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Having trouble with question involving percentiles and symmetry
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A large university will begin a $13$-day period during which students may register
for that semester’s courses. Of those $13$ days, the number of elapsed days
before a randomly selected student registers has a continuous distribution
with density function $f(t)$ that is symmetric about $t = 6.5$ and proportional
to $1 over t+1$ between days $0$ and $6.5.$
A student registers at the $60$th percentile of this distribution. Calculate the
number of elapsed days in the registration period for this student.
The constant of proportionality is $.5over ln7.5$,
So I thought you should do $$int^x_0.5over ln7.51over t+1dt=.6$$
$$.5over ln7.5ln(x+1)-.5over ln7.5ln(0+1)=.6$$
Which comes out to $x+1=e^left(.6ln7.5over.5right)$
However the real solution says you should do
$$int^13-x_0.5over ln7.51over t+1dt=.4$$
$$.5over ln7.5ln(13-x+1)-.5over ln7.5ln(0+1)=.4$$
Which comes out to $14-x=e^left(.4ln7.5over.5right)$ (a different number than my solution.)
If I understand correctly, the rationale behind the correct solution is that since there is a symmetry, the first 4oth percentile is the same as the last, which is the 60th percentile. I just don't understand why my way is not correct.
probability statistics
asked Aug 1 at 17:25
agblt
547
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up vote
0
down vote
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A large university will begin a $13$-day period during which students may register
for that semester’s courses. Of those $13$ days, the number of elapsed days
before a randomly selected student registers has a continuous distribution
with density function $f(t)$ that is symmetric about $t = 6.5$ and proportional
to $1 over t+1$ between days $0$ and $6.5.$
A student registers at the $60$th percentile of this distribution. Calculate the
number of elapsed days in the registration period for this student.
The constant of proportionality is $.5over ln7.5$,
So I thought you should do $$int^x_0.5over ln7.51over t+1dt=.6$$
$$.5over ln7.5ln(x+1)-.5over ln7.5ln(0+1)=.6$$
Which comes out to $x+1=e^left(.6ln7.5over.5right)$
However the real solution says you should do
$$int^13-x_0.5over ln7.51over t+1dt=.4$$
$$.5over ln7.5ln(13-x+1)-.5over ln7.5ln(0+1)=.4$$
Which comes out to $14-x=e^left(.4ln7.5over.5right)$ (a different number than my solution.)
If I understand correctly, the rationale behind the correct solution is that since there is a symmetry, the first 4oth percentile is the same as the last, which is the 60th percentile. I just don't understand why my way is not correct.
probability statistics
asked Aug 1 at 17:25
agblt
547
add a comment |Â
up vote
0
down vote
favorite
up vote
0
down vote
favorite
A large university will begin a $13$-day period during which students may register
for that semester’s courses. Of those $13$ days, the number of elapsed days
before a randomly selected student registers has a continuous distribution
with density function $f(t)$ that is symmetric about $t = 6.5$ and proportional
to $1 over t+1$ between days $0$ and $6.5.$
A student registers at the $60$th percentile of this distribution. Calculate the
number of elapsed days in the registration period for this student.
The constant of proportionality is $.5over ln7.5$,
So I thought you should do $$int^x_0.5over ln7.51over t+1dt=.6$$
$$.5over ln7.5ln(x+1)-.5over ln7.5ln(0+1)=.6$$
Which comes out to $x+1=e^left(.6ln7.5over.5right)$
However the real solution says you should do
$$int^13-x_0.5over ln7.51over t+1dt=.4$$
$$.5over ln7.5ln(13-x+1)-.5over ln7.5ln(0+1)=.4$$
Which comes out to $14-x=e^left(.4ln7.5over.5right)$ (a different number than my solution.)
If I understand correctly, the rationale behind the correct solution is that since there is a symmetry, the first 4oth percentile is the same as the last, which is the 60th percentile. I just don't understand why my way is not correct.
probability statistics
asked Aug 1 at 17:25
agblt
547
A large university will begin a $13$-day period during which students may register
for that semester’s courses. Of those $13$ days, the number of elapsed days
before a randomly selected student registers has a continuous distribution
with density function $f(t)$ that is symmetric about $t = 6.5$ and proportional
to $1 over t+1$ between days $0$ and $6.5.$
A student registers at the $60$th percentile of this distribution. Calculate the
number of elapsed days in the registration period for this student.
The constant of proportionality is $.5over ln7.5$,
So I thought you should do $$int^x_0.5over ln7.51over t+1dt=.6$$
$$.5over ln7.5ln(x+1)-.5over ln7.5ln(0+1)=.6$$
Which comes out to $x+1=e^left(.6ln7.5over.5right)$
However the real solution says you should do
$$int^13-x_0.5over ln7.51over t+1dt=.4$$
$$.5over ln7.5ln(13-x+1)-.5over ln7.5ln(0+1)=.4$$
Which comes out to $14-x=e^left(.4ln7.5over.5right)$ (a different number than my solution.)
If I understand correctly, the rationale behind the correct solution is that since there is a symmetry, the first 4oth percentile is the same as the last, which is the 60th percentile. I just don't understand why my way is not correct.
probability statistics
asked Aug 1 at 17:25
agblt
547
edited Aug 1 at 19:58
asked Aug 1 at 17:25
agblt
547
asked Aug 1 at 17:25
agblt
547
asked Aug 1 at 17:25
agblt
547
547
add a comment |Â
add a comment |Â
1 Answer
1
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1
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The symmetry of the pdf means that after between 6.5 and 13 days the pdf is actually ..
$$.5over ln7.51over (13-t)+1 $$
you could get the correct answer by noting that $x>6.5$
so
$$int^6.5_0.5over ln7.51over t+1dt + int^x_6.5.5over ln7.51over 14-tdt =.6
\ implies int^x_6.5.5over ln7.51over 14-tdt =.1 $$
answered Aug 1 at 18:45
WW1
6,2321712
Can you please explain a little more what you mean in your first sentence? Why is it not $1 over t+1$?
– agblt
Aug 1 at 19:57
1
$frac 1t+1$ is not symmetric about $t=6.5$, after 16.5 days the pdf for that regime is the reflection of $frac 1t+1$ about $t=6.5$ $$p( 6.5+x)=p(6.5-x)implies p(x)=p(13-x)$$
– WW1
Aug 1 at 20:52
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
The symmetry of the pdf means that after between 6.5 and 13 days the pdf is actually ..
$$.5over ln7.51over (13-t)+1 $$
you could get the correct answer by noting that $x>6.5$
so
$$int^6.5_0.5over ln7.51over t+1dt + int^x_6.5.5over ln7.51over 14-tdt =.6
\ implies int^x_6.5.5over ln7.51over 14-tdt =.1 $$
answered Aug 1 at 18:45
WW1
6,2321712
Can you please explain a little more what you mean in your first sentence? Why is it not $1 over t+1$?
– agblt
Aug 1 at 19:57
1
$frac 1t+1$ is not symmetric about $t=6.5$, after 16.5 days the pdf for that regime is the reflection of $frac 1t+1$ about $t=6.5$ $$p( 6.5+x)=p(6.5-x)implies p(x)=p(13-x)$$
– WW1
Aug 1 at 20:52
add a comment |Â
up vote
1
down vote
accepted
The symmetry of the pdf means that after between 6.5 and 13 days the pdf is actually ..
$$.5over ln7.51over (13-t)+1 $$
you could get the correct answer by noting that $x>6.5$
so
$$int^6.5_0.5over ln7.51over t+1dt + int^x_6.5.5over ln7.51over 14-tdt =.6
\ implies int^x_6.5.5over ln7.51over 14-tdt =.1 $$
answered Aug 1 at 18:45
WW1
6,2321712
Can you please explain a little more what you mean in your first sentence? Why is it not $1 over t+1$?
– agblt
Aug 1 at 19:57
1
$frac 1t+1$ is not symmetric about $t=6.5$, after 16.5 days the pdf for that regime is the reflection of $frac 1t+1$ about $t=6.5$ $$p( 6.5+x)=p(6.5-x)implies p(x)=p(13-x)$$
– WW1
Aug 1 at 20:52
add a comment |Â
up vote
1
down vote
accepted
up vote
1
down vote
accepted
The symmetry of the pdf means that after between 6.5 and 13 days the pdf is actually ..
$$.5over ln7.51over (13-t)+1 $$
you could get the correct answer by noting that $x>6.5$
so
$$int^6.5_0.5over ln7.51over t+1dt + int^x_6.5.5over ln7.51over 14-tdt =.6
\ implies int^x_6.5.5over ln7.51over 14-tdt =.1 $$
answered Aug 1 at 18:45
WW1
6,2321712
The symmetry of the pdf means that after between 6.5 and 13 days the pdf is actually ..
$$.5over ln7.51over (13-t)+1 $$
you could get the correct answer by noting that $x>6.5$
so
$$int^6.5_0.5over ln7.51over t+1dt + int^x_6.5.5over ln7.51over 14-tdt =.6
\ implies int^x_6.5.5over ln7.51over 14-tdt =.1 $$
answered Aug 1 at 18:45
WW1
6,2321712
answered Aug 1 at 18:45
WW1
6,2321712
answered Aug 1 at 18:45
WW1
6,2321712
answered Aug 1 at 18:45
WW1
6,2321712
6,2321712
Can you please explain a little more what you mean in your first sentence? Why is it not $1 over t+1$?
– agblt
Aug 1 at 19:57
1
$frac 1t+1$ is not symmetric about $t=6.5$, after 16.5 days the pdf for that regime is the reflection of $frac 1t+1$ about $t=6.5$ $$p( 6.5+x)=p(6.5-x)implies p(x)=p(13-x)$$
– WW1
Aug 1 at 20:52
add a comment |Â
Can you please explain a little more what you mean in your first sentence? Why is it not $1 over t+1$?
– agblt
Aug 1 at 19:57
1
$frac 1t+1$ is not symmetric about $t=6.5$, after 16.5 days the pdf for that regime is the reflection of $frac 1t+1$ about $t=6.5$ $$p( 6.5+x)=p(6.5-x)implies p(x)=p(13-x)$$
– WW1
Aug 1 at 20:52
Can you please explain a little more what you mean in your first sentence? Why is it not $1 over t+1$?
– agblt
Aug 1 at 19:57
Can you please explain a little more what you mean in your first sentence? Why is it not $1 over t+1$?
– agblt
Aug 1 at 19:57
1
1
$frac 1t+1$ is not symmetric about $t=6.5$, after 16.5 days the pdf for that regime is the reflection of $frac 1t+1$ about $t=6.5$ $$p( 6.5+x)=p(6.5-x)implies p(x)=p(13-x)$$
– WW1
Aug 1 at 20:52
$frac 1t+1$ is not symmetric about $t=6.5$, after 16.5 days the pdf for that regime is the reflection of $frac 1t+1$ about $t=6.5$ $$p( 6.5+x)=p(6.5-x)implies p(x)=p(13-x)$$
– WW1
Aug 1 at 20:52
add a comment |Â
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