Finding μ and σ from a normal cumulative distribution function

[Saracen Rue]

Homework Statement

The relationship between the expected value and the variance for a particular normal CDF is known to follow the rule $E(X)=arcsin(ln(Var(X)))$. Given that $Pr(0.32<Z<2.32)=Pr(12.9<X<74.275)$, determine the possible values of the mean and the standard deviation correct to 4 decimal places.

Homework Equations

$Z=\frac {X-μ} {σ}$
$Var(X)=σ^2$

The Attempt at a Solution

When expressed in the form $normCDf(lower limit, upper limit, σ, μ)$, $Pr(0.32<Z<2.32)=Pr(12.9<X<74.275)$ becomes $normCDf(0.32, 2.32, 1, 0)=normCDf(12.9, 74.275, σ, μ)$. After also taking into account $E(X)=arcsin(ln(Var(X)))$ and $Var(X)=σ^2$, we get $normCDf(0.32, 2.32, 1, 0)=normCDf(12.9, 74.275, σ, arcsin(ln(σ^2)))$. I attempted to solve this last equation on my calculator, but my calculator simply will not output a numerical answer. I'm not sure what to do from here, can anyone help?

Thank you for your time.

 

[haruspex]

I put all this into a spreadsheet, using the unknown mean as the variable, and played around with it until the probabilities matched to several significant figures. Only took a few minutes.

 

[Saracen Rue]

I put all this into a spreadsheet, using the unknown mean as the variable, and played around with it until the probabilities matched to several significant figures. Only took a few minutes.

I'm not familiar with how to use spreadsheets with probabilities. I doubt I am expected to do that though because we've never used spreadsheets in class.

 

[haruspex]

I'm not familiar with how to use spreadsheets with probabilities. I doubt I am expected to do that though because we've never used spreadsheets in class.

Ok.
If you plug in trial values for mu and compute Pr(12.9<X<74.275) for each on your calculator, it's the same, just a bit slower.
It will be a lot faster if you can pick a good starting point. A promising guess here would be to consider the X points 12.9 and 74.275 as being a direct mapping (by shifting the mean and scaling by variance) of the Z points 0.32 and 2.32. Indeed, it gives mu to within a couple of percent.

 

[Saracen Rue]

Ok.
If you plug in trial values for mu and compute Pr(12.9<X<74.275) for each on your calculator, it's the same, just a bit slower.
It will be a lot faster if you can pick a good starting point. A promising guess here would be to consider the X points 12.9 and 74.275 as being a direct mapping (by shifting the mean and scaling by variance) of the Z points 0.32 and 2.32. Indeed, it gives mu to within a couple of percent.

Ah thank you that helps a lot. I also just worked out if you rearrange $μ=arcsin(ln(σ^2))$ for σ, then you get $σ=e^\frac {sin(μ)} {2}$. Then when I solve $normCDf(0.32,2.32,1,0)=normCDf(12.9,74.275,e^\frac {sin(μ)} {2},μ$ my calculator actually gives me the values for μ; 12.555 and 74.510.

Anyway, thanks for your help :)

 

[haruspex]

Ah thank you that helps a lot. I also just worked out if you rearrange $μ=arcsin(ln(σ^2))$ for σ, then you get $σ=e^\frac {sin(μ)} {2}$. Then when I solve $normCDf(0.32,2.32,1,0)=normCDf(12.9,74.275,e^\frac {sin(μ)} {2},μ$ my calculator actually gives me the values for μ; 12.555 and 74.510.

Anyway, thanks for your help :)

The reason your calculator couldn't handle the arcsin form is that the arcsin function is defined only to return values in the range (−π/2, π/2]. mu here is outside that range. Switching to the sin form solved that.

 

[Saracen Rue]

The reason your calculator couldn't handle the arcsin form is that the arcsin function is defined only to return values in the range (−π/2, π/2]. mu here is outside that range. Switching to the sin form solved that.

Okay that's good to know. I'll remember that for the future. Thanks again for your help :)