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 550 and standard deviation 10. Students from UHV achieve scores which are normally distributed with mean 570 and standard deviation 13. Suppose X1,X2,...,X5

are the scores of 5 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 545 <X1< 560?

b) What is the probability that 545 < Xbar < 560?

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 5 UJJ students get scores of at least 550?

Explanation / Answer

a)  probability that 545 <  X1 < 560 =P(545<X<560)=P((545-550)/10<Z<(560-550)/10)=P(-0.5<Z<1)

=0.8413-0.3085 =0.5328

b)here mean score of 5 students of UJJ =550

and std error of mean =std deviation/(n)1/2 =10/(5)1/2 =4.472

probability that 545 < Xbar < 560=P(545<Xbar<560)=P((545-550)/4.472<Z<(560-550)/4.472)=P(-1.118<Z<2.236)

=0.9873-0.1318 =0.8556

c)standard deviation of the random variable Xbar Ybar =(102/5+132/3)1/2 =8.7369

d)expected value of Tbar =550-570=-20

e)  probability that Xbar > Ybar =P(Tbar>0)=P(Z>(0-(-20))/8.7369)=P(Z>2.2891)=0.0110

f)

probability that one  UJJ student get score more then 550 =P(X>550)=P(Z>(550-550)/10)=P(Z>0)=0.5

hence  probability that all 5 UJJ students get scores of at least 550 =(0.5)5 =0.03125

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