Chapter 9 Numerical Integration Numerical Integration Application: Normal Distributions Copyright © The McGraw-Hill Companies, Inc. Permission required

Chapter 9Numerical Integration

Numerical Integration Application:Normal Distributions

Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display.

Review: Numerical Integration

• When evaluating a definite integral, remember that we are simply finding the area under the curve between the lower and upper limits

• This can be done numerically by approximating the area with a series of trapezoids

Engineering Computation: An Introduction Using MATLAB and Excel

Review: Trapezoid Area

• A trapezoidal area is formed:


Other Numerical Integration Methods

• Higher-order functions allow fewer intervals to be used for similar accuracy

• For example, Simpson’s Rule uses 2nd order polynomial approximations to determine the area

• With the speed of modern computers, using these more efficient algorithms is usually not necessary – the trapezoid method with more intervals is sufficient


Application – Normal Distribution

• The standard deviation is a commonly-used measure of the variability of a data population or sample

• A higher standard deviation means that there is greater variability

• If we assume that the population or standard is distributed as a normal distribution, then we can obtain a lot of useful information about probabilities


Normal Distribution

• Many numerical populations can be approximated very closely by a normal distribution

• Examples:– Human heights and weights– Manufacturing tolerances– Measurement errors in scientific experiments

• The population is characterized by its mean and standard deviation


Normal Distribution

• The probability density function of the normal distribution can show a graphical view of how the population is distributed

• The probability density function (pdf) is given by this equation:


Example

• Consider a population with a mean of 10 and a standard deviation of 1. Here is the pdf:


Frequency values approach zero for x-values far away from the mean

Peak occurs at mean value of x (10)

Effect of Standard Deviation

• What if the standard deviation is 1.5 instead of 1? Here is the new pdf, compared with the original:


σ = 1.0

σ = 1.5

Note the “wider” shape, as values are more spread out from the mean when the standard deviation is greater

Notes About the Probability Density Function

• The curve is symmetric about the mean• Although the curve theoretically extends to infinity in

both directions, the function’s values approach zero and can be ignored for x-values far away from the mean

• The area under the curve is equal to one – this corresponds to 100% of the population having x-values greater than - and less than +

• The percentage of the population between other limits can be computed by integrating the function between those limits


The Standard Normal Distribution

• The pdf can be standardized by introducing a new variable Z, the standard normal random variable:

• The Z-value is simply the number of standard deviations from the mean. Negative values indicate that the data point is below the mean, positive values indicate that the data point is above the mean


Examples

• For a population with a mean of 30 and a standard deviation of 2, what is the Z-value for:

1. x = 27Z = -1.5 (1.5 standard deviations below the mean)

2. x = 34Z = +2.0 (2.0 standard deviations above the mean)

3. x = 30Z = 0


Standard Normal Distribution

• Using the Z-values, we can write the equation for the probability density function as:

• This form is especially useful in that the same probability density graph applies to all normal distributions (next slide)


Standard Normal Distribution


Integration of Standard Normal Distribution Function

• As we have stated, the area under the curve from - to + is equal to one

• Often, we want to find the area from - to a specific value of Z

• For example, suppose we want to find the percentage of values that are less than one standard deviation above the mean

• This will be


Integration of Standard Normal Distribution Function

• This integral is known as the cumulative standard normal distribution

• This integral does not have a closed-form solution, and must be evaluated numerically

• There are tables of the results in most statistics books


Typical Table of Integral Values


Z-value to one decimal place

Second decimal place of Z-value

Integral Value for Z = 1.00

• Integral = 0.841• This means that 84.1% of all values will be less

than (mean + 1 standard deviation)


Graphical Interpretation of Answer


In-Class Exercises

• Consider a sample of concrete test specimens with a mean compressive strength of 4.56 ksi and a standard deviation of 0.482 ksi

• Using tables of the cumulative standard normal distribution, find:1. The percentage of tests with strength values less than

3.50 ksi

2. The percentage of tests with strength values greater than 5.0 ksi

3. The strength value that 99.9% of tests will exceed


1. Percentage of Values Less Than 3.5 ksi

• Z = (3.5 – 4.56)/0.482 = -2.199 (round to -2.20)• From table:

• 0.0139, or 1.4% of values are less than 3.5 ksi.


2. Percentage of Values Greater Than 5 ksi

• Z = (5.0 – 4.56)/0.482 = 0.913 (round to 0.91)• From table:

• 0.8186 of the total values are less than 5 ksi, so• 1 - 0.8186 = 0.1814, or 18.1% of values are

greater than 3.5 ksi.


2. Percentage of Values Greater Than 5 ksi

• Note that Z = 0.913. Rather than rounding off, we can get a slightly more accurate answer by linear interpolation of the table values:

• Cumulative density =


3. Strength Value That 99.9% of Tests Will Exceed

• We want to find a value that 0.1% (0.001) of tests are below

• From the table, find the Z-value corresponding to a cumulative density of 0.001

• Z is approximately -3.10 (note that to 4 decimal places, Z = -3.08 and Z = -3.09 have the same value of 0.0010)


3. Strength Value That 99.9% of Tests Will Exceed

• Z = -3.10• So the value that 99.9% of tests will exceed is 3.10

standard deviations below the mean:

• Therefore, 99.9% of tests will exceed a value of 3.07 ksi.

• If we set this value as the lower limit, then an average of one of every 1000 batches of concrete will not meet the specification


Application with MATLAB

• Calculating the integral directly will eliminate the need to use the tables

• Since we cannot begin our integration at -, what value of lower limit should we use?

• How many intervals do we need to use to get acceptable precision for our answers?

• We will experiment with different values for the lower limit and number of intervals


Function normdist

function SUM = normdist(limit, k)

lower = -limit;

upper = limit;

inc = (upper-lower)/k;

SUM = 0;

x(1) = lower;

y(1) = 1/sqrt(2*pi)*exp(-x(1)^2/2);

for i = 2:(k+1)

x(i) = x(i-1)+inc;

y(i) = 1/sqrt(2*pi)*exp(-x(i)^2/2);

SUM = SUM + .5*(y(i) + y(i-1))*(x(i) -x(i-1));

end


Making “limit” an argument of the function allows us to examine the effect of using different values

k = number of increments

Evaluation of Limits, Number of Increments

>> format long

>> normdist(3,100)ans =0.997292229481189

>> normdist(3,1000)ans =0.997300124163755

>> normdist(6,1000)ans =0.999999998025951

>> normdist(6,10000)ans =0.999999998026819

>> normdist(6,100)ans =0.999999997940018

>> normdist(5,100)ans =0.999999414352763


Conclusions

• Limit = 6 (6 standard deviations) gives accuracy to eight decimal places

• Number of increments less important: similar results for 100, 1000, and 10000 increments

• Use lower limit of -6 standard deviations, 1000 increments


Modified Function normdist

function SUM = normdist(Z)

% Calculates the % of population with values less than Z

lower = -6;

upper = Z;

k = 1000;

inc = (upper-lower)/k;

SUM = 0;

x(1) = lower;

y(1) = 1/sqrt(2*pi)*exp(-x(1)^2/2);

for i = 2:(k+1)

x(i) = x(i-1)+inc;

y(i) = 1/sqrt(2*pi)*exp(-x(i)^2/2);

SUM = SUM + .5*(y(i) + y(i-1))*(x(i) -x(i-1));

end


Check

>> normdist(0)ans =0.5000Should be 0.50 since curve is symmetric

>> normdist(1)ans =0.8412Checks with value from table


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Chapter 9 Numerical Integration Numerical Integration Application: Normal Distributions Copyright © The McGraw-Hill Companies, Inc. Permission required