Statistical Statistical DistributionsDistributions

Uniform DistributionA R.V. is uniformly distributed on the

interval (a,b) if it probability functionFully defined by (a,b)

P(x) = 1/(b-a) for a <= x <= b = 0 otherwise

Uniform Distribution Probability Function

1 10

1

1/9

Probability that x is between 2 and 7.5? Probability that x = 8?

1 10

1

1/9

Uniform DistributionThe cumulative distribution of a

uniform RV is

F(x) = 0 for x < a = (x-a)/(b-a) for a <= x <= b

= 1 otherwise

Uniform Distribution Cumulative Function

1 10

1

Uniform DistributionDiscrete 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

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

Uniform - ExampleA 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.

Uniform Solution

5 10 15 20 25 30 40

1

1/30

X = RV, Uniform (0,30) -- i.e. 7:00 – 7:30Bus: 7:00, 7:20, 7:40Yellow Box <= 5 minute wait

A B C

P (x > 5) = A + C = 1 – B = 5/6

Arithmetic MeanGiven a set of measurements y1,

y2, y3,… yn

Mean = (y1+y2+…yn) / n

VarianceVariance 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)2i=1..n

Variance ExampleYi= 12, 10, 9, 8, 14, 7, 15, 6, 14, 10m = 10.5V= σ2 = (1/10) ((12-10.5)2 + (10-10.5)2 +

…. = (1/10) (1.52 + .52 + 1.52….) = (1/10) (88.5)

= 8.85Standard Deviation = σ = 2.975

Normal Distribution• Has 2 parameters

–Mean - μ–Variance – σ2

–Also, Standard deviation - σ

Normal Dist.

0-3 -2 -1 1 2 3

Mean +- n σ

.3413 .1359

.0215 .0013

Normal Distribution• Standard Normal Distribution has

– Mean = 0 StdDev = 1• Convert non-standard to standard

to use the tablesZ value = # of StdDev from the meanZ is value used for reading table Z = (x – m)

σ

Normal - ExampleThe 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 = - 0.5Z(90) = (90 – 75)/10 = 1.5.6915 - .5 = .1915 + .9332 - .5 = .4332

= .6247 or 62.47%

Exponential Distribution

A RV X is exponentially distributed with parameter > 0 if probability function

Mean = 1/Variance = 1 / 2 e = 2.71828182

e xP(x) =

For x >= 0

= 0 Otherwise

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

Exponential Rates• Engine fails every 3000 hours

– Mean: Average lifetime is 3000 hours– = 1/3000 = 0.00033333

• Arrivals are 5 every hour– Mean: Interarrival time is 12 minutes– = 1 / 5 = 0.2

• Mean = 1 /

Exponential Distribution

Probability Function

x

f(x)

See handout for various graphs.

Exponential Distribution

Cumulative FunctionGiven Mean = 1/ Variance = 1/ 2

F(x) = P (X <=x) = 1 – e - x

Exponential Distribution

Cumulative Function (<=)

x

1

F(x)

Forgetfulness PropertyGiven: 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.

Forgetfulness Example

• The lifetime of an electrical component is exponentially distributed with a mean of .

• What does this mean??

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.

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

How do we apply these?

1. We may be given the information that events occur according to a known distribution.

2. We may collect data and must determine if it conforms to a known distribution.