Likelihood function and Bayes? explanation required

[slingy]

I'm just having trouble with the last step here and I don't even know what to search for.
I've got this example function:

f(1240 | µ = 1100, σ/√n = 38.73 )
= f( 1240−1100 38.73 | 0,1 )
=f(3.61 | 0,1 )
=0.0006

The part I'm having trouble with is how it's calculated at the end, going from f(3.61 | 0,1 ) to 0.0006
Can someone explain what is done to get the answer of 0.0006 ?
Is there some other information missing?

 

[slingy]

I worked it out. Turns out the whole likelihood function is missing, the above is just setting it up as a z score to make it a bit easier to work out the likelihood.

 

[Harry_the_cat]

Can you please explain what the "f" means?

 

[slingy]

Can you please explain what the "f" means?

It's just to note that it's a function. It's a probability distribution function that represents a likelihood based on a sample mean and variance/SD, used for calculating posterior distributions.

 

[blamocur]

I'm just having trouble with the last step here and I don't even know what to search for.
I've got this example function:

f(1240 | µ = 1100, σ/√n = 38.73 )
= f( 1240−1100 38.73 | 0,1 )
=f(3.61 | 0,1 )
=0.0006

The part I'm having trouble with is how it's calculated at the end, going from f(3.61 | 0,1 ) to 0.0006
Can someone explain what is done to get the answer of 0.0006 ?
Is there some other information missing?

I cannot understand the problem's statement -- can you post the whole text?

 

[BigBeachBanana]

I'm just having trouble with the last step here and I don't even know what to search for.
I've got this example function:

f(1240 | µ = 1100, σ/√n = 38.73 )
= f( 1240−1100 38.73 | 0,1 )
=f(3.61 | 0,1 )
=0.0006

The part I'm having trouble with is how it's calculated at the end, going from f(3.61 | 0,1 ) to 0.0006
Can someone explain what is done to get the answer of 0.0006 ?
Is there some other information missing?

My interpretation:
f(1240 | µ = 1100, σ/√n = 38.73 ) means the Probability of a raw score less than or equal to 1240 with a mean of 1100 and a standard error of 38.73.
[math]\Pr(X \le 1240 |\mu =1100, \sigma=38.73)[/math]The next line standardizes the raw score into standard normal.
f( 1240−1100 38.73 | 0,1 ) -> [math]\Pr\left(\frac{X-\mu}{\sigma}\le\frac{1240-1100}{38.73}\right)=\Pr(Z\le 3.61)\approx .99985[/math]

 

[slingy]

My interpretation:
f(1240 | µ = 1100, σ/√n = 38.73 ) means the Probability of a raw score less than or equal to 1240 with a mean of 1100 and a standard error of 38.73.
[math]\Pr(X \le 1240 |\mu =1100, \sigma=38.73)[/math]The next line standardizes the raw score into standard normal.
f( 1240−1100 38.73 | 0,1 ) -> [math]\Pr\left(\frac{X-\mu}{\sigma}\le\frac{1240-1100}{38.73}\right)=\Pr(Z\le 3.61)\approx .99985[/math]

That's exactly what is going here.
In the problem statement the 1240 is a hypothesis, the 1100 is the mean of some sample data and 38.73 is the standard deviation of the sample data.

Then it's used in the limitation formula which is hard to type here from my phone but I'll try:

1/sqrt(2*pi*σ²) exp -( (X-µ)²/(2σ²))

Ends up as

1/sqrt(2*pi) * e^3.612/-2

=0.0006