Normal Distribution

Weibull

Lognormal

Exponential

Exam

Engineered Software Home Page

then the distribution of y becomes normal as n increases. If the random variables being summed are independent, the mean and variance of y are

When several random variables are averaged, the resulting average tends to normal regardless of the distribution of the variables being averaged. Mathematically, if

then the distribution of  y becomes normal as n increases. If the random variables being averaged have the same mean and variance then the mean of  y is equal to the mean of the individual variables being averaged, and the variance of  y is

where s² is the variance of the individual variables being averaged.

The tendency of sums and averages to become normally distributed as the number of variables being summed or averaged becomes large is known as the Central Limit Theorem or the Theory of Large Numbers. For distributions with little skewness, summing or averaging as few as 3 or 4 variables will result in a normal distribution. For highly skewed distributions, more than 30 variables may have to be summed or averaged to obtain a normal distribution.

The normal probability density function is

where m is the mean and s is the standard deviation.

The normal probability density function is not skewed, and is shown in the figure below.

The density function shown in the figure above is the standard normal probability density function. The standard normal probability density function has a mean of 0 and a standard deviation of 1. The normal probability density function cannot be integrated implicitly.  Because of this, historically, a transformation to the standard normal distribution is made, and the normal cumulative distribution function or reliability function is read from a table.  Of course with computers readily available, transforming a variable and looking-up values in a table is a waste of time.  To obtain the area under the normal probability density function to the left of x (the normal cumulative distribution function) use the expression =Normdist(x,m,s,1) in Microsoft Excel, or Normal(x,m,s) in Lotus 123.  If x is a normal random variable, it can be transformed to standard normal using the expression

Example:  The tensile strength of a metal extrusion is normally distributed with a mean 300 and a standard deviation of 5. What percentage of extrusions have a strength greater than 310?  What percentage of extrusions have a strength less than 295?  What percentage of extrusions have a strength between 295 and 310?

Solution:  The shaded area in the top graph in the figure below represents the probability of an extrusion being greater than 310. The shaded area in the bottom graph   represents the area under the standard normal distribution to the right of z = 2, which is the same as the probability of an extrusion being greater than 310. Transforming to standard normal,

z = (310-300)/5 = 2

=1-NORMDIST(310,300,5,1)

The shaded area in the top graph in the figure below represents the probability of an extrusion being less than 295. The shaded area in the bottom graph in the figure below represents the area under the standard normal distribution to the left of z = -1, which is the same as the probability of an extrusion being less than 295. Transforming to standard normal,

z = (295-300)/5 = -1

From a standard normal table the area to the left of  z = -1 is equal to 0.1587.   Using Microsoft Excel, this value is found from the expression =Normdist(295,300,5,1).  The equivalent expression in Lotus 123 is @Normal(295,300,5).  The probability of the strength being between 295 and 310 is 1-0.0228-0.1587 = 0.8185.  Click Here to download this solution in Microsoft Excel.

Example:  A type of battery is produced with an average voltage of 60 with a standard deviation of 4 volts.  If 9 batteries are selected at random, what is the probability that the total voltage of the 9 batteries is greater than 530?  What is the probability that the average voltage of the 9 batteries is less than 62?

Solution:  The expected total of the voltage of nine batteries is 540. The expected standard deviation of the voltage of the total of nine batteries is

Transforming to standard normal,

From a standard normal table, the area to the left of  z = -0.833 is 0.2024.  The area to the right of  z = -0.833 is 1-0.2024=0.7976.  This problem can also be solved using the expression =1-Normdist(530,540,12,1) in Microsoft Excel or the expression 1-@normal(530,540,12) in Lotus 123.

The probability density function of the voltage of the individual batteries and of the average of nine batteries is shown in the figure below. The distribution of the averages has less variance because the standard deviation of the averages is equal to the standard deviation of the individuals divided by the square root of the sample size.

The probability of the average voltage of 9 batteries being less than 62 is equal to the probability of a standard normal variable being less than  z = 1.5; this is shown in the expression below.

From a standard normal table, the area under the standard normal curve to the left of z = 1.6 is  0.9452, which is the probability of the average voltage of nine batteries being less than 62.  In Microsoft Excel, this value is found from the expression =Normdist(62,60,1.25,1).  In Lotus 123 the solution is given by the expression @Normal(62,60,1.25).  Click Here to download this solution in Microsoft Excel.

The normal hazard function is

where f(x) is the standard normal probability density function and F(x) is the standard normal cumulative distribution function.  The hazard function for the normal distribution is monotonically increasing; this is shown in the figure below.