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

Question

Suppose that on the CSSAT (Computer Science Achievent 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 UJ 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 Ybar? f) What is the probability that all 4 UJJ students get scores of at least 540?

Explanation / Answer

a)P(535<X<550)=P((535-540)/9<Z<(550-540)/9)=P(-0.5556<Z<1.1111)=0.8667-0.2893=0.5775

b)for std error of mean =9/(4)1/2 =4.5

P(535<Xbar<550)=P((535-540)/4.5<Z<(550-540)/4.5)=P(-1.1111<Z<2.2222)=0.9869-0.1333 =0.8536

c)

std deviation=(92/4+132/3)1/2 =8.7512

d) expected value=540-560=-20

e)P(Xbar>Ybar)=P(Z>(0-(-20))/8.7512)=P(Z>2.2854)=0.0111

f)

probability of one UJJ to get score at least 540=P(X>540)=P(Z>(540-540)/9)=P(Z>0)=0.5

hence probability that all 4 UJJJ get score at least 540=(0.5)4 =0.0625

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