1. Statistical Distributions

2. Uniform Distribution
A R.V. is uniformly distributed on the interval (a,b) if it probability function
Fully defined by (a,b)
P(x) = 1/(b-a) for a <= x <= b
= 0 otherwise

3. Uniform Distribution Probability Function1
1/9

4. Probability that x is between 2 and 7.5? Probability that x = 8?1
1/9

5. Uniform Distribution The cumulative distribution of a uniform RV is
F(x) = for x < a
= (x-a)/(b-a) for a <= x <= b
= otherwise

6. Uniform Distribution Cumulative Function1

7. Uniform Distribution Discrete vs. Continuous
Discrete RV
Number showing on a die
Continuous RV
Time of arrival
When programming, make it discrete to some number of decimal places

8. Uniform Distribution Mean = (a+b)/2 Variance = (b-a)2 /12
P (x < X < y) = F (y) – F (x)
= (y-a) - (x-a) = y – x – a + a = y - x
b-a b-a b – a b-a

9. Uniform - Example
A bus arrives at a bus stop every 20 minutes starting at 6:40 until 8:40. A passenger does not know the schedule but randomly arrives between 7:00 and 7:30 every morning. What is the probability the passenger waits more than 5 minutes.

10. Uniform Solution P (x > 5) = A + C = 1 – B = 5/6
X = RV, Uniform (0,30) -- i.e. 7:00 – 7:30
Bus: 7:00, 7:20, 7:40
Yellow Box <= 5 minute wait 1
1/30
A B C
P (x > 5) = A + C = 1 – B = 5/6

11. Arithmetic Mean Given a set of measurements y1, y2, y3,… yn
Mean = (y1+y2+…yn) / n

12. Variance
Variance of a set of measurements y1, y2, y3,… yn is the average of the deviations of the measurements about their mean (m).
V = σ2 = (1/n) Σ (yi – m)2
i=1..n

13. Variance Example V= σ2 = (1/10) ((12-10.5)2 + (10-10.5)2 +….
Yi= 12, 10, 9, 8, 14, 7, 15, 6, 14, 10
m = 10.5
V= σ2 = (1/10) (( )2 + ( )2 +….
= (1/10) ( ….)
= (1/10) (88.5)
= 8.85
Standard Deviation = σ = 2.975

14. Normal Distribution Has 2 parameters Mean - μ Variance – σ2
Also, Standard deviation - σ

15. Normal Dist.
.3413
.1359
.0215
.0013
Mean +- n σ

16. Normal Distribution Standard Normal Distribution has
Mean = StdDev = 1
Convert non-standard to standard to use the tables
Z value = # of StdDev from the mean
Z is value used for reading table
Z = (x – m) σ

17. Normal - Example
The scores on a college entrance exam are normally distributed with a mean of 75 and a standard deviation of 10. What % of scores fall between 70 & 90?
Z(70) = (70 – 75)/10 = - 0.5
Z(90) = (90 – 75)/10 = 1.5
= = .4332
= or 62.47%

18. Exponential Distribution
A RV X is exponentially distributed with parameter > 0 if probability function
Mean = 1/
Variance = 1 / e =
P(x) =
For x >= 0
= Otherwise

19. Exponential Distribution
Often used to model interarrival times when arrivals are random and those which are highly variable.
In these instances lambda is a rate
e.g. Arrivals or services per hour
Also models catastrophic component failure, e.g. light bulbs burning out

20. Exponential Rates Engine fails every 3000 hours
Mean: Average lifetime is 3000 hours
= 1/3000 =
Arrivals are 5 every hour
Mean: Interarrival time is 12 minutes
= 1 / 5 = 0.2
Mean = 1 /

21. Exponential Distribution Probability Function
f(x) x
See handout for various graphs.

22. Exponential Distribution Cumulative Function
Given Mean = 1/
Variance = 1/ 2
F(x) = P (X <=x) = 1 – e - x

23. Exponential Distribution Cumulative Function (<=)1
F(x) x

24. Forgetfulness Property
Given: the occurrence of events conforms to an exponential distribution:
The probability of an event in the next x-unit time frame is independent on the time since the last event.
That is, the behavior during the next x-units of time is independent upon the behavior during the past y-units of time.

25. Forgetfulness Example
The lifetime of an electrical component is exponentially distributed with a mean of .
What does this mean??

26. Forgetfulness Examples The following all have the same probability
Probability that a new component lasts the first 1000 hours.
Probability that a component lasts the next 1000 hours given that it has been working for 2500 hours.
Probability that a component lasts the next 1000 hours given that I have no idea how long it has been working.

27. Solution to Example
Suppose the mean lifetime of the component is 3000 hours.
= 1/3000
P(X >= 1000) = 1 – P(X <= 1000)
1 – (1-e -1/3* 1) = e -1/3 = .717

28. How do we apply these?
We may be given the information that events occur according to a known distribution.
We may collect data and must determine if it conforms to a known distribution.