Normality Tests, Binomial Distribution & R code

Binomial Distribution

• Can be used to describe the Binomial probability distribution and calculate probabilities
• Relates to a discrete random variable (nominal data) and is based on the Bernoulli process

Bernoulli Process

• Each trial or experiment has only two possible outcomes
• The probability of any and all outcomes remains fixed over time (constant probability)
• All trials or experiments are statistically independent

Binomial Formula

• P(r in n trials) = [n! / r!(n-r)!] * [p^r] * [q^(n-r)]
• p = probability of occurrence
• q = 1-p = probability of failure
• r = number of occurrences desired
• n = number of trials

Binomial Example

• A vendor frequently ships 2 bad parts out of 10
• Vendor ships 50 parts
• At least 9 parts out of 10 must be good
• It's possible to calculate the probability of receiving what was requested

Binomial Example (Calculations Shown)

• p = 0.80, q = 0.20, r = 45, n = 50
• P(45 in 50) = [50! / 45!(50-45)!] * [0.8^45] * [0.2^5] = 0.02953

Binomial Distribution in RStudio

• p = 0.80, q = 0.20, r = 45, n = 50
• Can be solved in RStudio as dbinom(x = 45, size = 50, prob = 0.8)
• The function ro(table.dist.binomial(n = 50, p = 0.80),5) can also be used

Binomial Example in ROIStat

• To calculate, first open ROI Stat
• Then go to Distributions > Binomial
• Enter in the value for p (π)
• Enter in the sample size (n)
• Select the Point (R) of Interest

Poisson Distribution

• Can be used to describe this probability distribution and calculate probabilities

Poisson Distribution Context

• Used for discrete random variables which can take integer (whole) values (ordinal data)
• Can apply to the number of parts produced during a 10 minute period, # of breakdowns per shift and the # of failures per 100 cycles

Poisson Formula

• P(X) = (λ^X / X!) * e^(-λ)
• P(X) = probability exactly X occurrences
• λ = Mean number of occurrences per time interval (or unit)
• e = 2.71828

Poisson Example

• λ = 25 parts produced per hour
• X = 10 parts produced in one hour

Poisson Example - Probability

• P(10) = (25^10 / 10!) * e^(-25) = 0.000365

Poisson Distribution in RStudio

• λ = 25 parts produced per hour
• X = 10 parts produced in one hour
• The calculation can be solved in code as dpois(x = 10, lambda = 25)
• ro(table.dist.poisson(lambda = 25),5) is an alternative function to use

Poisson Example in ROIStat

• Enter values for Open ROI Stat, Distributions > Poisson and Count (λ) and select the Point (R) of Interest

Testing for a Poisson Distribution

• All ratio discrete, count data do not necessarily conform to this probability distribution!

Testing in RStudio

• Use the follow function: poisson.dist.test(x = Discrete$DEFECTS)

Additional info for testing

• If the p-value is less than 0.05 then reject the hypothesis
• Data are likely from a Poisson distribution when the p (probability) is low
• If p is low, then reject H0

Results of Example of Testing

• chi.square = 44.173, degrees of freedom = 49, p-value = 0.6624
• alternative hypothesis: true chi.square is not equal to 49
• sample estimates:
  • chi.square sample variance 44.172840
  • sample mean = 3.240000

Ungrouped Testing

• function is hist.ungrouped(Discrete$DEFECTS)

Testing for a Poisson Distribution In ROIStat

• Open ROI Stat and Go to Distributions > Testing
• Select the data and reject if the p-value is < 0.05

Definition of Probability in quality control

• Used for monitoring the number of occurrences of a specified event in a specified inspection unit
• Inspection units can be length, area, number of parts, volume, or time

Considering Challenge

• For every 750 lines of code, there will be an average of 6 errors

What to Consider for The challenge

• A = (255 / 750)(6) = 2.04 errors
• λ = 6/750 = 0.008 errors per line x 255 lines; so
• 0.008 x 255 = 2.04 errors

Distribution Solution

• Produce the distribution for the relevant Poisson Distribution (λ = 2.04) with the function round.object(table.dist.poisson(2.04),4)

Finding P(0)

• round.object(table.dist.poisson(2.04),4)

  • x p.at.x eq.and.above eq.and.below
  • 0 0 0.1300 1.0000 0.1300
• 1 1 0.2653 0.8700 0.3953
• 2 2 0.2706 0.6047 0.6659
• 3 3 0.1840 0.3341 0.8498
• 4 4 0.0938 0.1502 0.9437
• 5 5 0.0383 0.0563 0.9819
• 6 6 0.0130 0.0181 0.9950
• 7 7 0.0038 0.0050 0.9988
• 8 8 0.0010 0.0012 0.9997
• 9 9 0.0002 0.0003 0.9999
• 10 10 0.0000 0.0001 1.0000
• 11 11 0.0000 0.0000 1.0000

Binomial practice activities

• What is the probability of finding exactly two nonconforming parts in the 50 parts and finding two are less nonforming parts in the 50 parts assuming a supplier ships 50 parts to a plant and a consistent 10% nonforming rate exists?
• You can use lolcat's 'table.dist.binomial()' function considering: π = 0.10, n = 50 and r = 2

Solution for previous

• ro(table.dist.binomial(n,p) [1:10,],4)
• The exact probability of x, or r = 2 can be obtained with the function: dbinom(x = 2,size = 50, prob = 0.1)
• pbinom(q = 2, size = 50, prob = 0.1) calculates the probability of 2 or fewer
• x p.at.x eq.and.above eq.and.below
• 0 0 0.0052 1.0000 0.0052
• 1 1 0.0286 0.9948 0.0338
• 2 2 0.0779 0.9662 0.1117
• 3 3 0.1386 0.8883 0.2503
• 4 4 0.1809 0.7497 0.4312
• 5 5 0.1849 0.5688 0.6161
• 6 6 0.1541 0.3839 0.7702
• 7 7 0.1076 0.2298 0.8779
• 8 8 0.0643 0.1221 0.9421
• 9 9 0.0333 0.0579 0.9755

You Try: Binomials

• Assume a product has a documented failure rate of 0.20 after 150 hours of use. If we were to place 30 randomly selected parts from this process in the field, what is the failure rate with 5 or fewer after 150 hours and then what is the probability of over 10 failing

Poisson Take away

• The number of OSHA-recordable safety accidents in a manufacturing plant has been running 4.2 accidents per 200,000 hours worked. What is the probability of having exactly two accidents in a 200,000-hour work period?

Facts:

• Given, X = 4.2 and X=2

You can use lolcat's'table.dist.poisson()' function to get the results (next slide) or directly with the R dpois() function, both demonstrated on the next slide:

• λ = 4.2 and X = 2

Example Solution

• ro(table.dist.poisson(lambda)[1:5,],4)
• x p.at.x eq.and.above eq.and.below
• 0 0 0.0150 1.0000 0.0150
• 1 1 0.0630 0.9850 0.0780
• 2 2 0.1323 0.9220 0.2102
• 3 3 0.1852 0.7898 0.3954
• 4 4 0.1944 0.6046 0.5898

Try A Poisson

• The average buckets of blanked saw chain cutters have 65 (入) and the output represents the Poisson function. What are the probability of creating 50 or more buckets per day?

NORMAL DISTRIBUTIONS

• A theoretical probability distribution for a continuous random variable and one of the most important distributions
• Mean = Median = Mode and symmetrical around μ
• Tails extend to ∞ but never touch the horizontal axis and Areas are always predictable
• Has a Y3 and Y4 of 0.00

Area in a curve

• 34.135% lies within 1 standard of Deviation
• 13.590% lies within 2 standard of Deviations
• 2.140% lies within 3 Standard of deviation
• The middle (μ) has a area of 68.27%
• Two standard distributions have an area of 95.45% and 3 have 99.73%

Z score

• The area corresponding to any score value may be found through the function
• Z = ( X – μ) / σ.
• Z is the number of standard deviation units from X to μ

Normal probability Question

• Tooling lasted 180 hours with a deviation of 5 hours and the probability will last less than 172 hours

Facts :

• Z = (X – μ) / σ
• Z = 172 - 180/5 = -1.60

Quick notes

• Can be coded in RStudio with function pnorm(q, mean, sd , lower.tail)or In distributions > Normal in ROI Stat

What's a follow up Normal Example

• center-to-center distance between the two holes has been an average (μ) of 5.20mm and a standard deviation (σ) of 0.05mm with an USL of 5.35mm and a LSL of 5.15mm.
• Find the amount out of specs?

Here's the solution

• Z= 5.15 - 5.20 /.05
• z = -1.00 and then Z= 5.35 - 5.20/.05 and Z= 3.00

How to test for Normality

• When n < 25, use the Anderson-Darling / Shapiro-Wilk tests for normality
• and when n ≥ 25, use Skewness Test, and Kurtosis Test (Moment Tests)

Testing Tips

• Probabilities ≥ 0.05 indicate that the data are normal and opposite if it's NOT normal

What RStudio

• anderson.darling.normality.test() or shapiro.wilk.normality.test() summary.continuous() is the solution
• Or find the EDA > Normality Tests functions in ROIStat

This quiz covers normality tests for a sample size of 30, interpretations of probability values, and characteristics of a Bernoulli process. It also covers the R function for the Shapiro-Wilk test, the binomial formula and calculating probabilities.