 randomly selected UJJ students. Let Xbar be the mean of these scores and let Xsum be the sum of these scores. Suppose Y1,Y2,Y3 are the scores of 3 randomly selected UHV students. Let Ybar be the mean of these scores and let Ysum be the sum of these scores.Then

a) What is the probability that 535 < X1< 550?  

b) What is the probability that 535 < Xbar < 550?  

c) What is the standard deviation of the random variable Xbar Ybar?  

d) Suppose Tbar = Xbar Ybar. What is the expected value of Tbar?  

e) What is the probability that Xbar > Ybar?  

f) What is the probability that all 4 UJJ students get scores of at least 540?

Explanation / Answer

Here if random score of a UJJ student is X where X ~ Normal (540,9)

and if random score of a UHV student is Y where Y ~ Normal (560,13)

(a) Pr (535 < X1 < 550) = Pr(X1 < 550) - Pr(X1 < 535) = (Z2) - (Z1)

where is normal standard cumulative standard distribution

Z2 = (550 - 540)/ 9 = 1.1111

Z1 = (535 - 540)/ 9 = -0.5555

Pr (535 < X1 < 550) = Pr(X1 < 550) - Pr(X1 < 535) = (Z2) - (Z1) =  (1.1111) - (-0.5555)

= 0.8667 - 0.2893 = 0.5774

(b) Here E(xbar)  = 540

standard deviation of mean where sample size is 4 = s/ sqrt(n) = 9/ sqrt(4) = 9/2 = 4.5

Pr (535 < x< 550) = Pr(X1 < 550) - Pr(X1 < 535) = (Z2) - (Z1)

where is normal standard cumulative standard distribution

Z2 = (550 - 540)/ 4.5 = 2.2222

Z1 = (535 - 540)/4.5 = -1.1111

Pr (535 < X1 < 550) = Pr(X1 < 550) - Pr(X1 < 535) = (Z2) - (Z1) =  (2.2222) - (-1.1111)

= 0.9867 - 0.1333 = 0.8534

(c) Here standard deviation of variable Xbar - Ybar

standard deviation se0 = sqrt [sx2 /n1 + sy2 /n2 ]= sqrt[92/4 + 132 /3] = 8.7512

(d) Expected value of Tbar = Xbar Ybar = 540 - 560 = -20

(e) Pr(Xbar > Ybar ) = Pr(Xbar - Ybar >0) = NORM (Tbar > 0 ; -20 ; 8.7512)

Z = (0 + 20) / 8.7512 = 2.285

so Pr(Xbar > Ybar ) = Pr(Z > 2.285) = 0.0111

(f) Here Pr(X1,X2,X3,X4 > 540) = [Pr(Xi > 540)] 4

Pr(Xi > 540) = 1 - Pr(Xi < 540) = 1- 0.5 = 0.5

Pr(X1,X2,X3,X4 > 540) = [Pr(Xi > 540)] 4 = 0.54 = 0.0625

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