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

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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