Carillon Technologies Limited Probability Distributions - Normal

Normal Distribution

1. Characteristics of the normal distribution

Any normal distribution can be expressed in a standard form.(We call this practice normalizing the data.)

Standard form:

Where:

x is the measurement point

mu () is the population mean

sigma () is the population standard deviation

Z is the standard normal deviate

and the probability density function is

 

2. Cumulative Normal Distribution

One Tailed Areas Instead of solving the equations analytically we use Tables of the Cumulative Normal Distribution [F(x)]. The tables provide us: Values that extend from approximately -4 to + 4 standard deviations. For example: What is the area under the normal curve for z < = 0.5
Solution:   The probability of a z less than .5 [P(z<=0.5)] = 0.6915 (from cumulative normal table)

Two Tailed Areas For example: A product line has over time established a process average of 51.0 lb with a process standard deviation of 1.0 lb. The current specification requirements for this process are 49.0 to 52.5 lbs. This process produces what percentage of non-conforming materials?

Solution: Problem Statement We want to determine what percentage of materials will fall outside of the limits. We write this problem as determining the probability of having materials below 49.0 pounds and above 52.5 pounds. P(x<=49.0 and x>= 52.5) Collect Information We know the limits 49.0 pounds and 52.5 pounds (x in the formula) We are given the average = 51.0 pounds and the standard deviation of 1.0 Generate a Solution Calculate the Zupper = (upper limit - the average)/standard deviation Then, determine the probability of having a value of less than the Zupper from the standard normal table. Calculate the Zlower = (lower limit - the average)/standard deviation Then, determine the probability of having a value of less than the Zlower. Subtract the Zlower probability from the Zupper probability to determine the fraction within tolerances. Finally, subtract the probability of being within tolerances from 1.0 to determine what fraction will be outside of tolerances; and convert to a %. Remember:The total probability for all possible outcomes is = 1.0 Implement a Solution Zupper = (52.5-51)/1.0 = 1.5 Probability from table = .9332 Zlower = (49.0 - 51.0)/1.0 = -2.0 Probability from table = .0228 Determine probability of parts within tolerance = .9332 - .0228 = .9104 or 91.04% Refine the Solution Another approach to solving this problem would be to directly determine the probability of being above and below the tolerance. (This can be a lot easier if the tables give us the % outside the limit.) Zupper = 1.5 Probability from table = .0668 Zlower = -2.0 Probability = .0228 Determine fraction outside of the specification requirements = 0.0668 + .0228 = 0.0896 Determine fraction inside of the specification requirements = 1- 0.0896 = 0.9104

Setting Specification Limits for a Specific % Out of Specification Working backwards to find the Z given that you have the probability. For example: A process in statistical control has been producing product with a process average of 14.5 mm and a standard deviation of 0.8mm. What specifications should be set such that <=.1% of the product will fall outside the chosen limit?

Solution: Problem Statement We want to determine what specification limits will give us a 1.0% maximum percentage of materials that will fall outside of the limits. We write this problem as determining the x's (Upper and Lower Specification Limits) that will given us a probability of 0.01 outside of the specification limits. P(USL>=x>=LSL) =0.01 Collect Information We know the desired % in tolerance = 99%. The % out of tolerance = 1% or 0.01 We are given the average = 14.5.0 mm and the standard deviation of 0.08 mm. Generate a Solution First, we will assume a normal distribution that is equally distributed about the mean. Our specification limits can be described as: Upper Specification limit = average + x Lower Specification limit = average - x With 99% within the specification limits, means that only 0.5% is outside the upper limit or 99.5 % falls below the upper limit and 0.5% falls below the lower specification limit. We must solve the Z equation (either upper or lower) where Zlower = [(average-x) - average]/std deviation for x or Zupper = [(average+x)-average]/std deviation for x Second, we know the average, the standard deviation and we can determine the Zupper or lower by entering the cumulative normal distribution table and look for the probability 0.995 or 0.005 if we are determing either the Zupper or Zlower. Third, once we have determined the Zupper upper or lower we can solve the equations for x. x = -Zlower * std deviation or x = Zupper * std deviation Fourth, the USL = the average + x and the LSL = the average - x

Implement a Solution For a Zlower the probability = 0.005 corresponding Z from table = -2.575 x = -(-2.575)*0.08mm = 0.206 or 0.21mm USL = 14.5mm+0.21mm=14.71mm LSL = 14.5m-0.21 = 14.29mm Refine the Solution For the Zupper the probability = 0.995 the corresponding Z from the table = 2.575 x=2.575*0.08mm = 0.206 or 0.21mm USL = 14.71mm, LSL = 14.2mm

Downloading the Example Excel Worksheets:

	Download Link to Cumulative Normal Distribution Worksheet
	Download Link to Normal Distribution Worksheet

Link to Analyzing Data Using Normal Probability Paper:

Analyzing Data Using Normal Probability Paper