Suppose that on the CSSAT (Computer Science Achievement Test) students from UJJ
ID: 3358616 • Letter: S
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?
g) Add any comments below.
Explanation / Answer
a)
Mean of X1 = 540 , SD of X1 = 9
P[535 < X1 < 550] = P[X < 550] - P[X < 535]
= P[Z < (550-540)/9] - P[Z < (535-540)/9]
= P[Z < 1.11] - P[Z < -0.55]
= 0.8665 - 0.2912 = 0.5753
b)
Mean of Xbar = 540 , SD of Xbar = 9/sqrt(4) = 4.5
P[535 < Xbar < 550] = P[Xbar < 550] - P[Xbar < 535]
= P[Z < (550-540)/4.5] - P[Z < (535-540)/4.5]
= P[Z < 2.22] - P[Z < -1.11]
= 0.9868 - 0.1335 = 0.8533
c)
SD of Xbar = 4.5
Var of Xbar = 4.52 = 20.25
Var of Ybar = 132 / 3 = 56.33
Var(Xbar Ybar) = Var(Xbar) + Var(Ybar) = 20.25 + 56.33 = 76.58
Standard deviation of (Xbar Ybar) = sqrt(76.58) = 8.75
d)
Xbar = 540, Ybar = 560
E[Xbar Ybar] = E[Xbar] - E[Ybar] = 540 - 560 = -20
So, the expected value of Tbar is -20
e)
Mean of Xbar-Ybar = -20 and SD of Xbar-Ybar = 8.75
P[Xbar > Ybar] = P[Xbar - Ybar > 0]
= P[Z > (0 - (-20) / 8.75)] = P[Z > 2.28]
= 0.0113
f)
P[X > 540] = P[Z > (540-540)/9]
= P[Z > 0] = 0.5
Probability that all 4 UJJ students get scores of at least 540 = 0.54 = 0.0625
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