Am I on the right track with all of these

[OptimusPrime]

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!

 

[HallsofIvy]

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)

Yes, that's the right definition (assuming you meant that y1 and y2 are between 0 and 1. What did you get when you actually integrated?

b)
Cov(Y1,Y2)=E(Y1Y2)-E(Y1)E(Y2)
Cov(Y1,Y2)=1/3-(17/30)(3/5)

I haven't checked your calculation but that looks good.

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.

OK

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%

When I check a table of "Normal Distribution" values (http://people.hofstra.edu/faculty/Stefan_Waner/RealWorld/normaltable.html is a good one), I get that the area under the curve, from 0 to 0.5 is 0.1915 and so this should be 69.15%, not 80.85%. You seem to have subtracted .1915 from 1 which would be the percentage of scores above 78+3= 81 or below 78.

b. find a score, S, such that (S - 78)/6 = 1.2815516.
S= 1.28*6+78

yes, that's what I get.

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

Yes. Again, 70% corresponds to z= 0.52 approx. What grade, S, is that?

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

The table tells me that the area under the normal curve from 0 to .675 (approx) is 0.25. By symmetry, the area from -infinity to -.675 is 0.25 so the first quartile corresponds to z= -.25. S= 78+ 6(-.25)= 76.5. 5 points above that is 81.5.

Any help would be appreciated
tHANKS!

Having typed all this in, I got an error message saying "The message you have entered is too short. Please lengthen your message to at least 10 characters."! Well, this is more than 10 characters right here, isn't it?