Continuous Probability Distributions

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Continuous Probability Distributions

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Continuous Probability Distributions And The Central Limit Theorem The Binomial Distribution From the Binomial Probability Tables we have the following: n x p .01 ... – PowerPoint PPT presentation

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Title: Continuous Probability Distributions


1
Continuous Probability Distributions And The
Central Limit Theorem
2
Continuous Distributions
In all of the previous examples that we have
studied so far, we have been dealing with samples
taken from a discrete set of population values
the possible values of the parameter of interest
were limited to a finite set of possibilities.
  • For example
  • The ages measured in years of Oscar-winning
    actors
  • The heights measured to the nearest inch of
    players in the NBA
  • The temperatures measured to the nearest degree
    of days in 2009
  • The grades expressed in integers on an exam
  • The weights expressed to the nearest tenth of a
    gram of dimes

In a discrete distribution every value in the set
of possible outcomes has a finite chance of
occurring
3
Continuous Distributions
Uniform Distribution
Suppose we discount our inability to measure
beyond a certain precision and assume that all
values of x in the range 1 ? x ? 5 are possible
outcomes for a measurement x. If we assume that
the measurement is equally likely to be anywhere
in this range, we depict the distribution as
follows
f(x)
x
4
Continuous Distributions
f(x) is called the probability density associated
with the point x. Since there are an infinite
number of real values in the line segment from 1
to 5, we cannot assign a non-zero probability to
any one value being selected. It only makes
sense to talk about the probability of x lying in
a range between two values x1 and x2
x1
x2
f(x)
x
5
Continuous Distributions
Step 1 Normalize the area in the distribution
(inside the rectangle) to 1
Area f(x) (5 1) 1 then f(x) ¼ for all
1 ? x ? 5
Step 2 Find the area between x1 2 and x2 3
P(2 ? x ? 3) f(x) (x2 x1) ¼ 1 ¼
x1
x2
f(x)
x
6
Continuous Distributions
In a continuous distribution
  • f(x) represents a probability density associated
    with a value x
  • The area under the curve f(x) is normalized to 1
  • The probability that a selected value of x lies
    in the region ?x x2 x1 is given by the
    formula P(x1 ? x ? x2) f(x) ?x

x1
x2
f(x)
x
7
Continuous Distributions
What do we mean by a probability density?
  • The probability density f(x) is the
    probability/linear length
  • The variable x may represent a value whose units
    are distance, time, weight, or some other measure
    like parts per hundred
  • The horizontal axis over which x is measured is
    marked out in gradations of the units of variable
    x. (for example if x represents weights of
    people in a population, the x-axis will be marked
    out in gradations of pounds or kilograms
  • For the uniform distribution, f(x) is a constant
    over the entire range of x, but in general, the
    probability density may vary over that range.
  • For a continuous distribution we can only find
    the probability that a selected value of x will
    lie in a range ?x between two other points.
  • P(x1 ? x ? x2) the area under the curve f(x)
    ?x
  • The probability that x is exactly equal to a
    single value x1, is 0.
  • A single point is a linear length of measure 0,
    and our formula
  • P(x1 ? x ? x2) the area under the curve f(x)
    ?x 0, since ?x 0

8
Continuous Distributions
The Normal Distribution
The most important example of a continuous
distribution is the normal distribution. The
normal distribution is
  • Bell shaped
  • Symmetrical
  • Unimodal
  • Unbounded
  • Has the property that the mean, median and mode
    are all equal

9
Continuous Distributions
Bell-shaped
10
Continuous Distributions
µ
68 of the values are within 1s of the mean
11
Continuous Distributions
µ
To convert a normal distribution with a mean of µ
and standard deviation ? into a standard normal
distribution with mean 0, standard deviation 1,
we convert all of the x-scores into corresponding
z-scores using the formula
z (x - ?)/?
12
Continuous Distributions
µ
x1 x2
Let us suppose that we have a normally
distributed population with a mean of 20 and a
standard deviation of 5 and we want to find the
probability that a selected member of this
population lies in the range of x1 21 to x2
24.5 represented by the shaded area in the above
diagram
13
Continuous Distributions
Step 1 Convert the problem of finding the
probability of selecting a value in the interval
between x1 and x2 in this normal distribution
into the probability of finding a value in the
corresponding interval in the standard normal
distribution
z2 (24.5 20) / 5 0.900
z1 (21 20) / 5 0.200
f(z)
z
0 z1 z2
14
Continuous Distributions
Step 2 Use the Standard Normal Distribution
Table on the back cover to find the probability
that a selected value of z lies in the shaded
region (same as the area under the curve inside
the shaded region of the previous diagram)
.00 .01 .02
.. .. .09
0.0 0.1 0.2 . 0.9 1.0
0.5000 0.5040 0.5080
0.5359 0.5398 0.5438
0.5478 0.5753
0.5793 0.5832 0.5871
0.6141
0.8159 0.8186 0.8212
0.8389 0.8413
Find z1 0.20 (table reads to nearest
hundredths)
15
Continuous Distributions
The area to the left of z1 (shaded blue) is
0.5793 from the table
The area to the left of z2 (shaded red) is
0.8159 from the table
The area in the interval between z1 and z2 (only
shaded red) 0.8159 0.5793
f(z)
z
0 z1 z2
P(21 ? x ? 24.5) P(0.20 ? z ? 0.90) 0.8159
0.5793 0.2366
16
The Central Limit Theorem
17
The Central Limit Theorem
Consider a population consisting of the numbers
1, 2, 3, 4, Assume there are an essentially
infinite number of each digit in the population
(sampling without replacement does not affect the
proportion of each value) and that ¼ of the
values correspond to each digit. The population
is depicted below
This distribution is clearly discrete (only 4
possible values), but we will use it to indicate
properties of the continuous sampling distribution
18
The Central Limit Theorem
First, calculate the population mean and standard
deviation of this population distribution.
? ?iP(xi) xi ¼ (1) ¼ (2) ¼ (3) ¼ ( 4)
2.5
??2 ?iP(xi) (xi - ?)2 ¼ ((1 2.5)2 (2
2.5)2 (3 2.5)2 (4 2.5)2 ) ¼
(2.25 0.25 0.25 2.25) 1.25
? ?1.25 1. 118
19
The Central Limit Theorem
Now, lets look at all possible samples of size 3
from this population
Let x the sample mean
(1,1,1) x 1 (1,1,2), (1,2,1),
(2,1,1) x 1.33 (1,1,3), (1,3,1), (3,1,1)
x 1.67 (1,1,4), (1,4,1), (4,1,1) x
2 (1,2,2), (2,1,2), (2,2,1) x
1.67 (1,3,3), (3,1,3), (3,3,1) x
2.33 (1,4,4), (4,1,4), (4,4,1) x
3 (1,2,3), (1,3,2), (2,1,3) (2,3,1), (3,1,2),
(3,2,1) x 2 (1,2,4), (1,4,2),
(2,1,4) (2,4,1), (4,1,2), (4,2,1) x
2.33 (1,3,4), (1,4,3), (3,1,4) (3,4,1), (4,1,3),
(4,3,1) x 2.67
(2,2,2) x
2 (2,3,4), (2,4,3), (3,2,4) (3,4,2), (4,2,3),
(4,3,2) x 3 (2,2,3), (2,3,2),
(3,2,2) x 2.33 (2,2,4), (2,4,2),
(4,2,2) x 2.67 (2,3,3), (3,2,3),
(3,3,2) x 2.67 (2,4,4), (4,2,4),
(4,4,2) x 3.33 (3,3,3)
x 3 (3,3,4), (3,4,3),
(4,3,3) x 3.33 (4,4,3), (4,3,4),
(3,4,4) x 3.67 (4,4,4)
x 4
20
The Central Limit Theorem
On the previous page we listed all 64 possible
samples that could be selected from our original
(discrete) distribution.
We now examine the distribution of sample means
Sample mean (x) number of occurrences
f(x) 1.00 1 1.33 3 1.67 6 2.00
10 2.33
12 2.67
12 3.00 10 3.33 6 3.67 3
4.00 1 64
21
The Central Limit Theorem
  • Given We are selecting samples of size n from
    a population (that may or may not be normal)
    with a population mean ? and a population
    standard deviation ?
  • Result
  • 1. The distribution of sample means, x, will,
    as the sample size n increases, approach a
    normal distribution.
  • The mean of all possible sample means for samples
    of size n will be the same as the population
    mean ?.
  • The standard deviation of all sample means is ?/?n

The Central Limit Theorem does not strictly apply
to samples of size lt 30 selected from a
non-Normal population, but the small sample size
(3) for our example allows us to enumerate all
possible samples from the original population and
will give approximate agreement to result 1
above, and exact agreement with results 2 and
3.
22
The Central Limit Theorem
Lets plot the frequency of occurrence of each of
the values for the sample means
12 10 8 6 4 2
The distribution is
discrete
23
The Central Limit Theorem
Now lets calculate the (population) mean and
standard deviation for the distribution of sample
means of size n 3.
?x (?xf(x) ? x)/N (1 ? 1 3 ?
1.33 6 ? 1.67 10 ? 2 12 ? 2.33
12 ? 2.67 10 ? 3 6 ? 3.33 3 ? 3.67 1 ?
4)/64 160/64 2.5
?x2 (?xf(x)/n ? (x)2 ?x2) (1 3
? 1.332 6 ? 1.672 10 ? 22 12 ? 2.332 12 ?
2.672 10 ? 32 6 ? 3.332 3 ?
3.672 42)/64 2.52 6.667 6.250 0.4167
?x ?0.4167 0.6455
24
The Central Limit Theorem
The Central Limit Theorem stipulates that ?x
? (the mean of the sampling distribution the
mean of the population from which the
samples are drawn) and ?x ? / ?n
In our example, we have found that
Mean of sampling distribution ?x 2.5 Mean of
population ? 2.5
Standard Deviation of sampling dist. ?x
0.6455 ?/?n 1.118 /?3 0.6455
25
The Central Limit Theorem
The distribution of sample means -- ?x 2.5,
?x 0.645
12 10 8 6 4 2
44/64 68.75 of values lie within ?? of mean
26
The Central Limit Theorem
The previous example serves to illustrate the
Central Limit Theorem
The sample size of n 3 was chosen so that ALL
of the samples in the sampling distribution could
be enumerated and the population of sample means
for sample size n 3 could be plotted and its
parameters calculated.
The sample size n 3 was too small to produce an
approximately normal distribution or
quasi-continuous distribution, BUT
  • Even for small samples from an original
    population that was not normal
  • The distribution of sample means is symmetric and
    somewhat bell-shaped
  • The mean of the sampling distribution equals the
    mean of the original population
  • The standard deviation of the sampling
    distribution equals the standard deviation of the
    original population divided by the square root of
    the sample size
  • The fraction of the values in the sampling
    distribution that are within 1 and 2 standard
    deviations from the mean are in substantial
    agreement with those fractions in a normal
    distribution

27
Central Limit Theorem
Suppose we change (skew) the original population
such that we have the following probabilities
x 1 p(x) 0.125 (1/8) x 2 p(x)
0.125 (1/8) x 3 p(x) 0.250 (1/4) x 4 p(x)
0.500 (1/2)
? 0.125 1 0.125 2 0.25 3 0.5 4
3.125 ?2 0.125 (1 3.125)2 0.125 (2
3.125)2 0.25 (3 3.125)2 0.5 (4
3.125)2 1.109375 ? ?1.109375 1.0533
28
Central Limit Theorem
Population mean 3.125
Population standard deviation 1.0533
29
Central Limit Theorem
The same 64 samples that can be selected from
this population now produce sample means with the
probabilities tabulated below
Sample mean (x) (x)2 p(x) x
p(x) (x)2 p(x) 1.00 1.0000
.00195 .00195 .00195 1.33 1.7778 .00586
.00781 .01042 1.67
2.7778 .01758
.02930 .04883 2.00 4.0000
.04883 .09765 .19530 2.33 5.4444
.08203 .19113 .44662 2.67 7.1111
.14063 .37498 .99995 3.00 9.0000
.20313 .60937 1.88280 3.33
11.1111 .18750 .62500
2.08333 3.67 13.4444 .18750 .68750
2.52080 4.00 16.0000
.12500 .50000 2.00000
3.12464 10.18948
?x 3.124
?2 10.18948 (3.12464)2 0.426103 ? 0.6527
30
Central Limit Theorem
Now graph the distribution of the sample means
31
Central Limit Theorem
When the initial population is not symmetric, the
distribution of sample means selected from the
original population does not produce as good an
agreement with the results of the central limit
theorem for samples of small size.
From the previous graph we see that the
distribution of sample means is not symmetric or
bell-shaped.
Population parameters ? 3.125 ? 1.0533
Distribution of sample means with n 3 ?x
3.124 ?x .6527
?x .6527 not a good fit for ?/?n
1.0533/ ?3 .608
32
The Binomial Distribution
  • Let
  • n number of trials
  • p probability of a success on any one trial
  • q probability of failure on any one trial
  • The outcome of each trial is independent of the
    outcome on preceding trials
  • x the number of successes in n trials

Then the probability of exactly x successes in n
trials is given by the term nCx pxqn-x in the
expansion of the binomial (p q)n
The value of this term for a given (n, x, p) can
be read from the Binomial Probability Tables in
the text.
33
The Binomial Distribution
Example In 1941, Ted Williams was batting
.39997(which rounded up to .400) on the last day
of the season. He was given the choice not to
play in the doubleheader scheduled for that day
and preserve his 400 average, but decided to play
anyway and got 6 hits in 8 (official) at bats to
finish the year hitting 406. What was the
probability that Ted would go 6 for 8 on that
date? What was the probability that he would end
up hitting 400 if he were to get 8 official at
bats on that day?
Assume the outcome of each at bat was independent
of any other
n 8 number of trials x 6 number of
successes (hits) p 0.40 probability of a hit
in any at bat q 1 p 0.60 probability of an
out in each at bat
34
The Binomial Distribution
From the Binomial Probability Tables we have the
following
.. 8 0 1 2
3 4 5 6
7 8
Probability of exactly 6 hits in 8 AB 0.041
0 0
.005 .046 .136
.232 .273 0 0
0 .009 .047 .124
.219 0 0 0
.001 .010 .041
.109 0 0 0
0 .001 .008
.031 0 0 0
0 0 .001
.004
35
The Normal Approximation of the Binomial
Distribution
The discrete Binomial Distribution may be
approximated using the continuous Normal
Distribution if the following are true
  • np ? 5, and
  • nq ? 5

Then the binomial distribution may be
approximated by a normal distribution with mean ?
np and standard deviation ? ?npq
36
The Normal Approximation of the Binomial
Distribution
Example Passenger Load on a Boeing 767-300 The
aircraft has a seating capacity of 213 seats.
When fully loaded, with passengers, crew, cargo,
and fuel, the pilot must verify that the gross
weight is below the maximum allowable limit, and
the weight must be properly distributed so that
balance of the aircraft is within safe acceptable
limits. Instead of weighing the passengers,
their weights are estimated according to FAA
rules.
Men have a mean weight of 172 pounds Women have a
mean weight of 143 pounds
Assume that if there are 122 or more male
passengers in the 213 seats, the load may have to
be adjusted. Assume also that the probability of
a man or woman booking a seat is 0.5. What is
the probability of having 122 or more male
passengers if the plane is fully booked?
Step 1 np nq (213)(0.5) 106.5 gt 5 Normal
approximation applies
37
The Normal Approximation of the Binomial
Distribution
Step 2 Find the mean and standard deviation of
the approximate normal distribution
  • np 106.5
  • ? ?npq ?(213)(0.5)(0.5) 7.297

Step 3 Represent the discrete value 122 by the
interval (121.5,122.5)
Step 4 Find probability that x ? 121.5 in this
distribution
Convert 121.5 to a z-score z (x - ?)/?
(121.5 106.5)/7.297 2.06
Step 5 From the Standard Normal Table P(z ?
2.06) 1 0.9803 0.0197
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