Download ECO 173 Chapter 8: Continuous Probability Distributions Lecture 3a

ECO 173
Brief review of Discrete Probability Distributions
&
Chapter 8: Continuous Probability Distributions
Lecture 2a
Instructor: Naveen Abedin
1
Recap from last class
• Random Variable: A function or rule that assigns a number to
each outcome of an experiment.
• Example: the dual coin toss experiment. The variable we are
interested to measure is the number of heads appearing in
each outcome of the experiment. The possible values of the
variable are 0, 1 and 2
Coin 1
Coin 2
Total Number of
Heads (X)
H
H
T
T
H
T
H
T
2
1
1
0
2
Probability Distribution
• A table, formula or graph that contains/describes
the values a random variable can take, and the
probability associated with these values is called
probability distribution.
Notation:
• X : symbol for random variable
• x : symbol for observation values that the
random variable can take
• P: symbol for probability. P(X = x) or P(x) denotes
the probability that the random variable X can
attain the value x.
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Probability Distribution (cont.)
• The probability distribution of the dual coin toss
experiment would be as such:
• Let X be the random variable that denoted the
number of heads obtained from the dual coin toss.
The possible observation values X can take are x = {0,
1, 2}
Probability Distribution of X (number of heads)
x
P (X = x)
0
0.25
1
0.5
2
0.25
Total
1
4
Probability Distribution (cont.)
• 2 fundamental requirements of a Probability
Distribution of a Discrete Random Variable:1. 0 ≤ P(x) ≤ 1, for all x
2. ∑ P(x) = 1 (sum of all probabilities must equal
to 1)
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Probability Distribution (cont.)
• Example 1: 116 million houses were surveyed. A list of the
number of household members, and the corresponding
number of houses are listed in the table below:
Number of Household Members
(X)
Number of Households (millions)
1
31.1
2
38.6
3
18.8
4
16.2
5
7.2
6
2.7
7 or more
1.4
Total
116
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Probability Distribution (cont.)
• Let X be the random variable that denotes the number of
household members. The values that X can take are x = {1, 2,
3, 4, 5, 6, 7 or more}. Probability distribution of X is as follows:
x
P(x)
1
31.1/116 = 0.268
2
38.6/116 = 0.333
3
0.162
4
0.140
5
0.062
6
0.023
7 or more
0.012
Total
1
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Expected Value & Mean
• The typical method for computing mean is the
formula:-
• However, when we are dealing with a very
large population, i.e. N is very large, this
method for computing average may not be
feasible
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Continuous Random Variable
• A continuous random variable is one that can
assume an uncountable number of values.
(Since there are infinite possible number of values the random variable
can take, we cannot make a list of values like in discrete random variable
case.)
• Since x can take an infinite possible number of
values, the probability of each individual value
is virtually 0 (e.g. P(X = 5) = 0). Hence, for
continuous probability distributions, we
calculate the probability of a range of values
(e.g. P(5 ≤ X ≤ 10) = ?) instead of a specific
value like P(X=5).
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Continuous Random Variable (cont.)
• In order to find probability of ranges of values
of a particular continuous random variable,
we need two things:1. Set fixed intervals/range
2. Calculate relative frequency within the range
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Probability Density Function
• Example 1: The following histogram records telephone
bills for long-distance calls. There are a total of 200
records.
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Probability Density Function (cont.)
• A probability distribution for $15 intervals can
be presented in a table:
Billing Interval ($)
Frequency
(Number of Houses)
Relative Frequency
(Probabilities)
0 ≤ X ≤ 15
71
71/200 = 0.355
15 < X ≤ 30
37
37/200 = 0.185
30 < X ≤ 45
13
13/200 = 0.065
45 < X ≤ 60
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9/200 = 0.045
60 < X ≤ 75
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10/200 = 0.05
75 < X ≤ 90
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18/200 = 0.09
90 < X ≤ 105
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28/200 = 0.14
105 < X ≤ 120
14
14/200 = 0.07
Total
200
1
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Probability Density Function (cont.)
• In order to make the area under the histogram equal to 1, we need to
rescale the vertical axis, by dividing the relative frequency by the interval,
15. This is going to make the area of each bar equal to the relative
frequency, and as shown in the table, the sum of relative frequencies is
equal to 1.
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Probability Density Function (cont.)
• This is helpful to find probabilities between
ranges of values that overlap intervals. Find P
(50 ≤ X ≤ 80).
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Probability Density Function (cont.)
• Suppose now that we increase the number of
intervals. This will cause the width of each
interval to shrink.
• This will help us gradually get an idea of a
smooth curve overlaying the histogram. This
curve is called the probability density function
(p.d.f). The area under p.d.f is equal to the
probability of an interval. Therefore, the total
area under the p.d.f curve is equal to 1.
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Probability Density Function (cont.)
• Requirements for probability density function,
whose range is from a ≤ x ≤ b.
1. f(x) ≥ 0 for all x between a and b, (i.e. relative
frequency/interval width cannot be negative)
2. Total area under the curve between a and b
must equal to 1.
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Normal Distribution
• The range of this function is from −∞ to ∞
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Normal Distribution (cont.)
• The normal distribution is primarily described through two
parameters – the mean (μ) and the standard deviation (σ).
• The mean, μ, is the value at the center of the normal
distribution. Hence, increasing the mean shifts the entire
curve to the right, and decreasing the mean shifts the entire
curve to the left.
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Normal Distribution (cont.)
• Standard deviation, σ, is the square root of variance,
measuring the degree of deviation from the mean.
Larger values of σ makes the curve wider, while
smaller values of σ makes the curve narrower.
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Normal Distribution (cont.)
• As before, to calculate the probability that a
normal random variable falls into any interval,
the area under the curve has to be calculated
for that interval.
• In order to calculate probabilities of random
variables following normal distribution, we
first need to standardize the random variable.
• A variable can be standardized by subtracting
the mean from the variable (X) and dividing it
by the standard deviation.
21
Normal Distribution (cont.)
• When a normal random variable is standardized, it is called a
standard normal random variable, and is denoted by the
symbol, Z.
• Once a normal random variable has been standardized, the
mean of the standard normal variable is 0, and the standard
deviation is equal to 1.
22
Normal Distribution (cont.)
• Example 4: Demand for gasoline is normally
distributed with a mean of 1000 gallons and
standard deviation of 100 gallons. The
manager noted that there is exactly 1100
gallons of regular gasoline in storage. The next
delivery is scheduled later today at the close
of business. The manager would like to know
the probability that he will have enough
regular gasoline to satisfy today’s demand.
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Normal Distribution (cont.)
• Through the process of standardization we
convert the distribution on the left to the one
on the right.
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Normal Distribution (cont.)
25
Normal Distribution (cont.)
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Probability ZA
• ZA represents the value z such that the area to its
right under the standard normal curve is A.
• To find A, we use the standard normal probabilities
table backwards – we locate the probability from the
table and identify the corresponding value of Z for
that probability.
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Standard Deviation
• Standard deviation is a measure that is used to
quantify the amount of variation or dispersion of a set
of data values. A low standard deviation indicates that
the data points tend to be close to the mean of the set,
while a high standard deviation indicates that the data
points are spread out over a wider range of values.
• Standard deviation is expressed in the same units as
the data.
• For example, each of the three populations {0, 0, 14,
14}, {0, 6, 8, 14} and {6, 6, 8, 8} has a mean of 7. Their
standard deviations are 7, 5, and 1, respectively. The
third population has a much smaller standard deviation
than the other two because its values are all close to
7.
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68-95-99.7 Rule
• The 68-96-99.7 rule is a shorthand to
remember the percentage of values that lie
within one, two and three standard
deviations, respectively, of the mean of a
normal distribution.
• The rule can be used as a simple test for
outliers. For example, it might be considered
in a study that data points that fall more than
3 standard deviations from the mean are likely
outliers.
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68-95-99.7 Rule (cont.)
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68-95-99.7 Rule (cont.)
• Example 5: Miss N’s scores in Chemistry this
semester were rather inconsistent: 100, 85,
55, 95, 75, 100. For this population, how many
scores are within one standard deviation of
the mean?
• Answer: All but 55 lie within 1 standard
deviation of the mean. This means 100, 85,
85, 75, and 100 lie within 68% of the mean
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