Continuous Probability Distributions

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Continuous Probability Distributions

• Chapter 8

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Introduction

• A continuous random variable has an uncountable
  infinite number of values in the interval (a,b).

• The probability that a continuous variable X will
  assume any particular value is zero.

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8.1 Probability Density Function

• To calculate probabilities of continuous random
  variables we define a probability density
  function f(x).

• The density function satisfies the following
  conditions
• f(x) is non-negative,
• The total area under the curve representing f(x)
  is equal to 1.

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Probability Density Function

• The probability that x falls between a and b
  is found by calculating the area under the graph
  of f(x) between a and b.

P(axb)
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Uniform Distribution

• A random variable X is said to be uniformly
  distributed if its density function is
• The expected value and the variance are

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Uniform Distribution

• The name is derived from the graph that
  describes this distribution

f(X)
X
a b
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Uniform Distribution

• Example 8.1
• The daily sale of gasoline is uniformly
  distributed between 2,000 and 5,000 gallons. Find
  the probability that sales are
• Between 2,500 and 3,500 gallon

f(x) 1/(5000-2000) 1/3000 for x 2000, 5000
P(2500X3000) (3000-2500)(1/3000) .1667
1/3000
x
2000
5000
2500
3000
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Uniform Distribution

• Example 8.1
• The daily sale of gasoline is uniformly
  distributed between 2,000 and 5,000 gallons. Find
  the probability that sales are

More than 4,000 gallons
f(x) 1/(5000-2000) 1/3000 for x 2000, 5000
P(X³4000) (5000-4000)(1/3000) .1333
1/3000
x
2000
5000
4000
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Uniform Distribution

• Example 8.1
• The daily sale of gasoline is uniformly
  distributed between 2,000 and 5,000 gallons. Find
  the probability that sales are

Exactly 2,500 gallons
f(x) 1/(5000-2000) 1/3000 for x 100,180
P(X2500) (2500-2500)(1/3000) 0
1/3000
x
2000
5000
2500
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8.2 Normal Distribution

• A random variable X with mean m and variance s2
  is normally distributed if its probability
  density function is given by

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The Shape of the Normal Distribution
m
The normal distribution is bell shaped, and
symmetrical around m.
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The effects of m and s
The effects of m and s
How does the standard deviation affect the shape
of f(x)?
s 2
s 3
s 4
How does the expected value affect the location
of f(x)?
m 10
m 11
m 12
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Calculating Normal Probabilities

• Two facts help calculate normal probabilities
• The normal distribution is symmetrical.

• Any normal variable with some m and s can be
  transformed into a specific normal variable with
  m 0 and s 1, calledSTANDARD NORMAL
  DISTRIBUTION

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Calculating Normal Probabilities

• Example 1
• The amount of time it takes to assemble a
  computer is normally distributed, with a mean of
  50 minutes and a standard deviation of 10
  minutes.
• What is the probability that a computer is
  assembled in between 45 and 60 minutes?

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Calculating Normal Probabilities

• Solution
• X denotes the assembly time of a computer.

• We seek the probability P(45ltXlt60).

• Express P(45ltXlt60) in terms of Z.

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Calculating Normal Probabilities

• Example 1 - continued

P(45ltXlt60) P( lt lt
)
P(-0.5ltZlt1)
To complete the calculation we need to compute
the probability under the standard normal
distribution
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Calculating Normal Probabilities

• Example 1 - continued

P(-.5ltZlt1)
We need to find the shaded area
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Calculating Normal Probabilities

• Example 1 - continued

The probability provided by the Z-Table covers
the area between -infinity and some z0.
z 0
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Calculating Normal Probabilities
Since we need to find the area between -0.5 and 1
(that is P(-.5ltZlt1)) well calculate the
difference between P(-infinityltZlt1) ltclickgt
and P(-infinityltZlt-.5) ltclickgt
P(Z lt 1)
P(Z lt -.5)
z 1
z -.5
P(Z lt 1) P(Zlt-.5)
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Usding the Normal Table

• Example 1 - continued

P(Zlt1
P(-.5ltZlt1)
.3413
0
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Calculating Normal Probabilities

• Example 1 - continued

- P(Zlt - .5)
.8413 - .3085
P(-.5ltZlt1)
P(Zlt1
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Calculating Normal Probabilities

• Example 2
• The rate of return (X) on an investment is
  normally distributed with mean of 10 and
  standard deviation of 5. What is the
  probability of losing money?
• Solution (i)

X
P(Xlt 0 )
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Calculating Normal Probabilities

• Solution(ii) Example 2 ( 8.2)

The curve for s 5 The curve for s 10
X
0 - 10 10
(ii) P(Xlt 0 ) P(Zlt )
P(Zlt - 1) .1587
Z
Comment When the standard deviation is 10
rather than 5, more values fall away from the
mean, so the probability of finding values at the
distribution tail increases from .0228 to .1587.
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Using Excel to Find Normal Probabilities

• For P(Xltk) enter in any empty cell
  normdist(k,m,s,True).
• Example Let m 50 and s 10.
• P(X lt 30) normdist(30,50,10,True)
• P(X gt 45) 1 - normdist(45,50,10,True)
• P(30ltXlt60) normdist(60,50,10,True)
  normdist(30,50,10,True).
• Using normsdist
• If the Z value is known you can
  useP(Zlt1.2234) normsdist(1.2234)

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Finding Values of Z

• Sometimes we need to find the value of Z for a
  given probability
• We use the notation zA to express a Z value for
  which P(Z gt zA) A

A
zA
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Finding Values of Z

• Example 3
• What percentage of the standard normal
  population is located to the right of
  z.10?Answer 10

• What percentage of the standard normal
  population is located to the left of
  z.30?Answer 70

• What percentage of the standard normal
  population is located between z.95 and z.40 55

Comment z.95 has a negative value
z.40
z.95
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Finding Values of Z

• Example 4
• Determine z exceeded by 5 of the population
• Solution
• z.05 is defined as the z value for which the
  upper tail of the distribution is .05. Thus the
  lower tail is .95!

.05
0.05
0.95
1.645
Z0.05
0
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Finding Values of Z

• Example 4
• Determine z not exceeded by 5 of the population
• Solution
• Note we look for the z exceeded by 95 of the
  population. Because of the symmetry of the normal
  distribution it is the negative value of z.05!

0.95
-1.645
-Z0.05
0
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8.5 Other Continuous Distribution

• Three new continuous distributions
• Student t-distribution
• Chi-squared distribution
• F distribution

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The Student t - Distribution

• The Student t density function
• n is the parameter of the student t
  distribution
• E(t) 0 V(t) n/(n 2)

(for n gt 2)
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The Student t - Distribution
n 3
n 10
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Determining Student t Values

• The student t distribution is used extensively in
  statistical inference.
• Thus, it is important to determine the
  probability for any given value of the variable
  t associated with a given number of degrees of
  freedom.
• We can do this using
• t tables
• Excel

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Using the t - Table
t
t
t
t

• The table provides the t values (tA) for which
  P(tn gt tA) A

The t distribution is Symmetrical around 0
tA
-1.812
1.812
t.100
t.05
t.025
t.01
t.005
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The Chi Squared Distribution

• The Chi Squared density function
• The parameter n is the number of degrees of
  freedom.

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The Chi Squared Distribution
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Determining Chi-Squared Values

• Chi squared values can be found from the chi
  squared table or from Excel.
• The c2-table entries are the c2 values of the
  right hand tail probability (A), for which P(c2n
  gt c2A) A.

A
c2A
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Using the Chi-Squared Table
To find c2 for which P(c2nltc2).01, lookup the
column labeledc21-.01 or c2.99
.05
A .99
c2.05
c2.995 c2.990 c2.05
c2.010 c2.005
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The F Distribution

• The density function of the F distributionn
  1 and n2 are the numerator and denominator
  degrees of freedom.

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The F Distribution

• This density function generates a rich family of
  distributions, depending on the values of n1 and
  n2

n1 5, n2 10 n1 50, n2 10
n1 5, n2 10 n1 5, n2 1
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Determining Values of F

• The values of the F variable can be found in the
  F table or from Excel.
• The entries in the table are the values of the F
  variable of the right hand tail probability (A),
  for which P(Fn1,n2gtFA) A.