find mean and standard deviation of normal distribution from pdf and CDF
$begingroup$
i have following problem,
X follows normal distribution $mathcal{N}(mu,sigma^2)$ with pdf f and cdf F. if $max_x f(x)=0.997356$ and $F(-1)+F(7)=1$. determine the expectation, standard deviation and $P(Xle 0)$.
thinking about it, i believe that expectation is the value of X when $f(x)=0.997356$.
can you please help?
normal-distribution
asked Dec 29 '18 at 13:40
NourNour
254
$endgroup$
add a comment |
$begingroup$
i have following problem,
X follows normal distribution $mathcal{N}(mu,sigma^2)$ with pdf f and cdf F. if $max_x f(x)=0.997356$ and $F(-1)+F(7)=1$. determine the expectation, standard deviation and $P(Xle 0)$.
thinking about it, i believe that expectation is the value of X when $f(x)=0.997356$.
can you please help?
normal-distribution
asked Dec 29 '18 at 13:40
NourNour
254
$endgroup$
add a comment |
$begingroup$
i have following problem,
X follows normal distribution $mathcal{N}(mu,sigma^2)$ with pdf f and cdf F. if $max_x f(x)=0.997356$ and $F(-1)+F(7)=1$. determine the expectation, standard deviation and $P(Xle 0)$.
thinking about it, i believe that expectation is the value of X when $f(x)=0.997356$.
can you please help?
normal-distribution
asked Dec 29 '18 at 13:40
NourNour
254
$endgroup$
i have following problem,
X follows normal distribution $mathcal{N}(mu,sigma^2)$ with pdf f and cdf F. if $max_x f(x)=0.997356$ and $F(-1)+F(7)=1$. determine the expectation, standard deviation and $P(Xle 0)$.
thinking about it, i believe that expectation is the value of X when $f(x)=0.997356$.
can you please help?
normal-distribution
normal-distribution
asked Dec 29 '18 at 13:40
NourNour
254
asked Dec 29 '18 at 13:40
NourNour
254
asked Dec 29 '18 at 13:40
NourNour
254
asked Dec 29 '18 at 13:40
NourNour
254
asked Dec 29 '18 at 13:40
NourNour
254
254
add a comment |
add a comment |
1 Answer
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$begingroup$
The distribution is symmetric wrt to $mu$ leading to $F(mu-x)+F(mu+x)=1$ for every $x$.
Keeping this in mind the equality $F(-1)+F(7)=1$ enables you to find $mu$.
Further $f(x)$ takes $frac1{sigmasqrt{2pi}}$ as maximum enabling you to find $sigma$.
Knowing $mu$ and $sigma$ you know the distribution so can find $P(Xleq0)$.
answered Dec 29 '18 at 14:19
drhabdrhab
103k545136
$endgroup$
$begingroup$
I was able to find mean as 3, but didnt get how to find $sigma$ using $frac1{sigmasqrt{2pi}}$. Would you clarify further?
$endgroup$
– Nour
Dec 30 '18 at 9:32
$begingroup$
Can you solve $sigma$ on base of: $0.997356=frac1{sigmasqrt{2pi}}$?
$endgroup$
– drhab
Dec 30 '18 at 9:36
$begingroup$
$sigma = frac1{0.997356sqrt{2pi}}$ which equals to 0.3999. i am not getting positive response for this answer, anything wrong i did?
$endgroup$
– Nour
Dec 31 '18 at 4:33
$begingroup$
I cannot find a mistake in what you did and have the same outcome: $0.4$
$endgroup$
– drhab
Dec 31 '18 at 8:23
add a comment |
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1 Answer
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1 Answer
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votes
$begingroup$
The distribution is symmetric wrt to $mu$ leading to $F(mu-x)+F(mu+x)=1$ for every $x$.
Keeping this in mind the equality $F(-1)+F(7)=1$ enables you to find $mu$.
Further $f(x)$ takes $frac1{sigmasqrt{2pi}}$ as maximum enabling you to find $sigma$.
Knowing $mu$ and $sigma$ you know the distribution so can find $P(Xleq0)$.
answered Dec 29 '18 at 14:19
drhabdrhab
103k545136
$endgroup$
$begingroup$
I was able to find mean as 3, but didnt get how to find $sigma$ using $frac1{sigmasqrt{2pi}}$. Would you clarify further?
$endgroup$
– Nour
Dec 30 '18 at 9:32
$begingroup$
Can you solve $sigma$ on base of: $0.997356=frac1{sigmasqrt{2pi}}$?
$endgroup$
– drhab
Dec 30 '18 at 9:36
$begingroup$
$sigma = frac1{0.997356sqrt{2pi}}$ which equals to 0.3999. i am not getting positive response for this answer, anything wrong i did?
$endgroup$
– Nour
Dec 31 '18 at 4:33
$begingroup$
I cannot find a mistake in what you did and have the same outcome: $0.4$
$endgroup$
– drhab
Dec 31 '18 at 8:23
add a comment |
$begingroup$
The distribution is symmetric wrt to $mu$ leading to $F(mu-x)+F(mu+x)=1$ for every $x$.
Keeping this in mind the equality $F(-1)+F(7)=1$ enables you to find $mu$.
Further $f(x)$ takes $frac1{sigmasqrt{2pi}}$ as maximum enabling you to find $sigma$.
Knowing $mu$ and $sigma$ you know the distribution so can find $P(Xleq0)$.
answered Dec 29 '18 at 14:19
drhabdrhab
103k545136
$endgroup$
$begingroup$
I was able to find mean as 3, but didnt get how to find $sigma$ using $frac1{sigmasqrt{2pi}}$. Would you clarify further?
$endgroup$
– Nour
Dec 30 '18 at 9:32
$begingroup$
Can you solve $sigma$ on base of: $0.997356=frac1{sigmasqrt{2pi}}$?
$endgroup$
– drhab
Dec 30 '18 at 9:36
$begingroup$
$sigma = frac1{0.997356sqrt{2pi}}$ which equals to 0.3999. i am not getting positive response for this answer, anything wrong i did?
$endgroup$
– Nour
Dec 31 '18 at 4:33
$begingroup$
I cannot find a mistake in what you did and have the same outcome: $0.4$
$endgroup$
– drhab
Dec 31 '18 at 8:23
add a comment |
$begingroup$
The distribution is symmetric wrt to $mu$ leading to $F(mu-x)+F(mu+x)=1$ for every $x$.
Keeping this in mind the equality $F(-1)+F(7)=1$ enables you to find $mu$.
Further $f(x)$ takes $frac1{sigmasqrt{2pi}}$ as maximum enabling you to find $sigma$.
Knowing $mu$ and $sigma$ you know the distribution so can find $P(Xleq0)$.
answered Dec 29 '18 at 14:19
drhabdrhab
103k545136
$endgroup$
The distribution is symmetric wrt to $mu$ leading to $F(mu-x)+F(mu+x)=1$ for every $x$.
Keeping this in mind the equality $F(-1)+F(7)=1$ enables you to find $mu$.
Further $f(x)$ takes $frac1{sigmasqrt{2pi}}$ as maximum enabling you to find $sigma$.
Knowing $mu$ and $sigma$ you know the distribution so can find $P(Xleq0)$.
answered Dec 29 '18 at 14:19
drhabdrhab
103k545136
edited Dec 29 '18 at 14:26
answered Dec 29 '18 at 14:19
drhabdrhab
103k545136
answered Dec 29 '18 at 14:19
drhabdrhab
103k545136
answered Dec 29 '18 at 14:19
drhabdrhab
103k545136
103k545136
$begingroup$
I was able to find mean as 3, but didnt get how to find $sigma$ using $frac1{sigmasqrt{2pi}}$. Would you clarify further?
$endgroup$
– Nour
Dec 30 '18 at 9:32
$begingroup$
Can you solve $sigma$ on base of: $0.997356=frac1{sigmasqrt{2pi}}$?
$endgroup$
– drhab
Dec 30 '18 at 9:36
$begingroup$
$sigma = frac1{0.997356sqrt{2pi}}$ which equals to 0.3999. i am not getting positive response for this answer, anything wrong i did?
$endgroup$
– Nour
Dec 31 '18 at 4:33
$begingroup$
I cannot find a mistake in what you did and have the same outcome: $0.4$
$endgroup$
– drhab
Dec 31 '18 at 8:23
add a comment |
$begingroup$
I was able to find mean as 3, but didnt get how to find $sigma$ using $frac1{sigmasqrt{2pi}}$. Would you clarify further?
$endgroup$
– Nour
Dec 30 '18 at 9:32
$begingroup$
Can you solve $sigma$ on base of: $0.997356=frac1{sigmasqrt{2pi}}$?
$endgroup$
– drhab
Dec 30 '18 at 9:36
$begingroup$
$sigma = frac1{0.997356sqrt{2pi}}$ which equals to 0.3999. i am not getting positive response for this answer, anything wrong i did?
$endgroup$
– Nour
Dec 31 '18 at 4:33
$begingroup$
I cannot find a mistake in what you did and have the same outcome: $0.4$
$endgroup$
– drhab
Dec 31 '18 at 8:23
$begingroup$
I was able to find mean as 3, but didnt get how to find $sigma$ using $frac1{sigmasqrt{2pi}}$. Would you clarify further?
$endgroup$
– Nour
Dec 30 '18 at 9:32
$begingroup$
I was able to find mean as 3, but didnt get how to find $sigma$ using $frac1{sigmasqrt{2pi}}$. Would you clarify further?
$endgroup$
– Nour
Dec 30 '18 at 9:32
$begingroup$
Can you solve $sigma$ on base of: $0.997356=frac1{sigmasqrt{2pi}}$?
$endgroup$
– drhab
Dec 30 '18 at 9:36
$begingroup$
Can you solve $sigma$ on base of: $0.997356=frac1{sigmasqrt{2pi}}$?
$endgroup$
– drhab
Dec 30 '18 at 9:36
$begingroup$
$sigma = frac1{0.997356sqrt{2pi}}$ which equals to 0.3999. i am not getting positive response for this answer, anything wrong i did?
$endgroup$
– Nour
Dec 31 '18 at 4:33
$begingroup$
$sigma = frac1{0.997356sqrt{2pi}}$ which equals to 0.3999. i am not getting positive response for this answer, anything wrong i did?
$endgroup$
– Nour
Dec 31 '18 at 4:33
$begingroup$
I cannot find a mistake in what you did and have the same outcome: $0.4$
$endgroup$
– drhab
Dec 31 '18 at 8:23
$begingroup$
I cannot find a mistake in what you did and have the same outcome: $0.4$
$endgroup$
– drhab
Dec 31 '18 at 8:23
add a comment |
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