1.Given the joint probability distribution of Y1 and Y2: 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!