# Statistical Distributions ```Statistical
Distributions
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)
= 0
for a &lt;= x &lt;= b
otherwise
Uniform Distribution
Probability Function
1
1/9
1
10
Probability that x is between 2 and 7.5?
Probability that x = 8?
1
1/9
1
10
Uniform Distribution
The cumulative distribution of a uniform
RV is
F(x) = 0
for x &lt; a
= (x-a)/(b-a) for a &lt;= x &lt;= b
= 1
otherwise
Uniform Distribution
Cumulative Function
1
1
10
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
Uniform Distribution
• Mean = (a+b)/2
• Variance = (b-a)2 /12
• P (x &lt; X &lt; y) = F (y) – F (x)
= (y-a) - (x-a) = y – x – a + a = y - x
b-a
b-a
b–a
b-a
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.
Uniform Solution
X = RV, Uniform (0,30) -- i.e. 7:00 – 7:30
Bus: 7:00, 7:20, 7:40
Yellow Box &lt;= 5 minute wait
1
1/30
A
5
B
10
15
C
20
25
30
P (x &gt; 5) = A + C = 1 – B = 5/6
40
Arithmetic Mean
Given a set of measurements y1, y2,
y3,… yn
Mean = (y1+y2+…yn) / n
Variance
Variance of a set of measurements y1,
y2, y3,… yn is the average of the
deviations of the measurements
V = σ2 = (1/n) Σ (yi – m)2
i=1..n
Variance Example
Yi= 12, 10, 9, 8, 14, 7, 15, 6, 14, 10
m = 10.5
V= σ2 = (1/10) ((12-10.5)2 + (10-10.5)2 +….
= (1/10) (1.52 + .52 + 1.52….)
= (1/10) (88.5)
= 8.85
Standard Deviation = σ = 2.975
Normal Distribution
• Has 2 parameters
– Mean - μ
– Variance – σ2
– Also, Standard deviation - σ
Normal Dist.
.3413
.1359
.0215
.0013
-3
-2
-1
0
1
Mean +- n σ
2
3
Normal Distribution
• Standard Normal Distribution has
– Mean = 0
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
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 &amp; 90?
Z(70) = (70 – 75)/10 = - 0.5
Z(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  &gt; 0 if probability function
Mean = 1/ 
Variance = 1 / 2
e = 2.71828182
P(x) =
=
e
0
 x
For x &gt;= 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

f(x)
x
See handout for various graphs.
Exponential Distribution
Cumulative Function
Given Mean = 1/ 
Variance = 1/  2
F(x) = P (X &lt;=x) = 1 – e -  x
Exponential Distribution
Cumulative Function (&lt;=)
1
F(x)
x
Forgetfulness Property
Given: the occurrence of events conforms to
an exponential distribution:
The probability of an event in the next xunit 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 &gt;= 1000) = 1 – P(X &lt;= 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.
```