Normal distribution, find mean and SD

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SUMMARY

The discussion focuses on calculating the mean (μ) and standard deviation (σ) of a normally distributed random variable X, given specific probabilities: P(X > 44.7) = 0.33 and P(X < 46) = 0.7123. The initial calculations yielded μ = 39.5 and σ = 13, which were incorrect due to premature rounding of z-scores. The accurate z-scores are z1 = 0.4399131657 and z2 = 0.5601164657, leading to the correct values of μ = 39.9 and σ = 10.8. The use of precise values throughout calculations is essential for accuracy.

PREREQUISITES
  • Understanding of normal distribution and z-scores
  • Familiarity with statistical equations for mean and standard deviation
  • Experience with matrix operations and reduced row echelon form (rref)
  • Knowledge of statistical software tools like Maple or Excel Solver
NEXT STEPS
  • Learn how to accurately use z-score tables for precise calculations
  • Explore statistical software options like Maple and Wolfram Alpha for advanced calculations
  • Study the implications of rounding in statistical computations
  • Practice solving problems involving normal distribution using different data sets
USEFUL FOR

Students in statistics, data analysts, and anyone involved in probability theory who seeks to improve their understanding of normal distribution calculations and the importance of precision in statistical analysis.

Phox
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Homework Statement


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 ?


Homework Equations





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
 
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Phox said:

Homework Statement


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 ?


Homework Equations





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

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!
 
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
 
Phox said:
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

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.
 

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