# Continuous Random Variables

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```					Continuous Random
Variables
Continuous Random Variable
 A continuous random variable is one
for which the outcome can be any value in
an interval of the real number line.
 Usually a measurement.
 Examples
– Let Y = length in mm
– Let Y = time in seconds
– Let Y = temperature in ºC
L. Wang, Department of Statistics
University of South Carolina; Slide 2
Continuous Random Variable
   We don’t calculate P(Y = y), we calculate
P(a < Y < b), where a and b are real
numbers.

   For a continuous random variable
P(Y = y) = 0.

L. Wang, Department of Statistics
University of South Carolina; Slide 3
Continuous Random Variables
   The probability density function (pdf) when
plotted against the possible values of Y forms a
curve. The area under an interval of the curve is
equal to the probability that Y is in that interval.

0.40
f(y)

a     b Y                L. Wang, Department of Statistics
University of South Carolina; Slide 4
The entire area under a probability
density curve for a continuous
random variable
A. Is   always greater than 1.
B. Is   always less than 1.
C. Is   always equal to 1.
D. Is   undeterminable.

L. Wang, Department of Statistics
University of South Carolina; Slide 5
Properties of a Probability Density Function (pdf)
1)   f(y) > 0 for all possible intervals of y.

2)
 f ( y)dy  1


3)   If y0 is a specific value of interest, then the
cumulative distribution function (cdf) is
y0

F ( y0 )  P(Y  y0 )     f ( y)dy


4)   If y1 and y2 are specific values of interest,
then                 y              2

P( y1  Y  y2 )        f ( y)dy  F ( y )  F ( y )
y1
2             1
L. Wang, Department of Statistics
University of South Carolina; Slide 6
Grams of lead per liter of gasoline
has the probability density function:
f(y) = 12.5y - 1.25
for 0.1 < y < 0.5
What is the probability that the next
liter of gasoline has less than 0.3

L. Wang, Department of Statistics
University of South Carolina; Slide 7
Suppose a random variable Y has
the following probability density
function: f(y) = y     if 0<y<1
2-y if 1 < y<2
0     if 2 < y.
Find the complete form of the
cumulative distribution function F(y)
for any real value y.
L. Wang, Department of Statistics
University of South Carolina; Slide 8
Expected Value for a Continuous
Random Variable
   Recall Expected Value for a discrete random
variable:

E(Y )     y  p( y)

   Expected value for a continuous random
variable:

E (Y )        

yf ( y)dy
L. Wang, Department of Statistics
University of South Carolina; Slide 9
Variance for Continuous Random
Variable
Recall: Variance for a discrete random
variable:

Var(Y )     ( y   ) p( y)
2                           2

Variance for a continuous random variable:

Var(Y )     ( y   ) f ( y)dy
2            2

L. Wang, Department of Statistics
           University of South Carolina; Slide 10
Difference between Discrete
and continuous random variables
•Possible values that can be assumed
•Probability distribution function
•Cumulative distribution function
•Expected value
•Variance

L. Wang, Department of Statistics
University of South Carolina; Slide 11
Times Between Industrial Accidents
   The times between accidents for a 10-
year period at a DuPont facility can be
modeled by the exponential distribution.
 y
f ( y )  e           y  0 and   0
where λ is the accident rate (the expected
number of accidents per day in this case)

L. Wang, Department of Statistics
University of South Carolina; Slide 12
Example of time between accidents
Let Y = the number of days between two
accidents.

Time
12 days   35 days   5 days

●         ●             ●●           ●

Accident     Accident        Accident
#1          #2               #3

L. Wang, Department of Statistics
University of South Carolina; Slide 13
Times Between Industrial Accidents
   Suppose in a 1000 day period there were
50 accidents.
λ = 50/1000 = 0.05 accidents per day

or
1/λ = 1000/50 = 20 days between accidents

L. Wang, Department of Statistics
University of South Carolina; Slide 14
What is the probability that this facility
will go less than 10 days between the
next two accidents?
Probability Density Function

0.06

?
0.05

0.04
f(y)

0.03

0.02

0.01

0
0   10   20   30      40       50     60        70       80       90      100
Y = Time between accidents

f(y) = 0.05e-0.05y                                                            L. Wang, Department of Statistics
University of South Carolina; Slide 15
Probability Density Function

?
0.06

0.05

0.04

f(y)
0.03

0.02

0.01

0
0   10   20       30      40       50       60      70      80       90      100
Y = Time between accidents

10
P(Y  10)  F (10)   0.05e 0.05 y dy
0

Recall:        e du  e
u             u

0.05 y 10
F (10 )  e                       0     |  0.39
L. Wang, Department of Statistics
University of South Carolina; Slide 16
In General…
y

P(Y  y )   e dt
 t

0

 t y                        y
P(Y  y )  F ( y )  e       |  1 e
0

 y
P(Y  y )  1  F ( y )  e

L. Wang, Department of Statistics
University of South Carolina; Slide 17
Exponential Distribution
Probability Density Function

0.06

0.05

0.04
f(y)

0.03

0.02

0.01

0
0   10   20   30      40       50     60        70       80       90      100
Y = Time between accidents

 y                                y
1 e                                       e                   L. Wang, Department of Statistics
University of South Carolina; Slide 18
If the time to failure for an electrical
component follows an exponential
distribution with a mean time to failure of
1000 hours, what is the probability that a
randomly chosen component will fail before
750 hours?
Hint: λ is the failure rate
(expected number of
failures per hour).

L. Wang, Department of Statistics
University of South Carolina; Slide 19
Mean and Variance for an
Exponential Random Variable

1
E (Y )     ye        y
dy 
0

                                   2
1   1
Var(Y )     y e
2           2    y
dy     2
0                         

Note: Mean = Standard Deviation
L. Wang, Department of Statistics
University of South Carolina; Slide 20
The time between accidents at a factory
follows an exponential distribution with
a historical average of 1 accident every
900 days. What is the probability that
that there will be more than 1200 days
between the next two accidents?

L. Wang, Department of Statistics
University of South Carolina; Slide 21
If the time between accidents follows an
exponential distribution with a mean of
900 days, what is the probability that
there will be less than 900 days between
the next two accidents?

L. Wang, Department of Statistics
University of South Carolina; Slide 22
Relationship between Exponential &
Poisson Distributions

 Recall that the Poisson distribution is used
to compute the probability of a specific
number of events occurring in a particular
interval of time or space.
 Instead of the number of events being the
random variable, consider the time or
space between events as the random
variable.
L. Wang, Department of Statistics
University of South Carolina; Slide 23
Relationship between Exponential &
Poisson
Exponential distribution models time
(or space) between Poisson events.

TIME

L. Wang, Department of Statistics
University of South Carolina; Slide 24
Exponential or Poisson Distribution?
   We model the number of industrial accidents
occurring in one year.

   We model the length of time between two
industrial accidents (assuming an accident
occurring is a Poisson event).

   We model the time between radioactive particles
passing by a counter (assuming a particle passing
by is a Poisson event).

    We model the number of radioactive particles
passing by a counter in one hour
L. Wang, Department of Statistics
University of South Carolina; Slide 25
Recall: For a Poisson Distribution
( t ) y e  t
P (Y  y )  p ( y )                         y = 0,1,2,…
y!
where λ is the mean number of events per base
unit of time or space and t is the number of base
units inspected.
The probability that no events occur in a span
of time (or space) is:

( t ) y e  t ( t ) 0 e  t     t
p ( 0)                                 e
y!              0!
L. Wang, Department of Statistics
University of South Carolina; Slide 26
Now let T = the time (or space)
until the next Poisson event.
 t
P(T  t )  e
In other words, the probability that the
length of time (or space) until the next
event is greater than some given time
(or space), t, is the same as the
probability that no events will occur in
time (or space) t.
L. Wang, Department of Statistics
University of South Carolina; Slide 27
 The arrival of radioactive particles at a
counter are Poisson events. So the
number of particles in an interval of time
follows a Poisson distribution. Suppose we
average 2 particles per millisecond.
 What is the probability that no particles
will pass the counter in the next 3
milliseconds?
 What is the probability that more than 3
millisecond will elapse before the next
particle passes?                L. Wang, Department of Statistics
University of South Carolina; Slide 28
Machine Failures
 If the number of machine failures in a
given interval of time follows a Poisson
distribution with an average of 1 failure
per 1000 hours, what is the probability
that there will be no failures during the
next 2000 hours?
 What is the probability that the time until
the next failure is more than 2000 hours?

L. Wang, Department of Statistics
University of South Carolina; Slide 29
   Number of failures in an interval of time
follows a Poisson distribution. If the mean
time to failure is 1000 hours, what is the
probability that more than 2500 hours will
pass before the next failure occurs?

A. e-4
B. 1 – e-4
C. e-2.5
D. 1 – e-2.5
L. Wang, Department of Statistics
University of South Carolina; Slide 30
Challenging questions

If ten of these components are used in
different devices that run independently,
what is the probability that at least one will
still be operating at 2500 hours?

What about he probability that exact 3 of
them will be still operating after 2500 hours?
L. Wang, Department of Statistics
University of South Carolina; Slide 31
Normal Distribution

f(y)

y

f(y) =       1      ( y   ) 2 / 2 2
e                      ,  y  
2 
E[Y] = μ      and        Var[Y] = σ2         L. Wang, Department of Statistics
University of South Carolina; Slide 32
Normal Distribution
   Characteristics
– Bell-shaped curve
– - < y < +
– μ determines distribution location and is
the highest point on curve
– Curve is symmetric about μ
– Curve has its points of inflection at μ + σ
L. Wang, Department of Statistics
University of South Carolina; Slide 33
Normal Distribution

σ       σ          σ

σ       σ

-4   -3    -2   -1   μ0       1          2             3             4

L. Wang, Department of Statistics
University of South Carolina; Slide 34
Normal Distribution
N(μ = 0, σ = 1)                  N(μ = 5, σ = 1)

f(y)

-4     -3   -2   -1   0   1   2   3    4      5        6        7       8

y
L. Wang, Department of Statistics
University of South Carolina; Slide 35
Normal Distribution

N(μ = 0,σ = 0.5)

f(y)                                   N(μ = 0,σ = 1)

-4   -3     -2   -1       0     1         2              3             4

y                L. Wang, Department of Statistics
University of South Carolina; Slide 36
Normal Distribution

N(μ = 5, σ = 0.5)
N(μ = 0, σ = 1)
f(y)

-4     -3   -2   -1   0   1       2   3   4          5         6        7         8

y
L. Wang, Department of Statistics
University of South Carolina; Slide 37
68-95-99.7 Rule
0.997
0.95
0.68

-4    µ-3σ
-3   µ-2σ
-2   µ-1σ
-1     µ
0     µ+1σ µ+2σ
1    2           µ+3σ
3               4

μ + 1σ covers         μ + 2σ covers         μ + 3σ covers
approximately 68%     approximately 95%     approximately99.7%
L. Wang, Department of Statistics
University of South Carolina; Slide 38
Earthquakes in a California Town
Since 1900, the magnitude of
earthquakes that measure 0.1 or
higher on the Richter Scale in a
certain location in California is
distributed approximately normally,
with μ = 6.2 and σ = 0.5, according
to data obtained from the United
States Geological Survey.
L. Wang, Department of Statistics
University of South Carolina; Slide 39

34%         34%

2.5%                                           2.5%
13.5%                     13.5%

-4   -3     5.2
-2   5.7
-1        6.2
0      6.7
1      7.2
2             3             4

57                 68%                             159
95%
L. Wang, Department of Statistics
University of South Carolina; Slide 40
Approximately what percent of the earthquakes
are above 5.7 on the Richter Scale?

34%         34%
2.5%                                           2.5%
13.5%                     13.5%
-4   -3     5.2
-2   5.7
-1        6.2
0      6.7
1      7.2
2             3             4

68%
95%

L. Wang, Department of Statistics
University of South Carolina; Slide 41
The highest an earthquake can read
and still be in the lowest 2.5% is _.

34%         34%

2.5%                                           2.5%
13.5%                     13.5%

-4   -3     5.2
-2   5.7
-1        6.2
0      6.7
1      7.2
2             3             4

68%
95%
L. Wang, Department of Statistics
University of South Carolina; Slide 42
The approximate probability an
earthquake is above 6.7 is ______.

34%         34%

2.5%                                           2.5%
13.5%                     13.5%

-4   -3     5.2
-2   5.7
-1        6.2
0      6.7
1      7.2
2             3             4

68%
95%
L. Wang, Department of Statistics
University of South Carolina; Slide 43
Standard Normal Distribution
   Standard normal distribution is the normal
distribution that has a mean of 0 and
standard deviation of 1.

N(µ = 0, σ = 1)

-4     -3   -2   -1   0   1   2           3           4

L. Wang, Department of Statistics
University of South Carolina; Slide 44
Z is Traditionally used as the Symbol
for a Standard Normal Random
Variable

Z    -4       -3    -2    -1     0     1       2             3             4

Y   4.7   5.2   5.7   6.2   6.7   7.2          7.7
L. Wang, Department of Statistics
University of South Carolina; Slide 45
Normal  Standard Normal
Any normally distributed random variable
can be converted to standard normal using
the following formula:
y
Z

We can compare observations from two
different normal distributions by converting
the observations to standard normal and
comparing the standardized observations.
L. Wang, Department of Statistics
University of South Carolina; Slide 46
What is the standard normal
value (or Z value) for a Richter
Recall Y ~ N(µ=6.2, σ=0.5)

L. Wang, Department of Statistics
University of South Carolina; Slide 47
Example
   Consider two towns in California. The distributions of
the Richter readings over 0.1 in the two towns are:

Town 1:        X ~ N(µ = 6.2, σ = 0.5)
Town 2:        Y ~ N(µ = 6.2, σ = 1).

- What is the probability that Town 1 has an
earthquake over 7 (on the Richter scale)?
- What is the probability that Town 2 has an
earthquake over 7?

L. Wang, Department of Statistics
University of South Carolina; Slide 48
Town 1                                            Town 2

0.212
0.055
X 
Z


Z   -4    -3    -2     -1    0     1     2      3    4   Z   -4   -3   -2      -1        0   1      2      3         4

X        4.7   5.2    5.7   6.2   6.7   7.2    7.7       Y        3.2 4.2     5.2    6.2     7.2 8.2     9.2

     7  6.2 
Town 1:            P( X  7.0)  P Z            P( Z  1.6)  0.055
       0.5 

     7  6.2 
Town 2:            P(Y  7.0)  P Z            P( Z  0.8)  0.212
       1.0         L. Wang, Department of Statistics
University of South Carolina; Slide 49
Standard Normal

0.10                          0.10
0.05                                 0.05
0.025
0.025
0.01                                             0.01
0.005                                                0.005

-4        -3           -2   -1   0   1   1.6452 2.326 3                     4
-2.326 -1.645
-2.576 -1.96 -1.282         1.282 1.96      2.576
L. Wang, Department of Statistics
University of South Carolina; Slide 50
 The thickness of a certain steel bolt that
continuously feeds a manufacturing
process is normally distributed with a
mean of 10.0 mm and standard deviation
of 0.3 mm. Manufacturing becomes
concerned about the process if the bolts
get thicker than 10.5 mm or thinner than
9.5 mm.
 Find the probability that the thickness of a
randomly selected bolt is > 10.5 or < 9.5
mm.                              L. Wang, Department of Statistics
University of South Carolina; Slide 51
Inverse Normal Probabilities
   Sometimes we want to answer a question which
is the reverse situation. Here we know the
probability, and want to find the corresponding
value of Y.

Area=0.025

-4   -3   -2    -1   0   1         2     3        4
X=?

y=?                             L. Wang, Department of Statistics
University of South Carolina; Slide 52
Inverse Normal Probabilities
   Approximately 2.5% of the bolts produced will
have thicknesses less than ______.

0.025

Z   -4    -3     -2   -1   0   1   2   3        4

Y                ?                        L. Wang, Department of Statistics
University of South Carolina; Slide 53
Inverse Normal Probabilities
   Approximately 2.5% of the bolts produced will
have thicknesses less than ______.

Y  10.0
2            y  9.4
0.3

L. Wang, Department of Statistics
University of South Carolina; Slide 54
Inverse Normal Probabilities
   Approximately 1% of the bolts produced will
have thicknesses less than ______.

0.01

Z   -4    -3    -2   -1   0   1   2   3        4

Y               ?                        L. Wang, Department of Statistics
University of South Carolina; Slide 55
Inverse Normal Probabilities
   Approximately 1% of the bolts produced will
have thicknesses less than ______.

Y  10.0
 2.326            y  9.3
0.3

L. Wang, Department of Statistics
University of South Carolina; Slide 56

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