Normal distribution, absolute values
$begingroup$
Let X be normally distributed with $E(X)=3$, and $Var(X)=4$, calculate $P(|X-3|<K)=0.76$.
I hope I did calculate it correctly? I would be glad if anyone could point out any mistakes... here is how I did it:
$$ P(|X-3| < K)=0.76= \$$
$$P(-(X-3) < K land X-3 < K)= \$$
$$ P(X-3 > -K land X-3 < K)\$$
$$ P(-K/2<Z<K/2)\$$
$$ Phi(K/2)-Phi(-K/2)=0.76\$$
$$2 Phi(K/2)=0.76$$
$$ K=2.35 $$
Many thanks.
probability
edited Dec 7 '18 at 19:42
Lucky
13815
asked Dec 7 '18 at 19:34
LillysLillys
778
$endgroup$
add a comment |
$begingroup$
Let X be normally distributed with $E(X)=3$, and $Var(X)=4$, calculate $P(|X-3|<K)=0.76$.
I hope I did calculate it correctly? I would be glad if anyone could point out any mistakes... here is how I did it:
$$ P(|X-3| < K)=0.76= \$$
$$P(-(X-3) < K land X-3 < K)= \$$
$$ P(X-3 > -K land X-3 < K)\$$
$$ P(-K/2<Z<K/2)\$$
$$ Phi(K/2)-Phi(-K/2)=0.76\$$
$$2 Phi(K/2)=0.76$$
$$ K=2.35 $$
Many thanks.
probability
edited Dec 7 '18 at 19:42
Lucky
13815
asked Dec 7 '18 at 19:34
LillysLillys
778
$endgroup$
add a comment |
$begingroup$
Let X be normally distributed with $E(X)=3$, and $Var(X)=4$, calculate $P(|X-3|<K)=0.76$.
I hope I did calculate it correctly? I would be glad if anyone could point out any mistakes... here is how I did it:
$$ P(|X-3| < K)=0.76= \$$
$$P(-(X-3) < K land X-3 < K)= \$$
$$ P(X-3 > -K land X-3 < K)\$$
$$ P(-K/2<Z<K/2)\$$
$$ Phi(K/2)-Phi(-K/2)=0.76\$$
$$2 Phi(K/2)=0.76$$
$$ K=2.35 $$
Many thanks.
probability
edited Dec 7 '18 at 19:42
Lucky
13815
asked Dec 7 '18 at 19:34
LillysLillys
778
$endgroup$
Let X be normally distributed with $E(X)=3$, and $Var(X)=4$, calculate $P(|X-3|<K)=0.76$.
I hope I did calculate it correctly? I would be glad if anyone could point out any mistakes... here is how I did it:
$$ P(|X-3| < K)=0.76= \$$
$$P(-(X-3) < K land X-3 < K)= \$$
$$ P(X-3 > -K land X-3 < K)\$$
$$ P(-K/2<Z<K/2)\$$
$$ Phi(K/2)-Phi(-K/2)=0.76\$$
$$2 Phi(K/2)=0.76$$
$$ K=2.35 $$
Many thanks.
probability
probability
edited Dec 7 '18 at 19:42
Lucky
13815
asked Dec 7 '18 at 19:34
LillysLillys
778
edited Dec 7 '18 at 19:42
Lucky
13815
asked Dec 7 '18 at 19:34
LillysLillys
778
edited Dec 7 '18 at 19:42
Lucky
13815
edited Dec 7 '18 at 19:42
Lucky
13815
edited Dec 7 '18 at 19:42
Lucky
13815
13815
asked Dec 7 '18 at 19:34
LillysLillys
778
asked Dec 7 '18 at 19:34
LillysLillys
778
asked Dec 7 '18 at 19:34
LillysLillys
778
778
add a comment |
add a comment |
2 Answers
2
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oldest
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$begingroup$
Since $Phi(-y)=1-Phi(y)$ you should actually get $2Phi(K/2)=1.76$, but you otherwise have the right strategy.
answered Dec 7 '18 at 19:47
J.G.J.G.
25.2k22539
$endgroup$
add a comment |
$begingroup$
Your work up to $Phi(K/2) - Phi(-K/2) = 0.76$ is good.
However, you made a mistake afterward, as $-Phi(-K/2) ne Phi(K/2)$. You may be thinking of $Phi(-K/2) = 1 - Phi(K/2)$.
answered Dec 7 '18 at 19:47
angryavianangryavian
40.6k23380
$endgroup$
$begingroup$
Thank you so much! I think I get it know!
$endgroup$
– Lillys
Dec 7 '18 at 19:53
add a comment |
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2 Answers
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2 Answers
2
active
oldest
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active
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votes
$begingroup$
Since $Phi(-y)=1-Phi(y)$ you should actually get $2Phi(K/2)=1.76$, but you otherwise have the right strategy.
answered Dec 7 '18 at 19:47
J.G.J.G.
25.2k22539
$endgroup$
add a comment |
$begingroup$
Since $Phi(-y)=1-Phi(y)$ you should actually get $2Phi(K/2)=1.76$, but you otherwise have the right strategy.
answered Dec 7 '18 at 19:47
J.G.J.G.
25.2k22539
$endgroup$
add a comment |
$begingroup$
Since $Phi(-y)=1-Phi(y)$ you should actually get $2Phi(K/2)=1.76$, but you otherwise have the right strategy.
answered Dec 7 '18 at 19:47
J.G.J.G.
25.2k22539
$endgroup$
Since $Phi(-y)=1-Phi(y)$ you should actually get $2Phi(K/2)=1.76$, but you otherwise have the right strategy.
answered Dec 7 '18 at 19:47
J.G.J.G.
25.2k22539
answered Dec 7 '18 at 19:47
J.G.J.G.
25.2k22539
answered Dec 7 '18 at 19:47
J.G.J.G.
25.2k22539
answered Dec 7 '18 at 19:47
J.G.J.G.
25.2k22539
25.2k22539
add a comment |
add a comment |
$begingroup$
Your work up to $Phi(K/2) - Phi(-K/2) = 0.76$ is good.
However, you made a mistake afterward, as $-Phi(-K/2) ne Phi(K/2)$. You may be thinking of $Phi(-K/2) = 1 - Phi(K/2)$.
answered Dec 7 '18 at 19:47
angryavianangryavian
40.6k23380
$endgroup$
$begingroup$
Thank you so much! I think I get it know!
$endgroup$
– Lillys
Dec 7 '18 at 19:53
add a comment |
$begingroup$
Your work up to $Phi(K/2) - Phi(-K/2) = 0.76$ is good.
However, you made a mistake afterward, as $-Phi(-K/2) ne Phi(K/2)$. You may be thinking of $Phi(-K/2) = 1 - Phi(K/2)$.
answered Dec 7 '18 at 19:47
angryavianangryavian
40.6k23380
$endgroup$
$begingroup$
Thank you so much! I think I get it know!
$endgroup$
– Lillys
Dec 7 '18 at 19:53
add a comment |
$begingroup$
Your work up to $Phi(K/2) - Phi(-K/2) = 0.76$ is good.
However, you made a mistake afterward, as $-Phi(-K/2) ne Phi(K/2)$. You may be thinking of $Phi(-K/2) = 1 - Phi(K/2)$.
answered Dec 7 '18 at 19:47
angryavianangryavian
40.6k23380
$endgroup$
Your work up to $Phi(K/2) - Phi(-K/2) = 0.76$ is good.
However, you made a mistake afterward, as $-Phi(-K/2) ne Phi(K/2)$. You may be thinking of $Phi(-K/2) = 1 - Phi(K/2)$.
answered Dec 7 '18 at 19:47
angryavianangryavian
40.6k23380
answered Dec 7 '18 at 19:47
angryavianangryavian
40.6k23380
answered Dec 7 '18 at 19:47
angryavianangryavian
40.6k23380
answered Dec 7 '18 at 19:47
angryavianangryavian
40.6k23380
40.6k23380
$begingroup$
Thank you so much! I think I get it know!
$endgroup$
– Lillys
Dec 7 '18 at 19:53
add a comment |
$begingroup$
Thank you so much! I think I get it know!
$endgroup$
– Lillys
Dec 7 '18 at 19:53
$begingroup$
Thank you so much! I think I get it know!
$endgroup$
– Lillys
Dec 7 '18 at 19:53
$begingroup$
Thank you so much! I think I get it know!
$endgroup$
– Lillys
Dec 7 '18 at 19:53
add a comment |
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