# Lognormal Distribution Help

1. Mar 5, 2013

### joemama69

1. The problem statement, all variables and given/known data

Q1) It has been observed that contamination level of water [measured in suitable units] in a given waterbody has lognormal distribution with mean = 23.58 units and variance = 3.85 sq units.

What % of samples would indicate contamination level between 20.65 units and 25.85 units ?

2. Relevant equations

Using the Z table and P((ln(x)-mean)/SD)

3. The attempt at a solution

P((ln(20.65)-23.58)/sqrt(3.85))<=Z<=((ln(20.65)-23.58)/sqrt(3.85)) = phi(-10.36)-phi(-10.47)

Values are way too large... whats going on.

2. Mar 5, 2013

### Ray Vickson

You are using the wrong mean and variance: 25.58 and 3.85 are the mean and variance of X, but you need the mean and variance of Y = ln(X). You will need to solve a set of coupled nonlinear equations to find the mean and variance of Y.

3. Mar 6, 2013

### joemama69

ok i tried these equations before but they didnt seem to pan out...

where u=mean & o=variance of X

U=e^(u+(o^2)/2) = e^(23.58+3.85/2) = 1.1931x10^11

O=(e^(o^2)-1)e^(2u+o^2)=(e^3.85 -1)e^(2(23.58 + 3.85)) = 6.5472x10^23

?????

4. Mar 6, 2013

### haruspex

I think you're driving those equations backwards. The given numbers are U and O, not μ and σ.

5. Mar 6, 2013

### Ray Vickson

As 'haruspex' has noted, you need to find the solution of
$$e^{m + \frac{1}{2}v} = 23.58, \: (e^v -1) e^{2m + v} = 3.85.$$

6. Mar 6, 2013

### joemama69

Ahhh... ok so hers what i got now...

(e^v -1)e^(2m+v) = 3.85

e^(2(m+v))-e^(2m+v) = 3.85

2m + 2v - 2m + v = ln 3.85

v = (ln3.85)/3 = 0.4494 = o

e^(m+v/2) = 23.58

m+v/2 = ln23.58

m = ln23.58 - v/2 = 2.9357 = u

I made a mistake somewhere cuz i got a z value of 7 something.

7. Mar 6, 2013

### haruspex

ln(a-b) is not ln(a)-ln(b).

You have $e^{m + \frac{1}{2}v} = 23.58, \: (e^v -1) e^{2m + v} = 3.85.$
What do you get if you square $e^{m + \frac{1}{2}v}$?

8. Mar 6, 2013

### joemama69

uh huhhhhh... didnt see that

so i squared that and cancel it out and get this...

e^v - 1 = 3.85/23.58^2, v = .006900

e^(2m+v) = 23.58^2... 2m+v = ln(23.58^2) = 6.3208 = 2m+.006900, m = 3.1569

soo my Z values are followed with x values 20.65 & 25.85

(ln(x)-.0069)/3.1569 = phi(1.0280) - phi(0.9569) = .8485-.8315 = 0.017

See any issues???

9. Mar 6, 2013

### Ray Vickson

Yes: read over what you have done and see where you went wrong.

Your m and v are OK, but then you went astray. The answer should be somewhere between 75% and 90% probability, but I won't say the exact figure.

10. Mar 11, 2013

### joemama69

ok it looks like i had the mean and variance flip flopped but still not getting it correct

so using (ln(x)-mean)/variance = (ln(20.65)-3.1569)/.0069 = -18.72??

also in the problem it says the variance is 3.85 sq units, so i have to sqrt that to find the actual variance, plug it into the previous equations, which then works out to a variance of .0035???

11. Mar 11, 2013

### haruspex

(ln(x)-mean)/SD; SD is sqrt of variance.

12. Mar 11, 2013

### joemama69

ahhhhh ok i think i got it now.

got 81.55%... thank you for your help