Calculate max of normal dist given std dev?

[JenniferMurphy]

What is the formula for calculating the maximum value of normal distribution given the std dev?

The tables below generate several normal distributions with different standard deviations. Each generates a different maximum.

The bigger the std dev, the smaller the maximum. This leads me to wonder if the maximum is caluclated so that the total area under the curve equals 1.

Thanks

Book1
	C	D	E	F	G
2	Mean	0	0	0	0
3	Std Dev	0.5	1	2	5
4	Max	0.7979	0.3989	0.1995	0.0798
5	-10	0.00000000	0.000000	0.000001	0.010798
6	-9	0.00000000	0.000000	0.000008	0.015790
7	-8	0.00000000	0.000000	0.000067	0.022184
8	-7	0.00000000	0.000000	0.000436	0.029945
9	-6	0.00000000	0.000000	0.002216	0.038837
10	-5	0.00000000	0.000001	0.008764	0.048394
11	-4	0.00000000	0.000134	0.026995	0.057938
12	-3	0.00000001	0.004432	0.064759	0.066645
13	-2	0.00026766	0.053991	0.120985	0.073654
14	-1	0.10798193	0.241971	0.176033	0.078209
15	0	0.79788456	0.398942	0.199471	0.079788
16	1	0.10798193	0.241971	0.176033	0.078209
17	2	0.00026766	0.053991	0.120985	0.073654
18	3	0.00000001	0.004432	0.064759	0.066645
19	4	0.00000000	0.000134	0.026995	0.057938
20	5	0.00000000	0.000001	0.008764	0.048394
21	6	0.00000000	0.000000	0.002216	0.038837
22	7	0.00000000	0.000000	0.000436	0.029945
23	8	0.00000000	0.000000	0.000067	0.022184
24	9	0.00000000	0.000000	0.000008	0.015790
25	10	0.00000000	0.000000	0.000001	0.010798
Sheet3

Cell Formulas
Range	Formula
D4:G4	D4	=MAX(D$5:D$25)
D5:G25	D5	=NORM.DIST($C5,D$2,D$3,FALSE)
C6:C25	C6	=C5+1

 

[KRice]

Hi Jennifer, yes...you're correct. The area under the standard normal distribution is 1. So you can do something like this:

=NORM.DIST(0,0,2,0)

where the 1st argument is x, the 2nd is the mean, the 3rd is the std dev, and the last says to return the probability distribution rather than the cumulative distribution. By setting the first two arguments equal to each other, you're positioning "x" at the mean and you'll get the maximum of the standard normal distribution for the standard deviation specified by the 3rd argument.

 

[JenniferMurphy]

Hi Jennifer, yes...you're correct. The area under the standard normal distribution is 1. So you can do something like this:

=NORM.DIST(0,0,2,0)

where the 1st argument is x, the 2nd is the mean, the 3rd is the std dev, and the last says to return the probability distribution rather than the cumulative distribution. By setting the first two arguments equal to each other, you're positioning "x" at the mean and you'll get the maximum of the standard normal distribution for the standard deviation specified by the 3rd argument.

Thanks.

I ended up using this formula, but I think it's the same thing.

 

[KRice]

Yes, that is the same thing. You're using the probability density function for the normal distribution, which has an [(x-mu)/sigma] term. And if you set x and mu equal (shown in your latest post as mu-mu), the entire expression reduces to what you've shown. That is equivalent to what I described using the NORM.DIST function with the first two terms set equal to each other.

 

[JenniferMurphy]

Thanks. I guess my expression would be more correct as:

 

[KRice]

I misspoke...in your expression, delete the "e", as that entire exponent goes to 0 (and e^0=1) when mu=x...so you'll have 1/(sigma*2*PI()).

 

[JenniferMurphy]

I misspoke...in your expression, delete the "e", as that entire exponent goes to 0 (and e^0=1) when mu=x...so you'll have 1/(sigma*2*PI()).

Duh, of course you are right. I discovered that when I put it to work and forgot to change the equation.

How about this:

 

[JenniferMurphy]

Kirk,

I hope you will have time to take a look at my latest thread about truncated normal distributions.