Mixing
Regression and percentile
Clash Royale CLAN TAG#URR8PPP
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On a midterm exam, the average is $50$ points (out of $100$) with an SD of $10$ points. On the final exam, the average was $100$ points (out of $200$) with an SD of 30 points. The correlation between the two is 0.6.
What percent of the people who got a 40 on the midterm improved their percentile on the final?
The solution given in this case is 31% but it doesn't seem right to me .
I started the question by first standardize the score of 40 on midterm:
$frac40-5010=-1$
Thus the same percentile of score on final would be $-1*30+100=70$
I then solve for the new average for the final score of the people who got 40 in the midterm, I get: $-1*0.6*30+100=82$
Finally I solve for the z score of the people who were 1 standard deviation below the new average and use normal approximation to get the percentage of people who improved on their final grade:
$frac70-82sqrt1-0.6^2*30 = -0.5$
-0.5 corresponds to 30.85% under the curve, we are interested in people who improved in their final score therefore we get: 100%-30.85%=69.15%
Is my solution correct in this case??
probability regression
edited 2 days ago
Key Flex
3,663422
asked Aug 4 at 3:29
pino231
3268
add a comment |Â
up vote
1
down vote
favorite
On a midterm exam, the average is $50$ points (out of $100$) with an SD of $10$ points. On the final exam, the average was $100$ points (out of $200$) with an SD of 30 points. The correlation between the two is 0.6.
What percent of the people who got a 40 on the midterm improved their percentile on the final?
The solution given in this case is 31% but it doesn't seem right to me .
I started the question by first standardize the score of 40 on midterm:
$frac40-5010=-1$
Thus the same percentile of score on final would be $-1*30+100=70$
I then solve for the new average for the final score of the people who got 40 in the midterm, I get: $-1*0.6*30+100=82$
Finally I solve for the z score of the people who were 1 standard deviation below the new average and use normal approximation to get the percentage of people who improved on their final grade:
$frac70-82sqrt1-0.6^2*30 = -0.5$
-0.5 corresponds to 30.85% under the curve, we are interested in people who improved in their final score therefore we get: 100%-30.85%=69.15%
Is my solution correct in this case??
probability regression
edited 2 days ago
Key Flex
3,663422
asked Aug 4 at 3:29
pino231
3268
add a comment |Â
up vote
1
down vote
favorite
up vote
1
down vote
favorite
On a midterm exam, the average is $50$ points (out of $100$) with an SD of $10$ points. On the final exam, the average was $100$ points (out of $200$) with an SD of 30 points. The correlation between the two is 0.6.
What percent of the people who got a 40 on the midterm improved their percentile on the final?
The solution given in this case is 31% but it doesn't seem right to me .
I started the question by first standardize the score of 40 on midterm:
$frac40-5010=-1$
Thus the same percentile of score on final would be $-1*30+100=70$
I then solve for the new average for the final score of the people who got 40 in the midterm, I get: $-1*0.6*30+100=82$
Finally I solve for the z score of the people who were 1 standard deviation below the new average and use normal approximation to get the percentage of people who improved on their final grade:
$frac70-82sqrt1-0.6^2*30 = -0.5$
-0.5 corresponds to 30.85% under the curve, we are interested in people who improved in their final score therefore we get: 100%-30.85%=69.15%
Is my solution correct in this case??
probability regression
edited 2 days ago
Key Flex
3,663422
asked Aug 4 at 3:29
pino231
3268
On a midterm exam, the average is $50$ points (out of $100$) with an SD of $10$ points. On the final exam, the average was $100$ points (out of $200$) with an SD of 30 points. The correlation between the two is 0.6.
What percent of the people who got a 40 on the midterm improved their percentile on the final?
The solution given in this case is 31% but it doesn't seem right to me .
I started the question by first standardize the score of 40 on midterm:
$frac40-5010=-1$
Thus the same percentile of score on final would be $-1*30+100=70$
I then solve for the new average for the final score of the people who got 40 in the midterm, I get: $-1*0.6*30+100=82$
Finally I solve for the z score of the people who were 1 standard deviation below the new average and use normal approximation to get the percentage of people who improved on their final grade:
$frac70-82sqrt1-0.6^2*30 = -0.5$
-0.5 corresponds to 30.85% under the curve, we are interested in people who improved in their final score therefore we get: 100%-30.85%=69.15%
Is my solution correct in this case??
probability regression
edited 2 days ago
Key Flex
3,663422
asked Aug 4 at 3:29
pino231
3268
edited 2 days ago
Key Flex
3,663422
edited 2 days ago
Key Flex
3,663422
edited 2 days ago
Key Flex
3,663422
3,663422
asked Aug 4 at 3:29
pino231
3268
asked Aug 4 at 3:29
pino231
3268
asked Aug 4 at 3:29
pino231
3268
3268
add a comment |Â
add a comment |Â
1 Answer
1
active
oldest
votes
up vote
0
down vote
accepted
I will solve this according to the least-squares regression line equation.
The linear relation between $2$ variables $x$ & $y$ is given by the equation for the regression line,$$ Y= beta_o+beta_1Χ+e.....(1)$$
Where $Y$ is the dependent variable
$Χ$ is the independent variabl
$beta_o$ is the y-intercept.It is the value of Υ for Χ=0
$beta_1$ is the slope of the line.It gives the amount of change in $Y$ for every unit change in $Χ$.
$e$ is the random error.
The predicted value of $Y$ for a given value of $Χ$ is denoted by $Y^prime$
$$Y^prime=a+bx.....(2)$$where $a= beta_0^prime=$ the estimated $y$-intercept of the regression line
and $b=beta_1^prime=$ the estimated slope of the regression line.
Regression line of $y$ on $x$ is $y-overliney=byx(x-overlinex).....overliney$ is the mean of $y,overlinex$ is the mean of $x.....(3)$
we consider $x$ is the midterm exam score
$y$ is the final exam score
Given midterm mean $=50=overlinex,overliney=100$
and the standard deviation for $x=10$ and $y=30$ $$byx=rtimesdfracsigmasigma x=0.6timesdfrac3010=1.8$$
Plug in these values in $(3)$, we get
$$y-50=1.8(x-50)$$
$$y=1.8x+10$$$beta_0$ is the $y$-intercept $=10$
$beta_1$ is the slope of the line $=1.8$
Now, plug these in $(1)$, we get
If $X=40$ then
$Y^prime(40)=82$
Therefore, the predicted final exam score is $82$
answered 2 days ago
Key Flex
3,663422
should not the adjusted SD for $y$ be $30cdot frac100200=15$?
– farruhota
2 days ago
@farruhota I didn't get where you pointed the mistake, but thanks, I corrected few things where I took wrong mean and SD of final.
– Key Flex
2 days ago
in that case, note $0.6cdot frac3010=1.8$ and $y=1.8x+10$, so $y(40)=82$ just like the OP found.
– farruhota
2 days ago
@farruhota Thanks, for correcting me.
– Key Flex
2 days ago
you are welcome. Originally, when you used $barx=50, bary=50$, I thought you were converting to percentage mark (adjusting to 100), nevermind.
– farruhota
2 days ago
 |Â
show 4 more comments
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
accepted
I will solve this according to the least-squares regression line equation.
The linear relation between $2$ variables $x$ & $y$ is given by the equation for the regression line,$$ Y= beta_o+beta_1Χ+e.....(1)$$
Where $Y$ is the dependent variable
$Χ$ is the independent variabl
$beta_o$ is the y-intercept.It is the value of Υ for Χ=0
$beta_1$ is the slope of the line.It gives the amount of change in $Y$ for every unit change in $Χ$.
$e$ is the random error.
The predicted value of $Y$ for a given value of $Χ$ is denoted by $Y^prime$
$$Y^prime=a+bx.....(2)$$where $a= beta_0^prime=$ the estimated $y$-intercept of the regression line
and $b=beta_1^prime=$ the estimated slope of the regression line.
Regression line of $y$ on $x$ is $y-overliney=byx(x-overlinex).....overliney$ is the mean of $y,overlinex$ is the mean of $x.....(3)$
we consider $x$ is the midterm exam score
$y$ is the final exam score
Given midterm mean $=50=overlinex,overliney=100$
and the standard deviation for $x=10$ and $y=30$ $$byx=rtimesdfracsigmasigma x=0.6timesdfrac3010=1.8$$
Plug in these values in $(3)$, we get
$$y-50=1.8(x-50)$$
$$y=1.8x+10$$$beta_0$ is the $y$-intercept $=10$
$beta_1$ is the slope of the line $=1.8$
Now, plug these in $(1)$, we get
If $X=40$ then
$Y^prime(40)=82$
Therefore, the predicted final exam score is $82$
answered 2 days ago
Key Flex
3,663422
should not the adjusted SD for $y$ be $30cdot frac100200=15$?
– farruhota
2 days ago
@farruhota I didn't get where you pointed the mistake, but thanks, I corrected few things where I took wrong mean and SD of final.
– Key Flex
2 days ago
in that case, note $0.6cdot frac3010=1.8$ and $y=1.8x+10$, so $y(40)=82$ just like the OP found.
– farruhota
2 days ago
@farruhota Thanks, for correcting me.
– Key Flex
2 days ago
you are welcome. Originally, when you used $barx=50, bary=50$, I thought you were converting to percentage mark (adjusting to 100), nevermind.
– farruhota
2 days ago
 |Â
show 4 more comments
up vote
0
down vote
accepted
I will solve this according to the least-squares regression line equation.
The linear relation between $2$ variables $x$ & $y$ is given by the equation for the regression line,$$ Y= beta_o+beta_1Χ+e.....(1)$$
Where $Y$ is the dependent variable
$Χ$ is the independent variabl
$beta_o$ is the y-intercept.It is the value of Υ for Χ=0
$beta_1$ is the slope of the line.It gives the amount of change in $Y$ for every unit change in $Χ$.
$e$ is the random error.
The predicted value of $Y$ for a given value of $Χ$ is denoted by $Y^prime$
$$Y^prime=a+bx.....(2)$$where $a= beta_0^prime=$ the estimated $y$-intercept of the regression line
and $b=beta_1^prime=$ the estimated slope of the regression line.
Regression line of $y$ on $x$ is $y-overliney=byx(x-overlinex).....overliney$ is the mean of $y,overlinex$ is the mean of $x.....(3)$
we consider $x$ is the midterm exam score
$y$ is the final exam score
Given midterm mean $=50=overlinex,overliney=100$
and the standard deviation for $x=10$ and $y=30$ $$byx=rtimesdfracsigmasigma x=0.6timesdfrac3010=1.8$$
Plug in these values in $(3)$, we get
$$y-50=1.8(x-50)$$
$$y=1.8x+10$$$beta_0$ is the $y$-intercept $=10$
$beta_1$ is the slope of the line $=1.8$
Now, plug these in $(1)$, we get
If $X=40$ then
$Y^prime(40)=82$
Therefore, the predicted final exam score is $82$
answered 2 days ago
Key Flex
3,663422
should not the adjusted SD for $y$ be $30cdot frac100200=15$?
– farruhota
2 days ago
@farruhota I didn't get where you pointed the mistake, but thanks, I corrected few things where I took wrong mean and SD of final.
– Key Flex
2 days ago
in that case, note $0.6cdot frac3010=1.8$ and $y=1.8x+10$, so $y(40)=82$ just like the OP found.
– farruhota
2 days ago
@farruhota Thanks, for correcting me.
– Key Flex
2 days ago
you are welcome. Originally, when you used $barx=50, bary=50$, I thought you were converting to percentage mark (adjusting to 100), nevermind.
– farruhota
2 days ago
 |Â
show 4 more comments
up vote
0
down vote
accepted
up vote
0
down vote
accepted
I will solve this according to the least-squares regression line equation.
The linear relation between $2$ variables $x$ & $y$ is given by the equation for the regression line,$$ Y= beta_o+beta_1Χ+e.....(1)$$
Where $Y$ is the dependent variable
$Χ$ is the independent variabl
$beta_o$ is the y-intercept.It is the value of Υ for Χ=0
$beta_1$ is the slope of the line.It gives the amount of change in $Y$ for every unit change in $Χ$.
$e$ is the random error.
The predicted value of $Y$ for a given value of $Χ$ is denoted by $Y^prime$
$$Y^prime=a+bx.....(2)$$where $a= beta_0^prime=$ the estimated $y$-intercept of the regression line
and $b=beta_1^prime=$ the estimated slope of the regression line.
Regression line of $y$ on $x$ is $y-overliney=byx(x-overlinex).....overliney$ is the mean of $y,overlinex$ is the mean of $x.....(3)$
we consider $x$ is the midterm exam score
$y$ is the final exam score
Given midterm mean $=50=overlinex,overliney=100$
and the standard deviation for $x=10$ and $y=30$ $$byx=rtimesdfracsigmasigma x=0.6timesdfrac3010=1.8$$
Plug in these values in $(3)$, we get
$$y-50=1.8(x-50)$$
$$y=1.8x+10$$$beta_0$ is the $y$-intercept $=10$
$beta_1$ is the slope of the line $=1.8$
Now, plug these in $(1)$, we get
If $X=40$ then
$Y^prime(40)=82$
Therefore, the predicted final exam score is $82$
answered 2 days ago
Key Flex
3,663422
I will solve this according to the least-squares regression line equation.
The linear relation between $2$ variables $x$ & $y$ is given by the equation for the regression line,$$ Y= beta_o+beta_1Χ+e.....(1)$$
Where $Y$ is the dependent variable
$Χ$ is the independent variabl
$beta_o$ is the y-intercept.It is the value of Υ for Χ=0
$beta_1$ is the slope of the line.It gives the amount of change in $Y$ for every unit change in $Χ$.
$e$ is the random error.
The predicted value of $Y$ for a given value of $Χ$ is denoted by $Y^prime$
$$Y^prime=a+bx.....(2)$$where $a= beta_0^prime=$ the estimated $y$-intercept of the regression line
and $b=beta_1^prime=$ the estimated slope of the regression line.
Regression line of $y$ on $x$ is $y-overliney=byx(x-overlinex).....overliney$ is the mean of $y,overlinex$ is the mean of $x.....(3)$
we consider $x$ is the midterm exam score
$y$ is the final exam score
Given midterm mean $=50=overlinex,overliney=100$
and the standard deviation for $x=10$ and $y=30$ $$byx=rtimesdfracsigmasigma x=0.6timesdfrac3010=1.8$$
Plug in these values in $(3)$, we get
$$y-50=1.8(x-50)$$
$$y=1.8x+10$$$beta_0$ is the $y$-intercept $=10$
$beta_1$ is the slope of the line $=1.8$
Now, plug these in $(1)$, we get
If $X=40$ then
$Y^prime(40)=82$
Therefore, the predicted final exam score is $82$
answered 2 days ago
Key Flex
3,663422
edited 2 days ago
answered 2 days ago
Key Flex
3,663422
answered 2 days ago
Key Flex
3,663422
answered 2 days ago
Key Flex
3,663422
3,663422
should not the adjusted SD for $y$ be $30cdot frac100200=15$?
– farruhota
2 days ago
@farruhota I didn't get where you pointed the mistake, but thanks, I corrected few things where I took wrong mean and SD of final.
– Key Flex
2 days ago
in that case, note $0.6cdot frac3010=1.8$ and $y=1.8x+10$, so $y(40)=82$ just like the OP found.
– farruhota
2 days ago
@farruhota Thanks, for correcting me.
– Key Flex
2 days ago
you are welcome. Originally, when you used $barx=50, bary=50$, I thought you were converting to percentage mark (adjusting to 100), nevermind.
– farruhota
2 days ago
 |Â
show 4 more comments
should not the adjusted SD for $y$ be $30cdot frac100200=15$?
– farruhota
2 days ago
@farruhota I didn't get where you pointed the mistake, but thanks, I corrected few things where I took wrong mean and SD of final.
– Key Flex
2 days ago
in that case, note $0.6cdot frac3010=1.8$ and $y=1.8x+10$, so $y(40)=82$ just like the OP found.
– farruhota
2 days ago
@farruhota Thanks, for correcting me.
– Key Flex
2 days ago
you are welcome. Originally, when you used $barx=50, bary=50$, I thought you were converting to percentage mark (adjusting to 100), nevermind.
– farruhota
2 days ago
should not the adjusted SD for $y$ be $30cdot frac100200=15$?
– farruhota
2 days ago
should not the adjusted SD for $y$ be $30cdot frac100200=15$?
– farruhota
2 days ago
@farruhota I didn't get where you pointed the mistake, but thanks, I corrected few things where I took wrong mean and SD of final.
– Key Flex
2 days ago
@farruhota I didn't get where you pointed the mistake, but thanks, I corrected few things where I took wrong mean and SD of final.
– Key Flex
2 days ago
in that case, note $0.6cdot frac3010=1.8$ and $y=1.8x+10$, so $y(40)=82$ just like the OP found.
– farruhota
2 days ago
in that case, note $0.6cdot frac3010=1.8$ and $y=1.8x+10$, so $y(40)=82$ just like the OP found.
– farruhota
2 days ago
@farruhota Thanks, for correcting me.
– Key Flex
2 days ago
@farruhota Thanks, for correcting me.
– Key Flex
2 days ago
you are welcome. Originally, when you used $barx=50, bary=50$, I thought you were converting to percentage mark (adjusting to 100), nevermind.
– farruhota
2 days ago
you are welcome. Originally, when you used $barx=50, bary=50$, I thought you were converting to percentage mark (adjusting to 100), nevermind.
– farruhota
2 days ago
 |Â
show 4 more comments
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