# Normal distribution, find mean and SD

1. Mar 14, 2013

### Phox

1. The problem statement, all variables and given/known data
Suppose X is a normally distributed random variable. Suppose also that P ( X > 44.7 ) = 0.33 and P ( X < 46 ) = 0.7123. What is the mean and standard deviation of X ?

2. Relevant equations

3. The attempt at a solution
P(X<44.7) = 1-P(X>44.7) = 1-.33 = .67
P(X<44.7) = P((x-μ)/σ < (44.7-μ)/σ) = .67
P(Z < (44.7-μ)/σ) = .67

P(X<46) = P((x-μ)/σ < (46-μ)/σ) = P(Z < (46-μ)/σ) = .712

Looked up corresponding z-scores:
Z1 = .4
Z2 = .5
It's possible these are wrong

Setting up system of equations.

(44.7-μ)/σ = .4
(46-μ)/σ = .5

Solved using matrix -> rref

μ = 39.5
σ = 13

These answers aren't correct. What am I doing wrong here?

Thanks

2. Mar 14, 2013

### Ray Vickson

The problem is premature rounding. Just because data is given to a small number of significant figures does NOT mean that you should work through the problem with a small number of figures. The accurate values of z1 and z2 are
z1 = .4399131657 and z2 = .5601164657. If you use these you will get
μ = 39.94233436 ≈ 39.9 and σ = 10.81501090 ≈ 10.8 . Note that we round off at the end, after doing all the calculations!

3. Mar 14, 2013

### Phox

Thank you!

I guess I'm not really sure how to use my z-score table correctly. I don't know how you get z1 and z2 to that many decimal places of accuracy

4. Mar 14, 2013

### Ray Vickson

I just used a computer package (Maple in my case). You can use the EXCEL spreadsheet (Solver tool) or the on-line program Wolfram Alpha. Some scientific hand-held calculators give similar accuracy.