Probability over an interval in a Normal Distribution?

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SUMMARY

The discussion centers on calculating the probability of a normally distributed random variable, specifically the developing time of photographic prints, which has a mean of 16.28 seconds and a standard deviation of 0.12 seconds. Participants outline the necessity of using the Cumulative Distribution Function (CDF) to find the area under the curve between 16.00 and 16.50 seconds. They emphasize the need to convert the values to the standard normal distribution and utilize CDF tables or software tools like Excel for calculations, as the integral cannot be solved explicitly.

PREREQUISITES
  • Understanding of Gaussian Distribution and its properties
  • Knowledge of Cumulative Distribution Function (CDF)
  • Familiarity with standard normal distribution conversion
  • Basic proficiency in using statistical software or tools like Excel
NEXT STEPS
  • Learn how to use CDF tables for the standard normal distribution
  • Explore the error function (erf) and its applications in probability
  • Practice calculating probabilities using Excel's built-in functions
  • Study the properties of normal distributions and their applications in real-world scenarios
USEFUL FOR

This discussion is beneficial for students studying statistics, data analysts, and anyone involved in probability theory, particularly those working with normal distributions in practical applications.

adamwitt
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Homework Statement


In a photographic process, the developing time of prints may be looked upon as
a random variable which is normally distributed with a mean of 16.28 seconds
and a standard deviation of 0.12 second. Find the probability that it will take
anywhere from 16.00 to 16.50 seconds to develop one of the prints.


Homework Equations


130a296efe2b12deae42cd9200ab04f2.png


sigma = std dev = 0.12
mu = mean = 16.28

The Attempt at a Solution



I *think* I know what I need to do, just don't know how to do it:

So I have the Guassian Distribution formula (with std dev & mean plugged in) as my Probability Density Function.
I need to find the area under the Cumulative Distribution Function over the interval 16 to 16.50.
Because the integrals in CDFs are evaluated from -inf to a, I need to subtract the (integral of CDF from -inf to 16) from the (integral of CDF from -inf to 16.50), and that will be my answer.

But I don't know how to (A) Get the CDF, (B) Evaluate the CDF integral.

Ive tried reading up on the net but I'm not following the theory, can someone please show me how to do this? Thank you!
 
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Yes, that is entirely correct. However, you cannot solve the CDF integral explicitly or find the CDF, since the integral involved is unsolvable.

What you need to do is find the CDF is tables or using computer programs. Are you following a course in probability? Then you should have been given a table with the CDF of the standard normal distribution. Otherwise, see http://www.pavementinteractive.org/index.php?title=Normal_Distribution where a table has been given at the bottom of the page...

Of course, these tables and many of the tables you know are only of the standard normal distribution (that is: mean=0 and stdev=1). So you'll need to convert your probability to a probability involving the standard normal distribution...
 
adamwitt said:

Homework Statement


In a photographic process, the developing time of prints may be looked upon as
a random variable which is normally distributed with a mean of 16.28 seconds
and a standard deviation of 0.12 second. Find the probability that it will take
anywhere from 16.00 to 16.50 seconds to develop one of the prints.


Homework Equations


130a296efe2b12deae42cd9200ab04f2.png


sigma = std dev = 0.12
mu = mean = 16.28

The Attempt at a Solution



I *think* I know what I need to do, just don't know how to do it:

So I have the Guassian Distribution formula (with std dev & mean plugged in) as my Probability Density Function.
I need to find the area under the Cumulative Distribution Function over the interval 16 to 16.50.
Because the integrals in CDFs are evaluated from -inf to a, I need to subtract the (integral of CDF from -inf to 16) from the (integral of CDF from -inf to 16.50), and that will be my answer.

But I don't know how to (A) Get the CDF, (B) Evaluate the CDF integral.

Ive tried reading up on the net but I'm not following the theory, can someone please show me how to do this? Thank you!

Yes, that's correct. The CDF is the integral:

F(x, \mu, \sigma^2) = \int \limits_{-\infty} ^ x f(t, \mu, \sigma^2)dt = \Phi \left( \frac{x - \mu}{\sigma} \right) = \frac{1}{2} \left( 1 + \mbox{erf} \left( \frac{x - \mu}{\sqrt{2}\sigma} \right) \right)

There should be some table of value at the back of your book of \Phi, or erf function.

Or you can even get it online:
  • http://www.geophysik.uni-muenchen.de/~malservisi/GlobaleGeophysik2/erf_tables.pdf"
  • http://www.math.unb.ca/~knight/utility/NormTble.htm"

Cheers,
 
Last edited by a moderator:
You need access to tables or computer-generated values of the normal cumulative. Besides the sources listed in the other responses, you can find the normal cumulative on some of the better hand-held calculators. It is also a built-in function in EXCEL and other, similar spreadsheets.

R.G. Vickson
 
Ray Vickson said:
You need access to tables or computer-generated values of the normal cumulative. Besides the sources listed in the other responses, you can find the normal cumulative on some of the better hand-held calculators. It is also a built-in function in EXCEL and other, similar spreadsheets.

R.G. Vickson

Another refugee from sci.math, eh? Welcome to PF Ray (I assume that's you).

--Lynn
 

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