public void Probability()
{
/*
* An average light bulb manufactured by the Acme Corporation lasts 300 days with a standard deviation of 50 days.
* Assuming that bulb life is normally distributed, what is the probability that an Acme light bulb will last at most 365 days?
*
* Solution: Given a mean score of 300 days and a standard deviation of 50 days, we want to find the cumulative probability
* that bulb life is less than or equal to 365 days. Thus, we know the following:
*
* The value of the normal random variable is 365 days.
* The mean is equal to 300 days.
* The standard deviation is equal to 50 days.
*
* We enter these values into the Normal Distribution Calculator and compute the cumulative probability.
* The answer is: P( X < 365) = 0.90. Hence, there is a 90% chance that a light bulb will burn out within 365 days.
*/
NormalDistribution1 distribution = new NormalDistribution1(300, 50);
Assert.AreEqual(0.90, Math.Round(distribution.CDF(365), 2));
/*
* Suppose scores on an IQ test are normally distributed. If the test has a mean of 100 and a standard deviation of 10,
* what is the probability that a person who takes the test will score between 90 and 110?
*
* Solution: Here, we want to know the probability that the test score falls between 90 and 110.
* The "trick" to solving this problem is to realize the following:
*
* P( 90 < X < 110 ) = P( X < 110 ) - P( X < 90 )
*
* We use the Normal Distribution Calculator to compute both probabilities on the right side of the above equation.
*
* To compute P( X < 110 ), we enter the following inputs into the calculator:
* The value of the normal random variable is 110, the mean is 100, and the standard deviation is 10. We find that P( X < 110 ) is 0.84.
*
* To compute P( X < 90 ), we enter the following inputs into the calculator:
* The value of the normal random variable is 90, the mean is 100, and the standard deviation is 10. We find that P( X < 90 ) is 0.16.
*
* We use these findings to compute our final answer as follows:
*
* P( 90 < X < 110 ) = P( X < 110 ) - P( X < 90 )
* P( 90 < X < 110 ) = 0.84 - 0.16
* P( 90 < X < 110 ) = 0.68
*
* Thus, about 68% of the test scores will fall between 90 and 110
*/
distribution = new NormalDistribution1(100, 10);
Assert.AreEqual(0.84, Math.Round(distribution.CDF(110), 2));
Assert.AreEqual(0.16, Math.Round(distribution.CDF(90), 2));
}
public void ZNormalDistribution()
{
NormalDistribution1 distribution = new NormalDistribution1(0.0, 1.0);
Assert.AreEqual(0.0, distribution.Mean);
Assert.AreEqual(1.0, distribution.Variance);
Assert.AreEqual(1.0, distribution.Sigma);
Assert.AreEqual(0.00005, Math.Round(distribution.CDF(-3.9), 5));
Assert.AreEqual(0.00034, Math.Round(distribution.CDF(-3.4), 5));
Assert.AreEqual(0.00187, Math.Round(distribution.CDF(-2.9), 5));
Assert.AreEqual(0.00820, Math.Round(distribution.CDF(-2.4), 5));
Assert.AreEqual(0.02872, Math.Round(distribution.CDF(-1.9), 5));
Assert.AreEqual(0.08076, Math.Round(distribution.CDF(-1.4), 5));
Assert.AreEqual(0.18406, Math.Round(distribution.CDF(-0.9), 5));
Assert.AreEqual(0.34458, Math.Round(distribution.CDF(-0.4), 5));
Assert.AreEqual(0.50000, Math.Round(distribution.CDF(0.0), 5));
Assert.AreEqual(0.65542, Math.Round(distribution.CDF(0.4), 5));
Assert.AreEqual(0.81594, Math.Round(distribution.CDF(0.9), 5));
Assert.AreEqual(0.91924, Math.Round(distribution.CDF(1.4), 5));
Assert.AreEqual(0.97128, Math.Round(distribution.CDF(1.9), 5));
Assert.AreEqual(0.99180, Math.Round(distribution.CDF(2.4), 5));
Assert.AreEqual(0.99813, Math.Round(distribution.CDF(2.9), 5));
Assert.AreEqual(0.99966, Math.Round(distribution.CDF(3.4), 5));
Assert.AreEqual(0.99995, Math.Round(distribution.CDF(3.9), 5));
}