Continuous Random Variables

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Continuous Random Variables
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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

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Continuous Random Variable

• We dont calculate P(Y y), we calculate P(a lt Y
  lt b), where a and b are real numbers.
• For a continuous random variable
• P(Y y) 0.

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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
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The entire area under a probability density curve
for a continuous random variable

• Is always greater than 1.
• Is always less than 1.
• Is always equal to 1.
• Is undeterminable.

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Properties of a Probability Density Function (pdf)

• f(y) gt 0 for all possible intervals of y.
• If y0 is a specific value of interest, then the
  cumulative distribution function (cdf) is
• If y1 and y2 are specific values of interest,
  then

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Grams of lead per liter of gasoline has the
probability density function f(y) 12.5y -
1.25for 0.1 lt y lt 0.5What is the probability
that the next liter of gasoline has less than 0.3
grams of lead?
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Suppose a random variable Y has the following
probability density function f(y) y if
0ltylt1 2-y if 1 lt ylt2
0 if 2 lt y.Find the
complete form of the cumulative distribution
function F(y) for any real value y.
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Expected Value for a Continuous Random Variable

• Recall Expected Value for a discrete random
  variable
• Expected value for a continuous random variable

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Variance for Continuous Random Variable
Recall Variance for a discrete random variable
Variance for a continuous random variable
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Difference between Discreteand continuous random
variables

• Possible values that can be assumed
• Probability distribution function
• Cumulative distribution function
• Expected value
• Variance

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Times Between Industrial Accidents

• The times between accidents for a 10-year period
  at a DuPont facility can be modeled by the
  exponential distribution.

where ? is the accident rate (the expected number
of accidents per day in this case)
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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

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Times Between Industrial Accidents

• Suppose in a 1000 day period there were 50
  accidents.

• ? 50/1000 0.05 accidents per day

or
1/? 1000/50 20 days between accidents
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What is the probability that this facility will
go less than 10 days between the next two
accidents?
?
f(y) 0.05e-0.05y
16
?
Recall
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In General
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Exponential Distribution
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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 ? is the failure rate (expected number of
failures per hour).
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Mean and Variance for an Exponential Random
Variable
Note Mean Standard Deviation
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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?
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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?
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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.

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Relationship between Exponential Poisson
Exponential distribution models time (or space)
between Poisson events.
TIME
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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

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Recall For a Poisson Distribution
y 0,1,2,
where ? 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
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Now let T the time (or space) until the next
Poisson event.
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.
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Radioactive Particles

• 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?

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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?

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• 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?

• e-4
• 1 e-4
• e-2.5
• 1 e-2.5

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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?
Challenging questions
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Normal Distribution

• f(y)
• EY µ and VarY s2

f(y)
y
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Normal Distribution

• Characteristics
• Bell-shaped curve
• -? lt y lt ?
• µ determines distribution location and is the
  highest point on curve
• Curve is symmetric about µ
• s determines distribution spread
• Curve has its points of inflection at µ s

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Normal Distribution
s
s
s
s
s

µ

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Normal Distribution
N(µ 5, s 1)
N(µ 0, s 1)
f(y)
y
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Normal Distribution
N(µ 0,s 0.5)
f(y)
N(µ 0,s 1)
y
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Normal Distribution
N(µ 5, s 0.5)
N(µ 0, s 1)
f(y)
y
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68-95-99.7 Rule
0.997
0.95
0.68
µ
µ1s
µ2s
µ3s
µ-1s
µ-2s
µ-3s
µ 1s covers approximately 68
µ 2s covers approximately 95
µ 3s covers approximately99.7
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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 s 0.5,
  according to data obtained from the United States
  Geological Survey.

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Earthquake Richter Scale Readings
34
34
2.5
2.5
13.5
13.5
6.2
5.7
6.7
7.2
5.2
68
159
57
95
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Approximately what percent of the earthquakes are
above 5.7 on the Richter Scale?
34
34
2.5
2.5
13.5
13.5
6.2
5.7
6.7
7.2
5.2
68
95
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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
6.2
5.7
6.7
7.2
5.2
68
95
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The approximate probability an earthquake is
above 6.7 is ______.
34
34
2.5
2.5
13.5
13.5
6.2
5.7
6.7
7.2
5.2
68
95
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Standard Normal Distribution

• Standard normal distribution is the normal
  distribution that has a mean of 0 and standard
  deviation of 1.

N(µ 0, s 1)
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Z is Traditionally used as the Symbol for a
Standard Normal Random Variable
Z
6.2
6.7
7.2
7.7
5.7
5.2
4.7
Y
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Normal ? Standard Normal
Any normally distributed random variable can be
converted to standard normal using the following
formula
We can compare observations from two different
normal distributions by converting the
observations to standard normal and comparing the
standardized observations.
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What is the standard normal value (or Z value)
for a Richter reading of 6.5?Recall Y N(µ6.2,
s0.5)
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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, s 0.5)
• Town 2 Y N(µ 6.2, s 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?

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• Town 1 Town 2
• Town 1
• Town 2

0.212
0.055
Z
Z
X
Y
4.7 5.2 5.7 6.2 6.7 7.2 7.7
3.2 4.2 5.2 6.2 7.2 8.2 9.2
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Standard Normal
0.10
0.10
0.05
0.05
0.025
0.025
0.01
0.01
0.005
0.005
1.645
2.326
-1.645
-2.326
1.96
1.282
2.576
-2.576
-1.96
-1.282
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• 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 gt 10.5 or lt 9.5 mm.

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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.

Area0.025
y ?
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Inverse Normal Probabilities

• Approximately 2.5 of the bolts produced will
  have thicknesses less than ______.

0.025
Z
Y
?
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Inverse Normal Probabilities

• Approximately 2.5 of the bolts produced will
  have thicknesses less than ______.

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Inverse Normal Probabilities

• Approximately 1 of the bolts produced will have
  thicknesses less than ______.

0.01
Z
Y
?
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Inverse Normal Probabilities

• Approximately 1 of the bolts produced will have
  thicknesses less than ______.