1.Given the joint probability distribution of Y1 and Y2:(adsbygoogle = window.adsbygoogle || []).push({});

f(y1,y2)= (2/5)*(2y1+3y2), 0

0, elsewhere

find

a) f(y1) and f(y2) the marginal distributions

b) Given: E(y1y2) = 1/3 E(y1)=17/30, E(y2)=3/5

then Cov(y1,y2)=?

c) If E(y1^2) = 7/18 and E(y2^2) = 4/9 and E(y1)=17/30 and E(y2)=3/5

Find the correlation coeficcient. Comment on the strength of the correlation coefficient.

r(y1,y2)= Cov(y1,y2) / [Sd(y1) * SD(y2)]

So am I on the right track. Can anyone help me please?

a)

f(y1)=int(f(y1,y2),y2,0,1)

f(y2)=int(f(y1,y2),y1,0,1)

b)

Cov(Y1,Y2)=E(Y1Y2)-E(Y1)E(Y2)

Cov(Y1,Y2)=1/3-(17/30)(3/5)

c)

need SD(Y1) and SD(Y2)

V(Y1)=E(Y1^2)-[E(Y1)]^2

V(Y2)=E(Y2^2)-[E(Y2)]^2

so the correlation is weak.

2.Scores on an exam are assumed to be normally distrubuted with a mean of 78 and variance of 36

a) What is the probability that a person taking the exam scores higher than 75?

b) Suppose the student socring in the top 10% of this distribution are to receive an A grade, what is the minimum score that a student must achieve to earn an A grade?

c) What must be the cut off point for passing the exam if the examiner wants only 30% of all scores to be passing?

d) Approximately, what proportion of the students have scores 5 or more points above the score that cuts off the lowest 25 %?

This is what I did.

All: Mean = 78

All: Variance = 36, then Standard Deviation = 6

a. (75-78)/6 = -1/2 -- That is one-half standard deviation below the mean. I get 80.85%

b. find a score, S, such that (S - 78)/6 = 1.2815516.

S= 1.28*6+78

c)have to Find the 70th percentile of the standard normal distribution and translate to a grade.

d) have to find the first quartile (25th percentile) of the standard normal distribution, translate to a grade, and add 5 points

Any help would be appreciated

tHANKS!

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# Am I on the right track with all of these

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