Continuous Random Variables
Consider the following table of sales,
divided into intervals of 1000 units each,
interval
(0,1000]
(1000,2000]
(2000,3000]
(3000,4000]
(4000,5000]
(5000,6000]
(6000,7000]
and the relative frequency of each interval.
interval
(0,1000]
(1000,2000]
(2000,3000]
(3000,4000]
(4000,5000]
(5000,6000]
(6000,7000]
relative
freq.
0
0.05
0.25
0.30
0.25
0.10
0.05
1.00
We’re going to divide the relative frequencies
by the width of the cells (which here is 1000).
This will make the graph have an area of 1.
interval
(0,1000]
(1000,2000]
(2000,3000]
(3000,4000]
(4000,5000]
(5000,6000]
(6000,7000]
relative f(x)  relativefreq.
freq.
cell width
0
0.05
0.25
0.30
0.25
0.10
0.05
0
0.00005
0.00025
0.00030
0.00025
0.00010
0.00005
Graph
interval
f(x) 
relativefreq.
cell width
f(x) = p(x)
0.00030
(0,1000]
0
(1000,2000]
0.00005
(2000,3000]
0.00025
(3000,4000]
0.00030
0.00015
(4000,5000]
0.00025
0.00010
(5000,6000]
0.00010
(6000,7000]
0.00005
0.00025
0.00020
0.00005
0
0
1000 2000 3000 4000 5000 6000 7000
sales
The area of each bar is the frequency of the category,
so the total area is 1.
Graph
interval
f(x) 
relativefreq.
cell width
f(x) = p(x)
0.00030
(0,1000]
0
(1000,2000]
0.00005
(2000,3000]
0.00025
(3000,4000]
0.00030
0.00015
(4000,5000]
0.00025
0.00010
(5000,6000]
0.00010
(6000,7000]
0.00005
0.00025
0.00020
0.00005
0
0
1000 2000 3000 4000 5000 6000 7000
sales
Here is the frequency polygon.
If we make the intervals 500 units instead of 1000,
the graph would probably look something like this:
f(x) = p(x)
The height of the
bars increases and
decreases more
gradually.
sales
If we made the intervals infinitesimally small, the
bars and the frequency polygon would become
smooth, looking something like this:
f(x) = p(x)
This what the distribution
of a continuous random
variable looks like.
This curve is denoted f(x)
or p(x) and is called the
probability density
function.
sales
pmf versus pdf
For a discrete random variable, we had a
probability mass function (pmf).
The pmf looked like a bunch of spikes, and
probabilities were represented by the
heights of the spikes.
For a continuous random variable, we have
a probability density function (pdf).
The pdf looks like a curve, and probabilities
are represented by areas under the curve.
Pr(a < X < b)
f(x) = p(x)
a
b
sales
A continuous random variable has an infinite
number of possible values & the probability
of any one particular value is zero.
If X is a continuous random variable,
which of the following probabilities is largest?
(Hint: This is a trick question.)
1.
2.
3.
4.
Pr(a < X < b)
Pr(a ≤ X < b)
Pr(a < X ≤ b)
Pr(a ≤ X ≤ b)
They’re all equal.
They differ only in whether they include the
individual values a and b, and any one particular
value has zero probability!
Properties of
probability density functions (pdfs)
1. f(x) ≥ 0 for values of x
This means that when we draw the pdf
curve, while it may be on the left side of
the vertical axis (have negative values of
x), it can not go below the horizontal axis,
where f would be negative.
Pr( - ∞ < X < ∞) = 1
The total area under the pdf curve, which
corresponds to the total probability, is 1.
Example
f(x) = 2
f(x) = 0
if 1 ≤ x ≤ 1.5
otherwise
and
This function satisfies both the properties of pdfs.
First, it’s never negative.
f(x)
Second, the total area under the curve is (1/2) (2) = 1.
2.0
0
1.0
1.5
x
Cumulative Distribution Function for a
Continuous Random Variable
F(x) = Pr(X ≤ x) = area under the f(x) curve
up to where X=x.
Rectangle Example: What is F(1.2)?
F(1.2) = Pr(X ≤ 1.2)
= the area under the pdf up to where x is 1.2.
f(x)
2.0
0
1.0
1.5
x
Rectangle Example: What is F(1.2)?
F(1.2) = Pr(X ≤ 1.2)
= the area under the pdf up to where x is 1.2.
= (0.2) (2.0)
f(x)
= 0.4
2.0
0
1.0 1.2
1.5
x
Continuous Uniform Distributions.
Distributions like our rectangle example are called uniform
distributions.
Notice that there are both discrete uniform distributions
that we discussed earlier and continuous uniform
distributions.
The continuous uniform distribution has the following form:
1
f ( x) 
ba
if a  x  b
f ( x)  0
otherwise
f ( x)
Notice that the area of the rectangle
will always be 1 because
1
ba
0
area = length • width
 1 
 (b  a ) 
 1
ba
a
b
x
Mean, variance, and standard deviation
of the continuous uniform distribution
mean:
ab

2
variance:
2
(
b

a
)
2 
12
standard deviation:
(b  a )2

12
The most famous distribution is the
Normal or Gaussian distribution.
Its probability density function (pdf) is
f ( x) 
1
 2
e
1
 [( x   ) /  ]2
2
for -   x  
 is the mean of the distribution,  is the standard deviation,
e  2.718, and   3.14 .
It is sometimes denoted N (, 2), which means the normal
distribution with a mean of  and a variance of 2.
If you have three normal distributions with
the same standard deviation (same spread),
but different means (different averages),
they would look like this:
1
2
3
If you had the same mean but different standard
deviations, it would look like this:
large standard deviation
If you had the same mean but different standard
deviations, it would look like this:
medium standard deviation
large standard deviation
If you had the same mean but different standard
deviations, it would look like this:
small standard deviation
middle standard deviation
largest standard deviation
Keep in mind that the
areas are all the same,
since they all equal 1.
Recall that if a random variable has mean 
and standard deviation , then (X-)/ has
mean 0 and standard deviation 1.
If X is normally distributed, then (X-)/ will
be standard normal, N(0,1), normal with
mean 0 and variance 1.
This theorem is extremely useful.
It means that we don’t need to use the
messy normal formula.
We can standardize any normal distribution
and look up probabilities in tables for the
standard normal distribution.
Using the standard normal table is not difficult,
but it takes practice to get accustomed to it.
The table in your textbook gives probabilities that
the standard normal (often called Z) is less than
a particular number, that is Pr(Z ≤ a).
Some tables are set up differently, so you need to
notice how a table is computed when you use it.
For example, a book we used previously gave
probabilities that Z is between zero and a
positive number, that is, Pr(0 ≤ Z ≤ a).
Given any setup, you can always calculate the
probabilities that you need.
Z table: You get the integer part & the
1st decimal from the left column & the
second decimal from the top row.
Example: Pr(Z ≤ 2.63) = 0.9957
0.9957
?
0
2.63
Z
z .00 .01 .02 .03 .04 .05 .06 .07 .08 .09
0.0
0.1
0.2
2.6
3.0
.9957
Do not memorize a lot of rules.
You just need to remember 2 easy facts.
1. The graph is symmetric about 0.
2. The total area under the curve is 1.
0
Z
Example
Pr(Z < 1.85)
0
1.85
Z
Example
Pr(Z < 1.85) = 0.9678
0.9678
0
1.85
Z
Example
Pr(0 < Z < 1.85)
0.9678
0
1.85
Z
Example
Pr(0 < Z < 1.85) = 0.9678 – 0.5 = 0.4678
0.5
0.4678
0.9678
0
1.85
Z
Example
Pr(Z > 1.85)
0.9678
0
?
1.85
Z
Example
Pr(Z > 1.85) = 1 - 0.9678 = 0.0322
0.9678
0.0322
0
1.85
Z
Example
Pr(Z < -1.85)
There are two ways to do this.
0.9678
0.0322
-1.85
0
1.85
Z
Pr( Z < -1.85) = Pr(Z > 1.85) = 0.0322
Example
The second way to determine Pr(Z < -1.85)
is directly from the negative part of the Z table.
-1.85
0
Pr( Z < -1.85) = 0.0322
Z
Example
Pr(Z > -1.85)
-1.85
0
1.85
Z
Example
Pr(Z > -1.85) = Pr(Z < 1.85) = 0.9678
-1.85
0
1.85
Z
Example
Pr(-1< Z < 2)
-1.00
0
2.00
Z
Example
Pr(-1< Z < 2) = 0.9772 - 0.1587 = 0.8185
0.1587
0.9772
-1.00
0
2.00
Z
Example
Pr(1< Z < 2)
?
0
1
2
Z
Example
Pr(1< Z < 2) = 0.9772 - 0.8413 = 0.1359
0.8413
0.9772
0
1
2
Z
Example
Pr( Z < 7)
0
7
Z
Example
Pr( Z < 7) = 1.0000
(to 4 decimal places)
0
7
Z
Example
Pr( Z > 7)
0
7
Z
Example
Pr( Z > 7) = 0.0000
(to 4 decimal places)
0
7
Z
Example
Pr(0 < Z < 7)
0
7
Z
Example
Pr(0 < Z < 7) = 0.5000
0
(to 4 decimal places)
7
Z
Example
What is the value of a such that Pr(Z < a) = 0.9207 ?
0.9207
0
a
Z
Example
What is the value of a such that Pr(Z < a) = 0.9207 ?
0.9207
0
1.41
Z
Example
What is the value of b such that Pr(Z > b) = 0.0250 ?
0.0250
0
b
Z
Example
What is the value of b such that Pr(Z > b) = 0.0250 ?
1 – 0.0250 = 0.9750
0
0.0250
b
Z
Example
What is the value of b such that Pr(Z > b) = 0.0250 ?
1 – 0.0250 = 0.9750
0
0.0250
1.96
Z
Example
What is the value of k such that Pr(0 < Z < k) = 0.4750 ?
0.4750
0.5 + 0.4750 = 0.9750
0
k
Z
Example
What is the value of k such that Pr(0 < Z < k) = 0.4750 ?
0.4750
0.5 + 0.4750 = 0.9750
0
1.96
Z
Example:
If X is N(2, 9), determine Pr(X ≤ 5).
Pr(X  5)
 X- 5 
 Pr 


 
 
 X- 52
 Pr 


3 
 
 Pr(Z  1)
 0.8413
0.8413
0
1.00
Z
Useful Fact
The distribution of the individual observation is the
same as the distribution of the population from
which it was drawn.
For example, if the mean height of a population of
men is 70 inches, then the expected value or
mean of a randomly selected man will be 70
inches.
Also, if 5% of the population of men is over 78
inches, then the probability that a randomly
selected man will be over 78 inches tall is 5%.
Example: Suppose that women’s heights are
normally distributed with mean 64 inches & standard
deviation 3 inches. What is the probability that a
randomly selected woman is under five feet tall?
Pr(X  60)
 X -  60   
 Pr 


 
 
 X -  60  64 
 Pr 


3 
 
 Pr(Z  1.33)  0.0918
-1.33
0
Z
Sample Distribution of X
the probability distribution of all possible values of
that could occur when a sample of size n is taken
from some specified population
x
Example: Suppose we have a population of chips,
1/3 of which have a 1 on them, 1/3 have a 2, & 1/3
have a 3.
(a) Show in table form the distribution of the sample mean
(with n=2, sampled with replacement).
(b) Graph the distribution of the sample mean.
(c) Graph the distribution of the original population of chips.
(d) What are the mean & variance of the original population?
(e) What are the mean & variance of the sample mean?
Show in table form the distribution of the sample
mean (with n=2, sampled with replacement).
sample
1,1
1,2
2,1
1,3
3,1
2,2
2,3
3,2
3,3
x
sample mean
1.0
1.5
1.5
2.0
2.0
2.0
2.5
2.5
3.0
x
sample mean
1.0
1.5
2.0
2.5
3.0
probability
1/9
2/9
3/9
2/9
1/9
Graph the distribution of the sample mean.
x
sample mean
1.0
1.5
2.0
2.5
3.0
Probability
probability
1/9
2/9
3/9
2/9
1/9
3/9
2/9
1/9
0
0.5
1.0
1.5
2.0
2.5
sample mean
x
3.0
Graph the distribution of the original population of chips.
Since there are three equally likely values (1, 2, and 3),
the distribution looks like this.
Probability
1/3
0
1
2
3
x
What are the mean & variance
of the original population?
x
1
2
3
p(x)
1/3
1/3
1/3
xp(x)
1/3
2/3
3/3
=6/3=2
What are the mean & variance
of the original population?
x
1
p(x)
1/3
xp(x)
1/3
2
1/3
2/3
3
1/3
x2
x2p(x)
1
1/3
4
4/3
9
9/3
E(X2) =14/3
3/3
V(X) = E(X2)- [E(X)]2
2
=
14/3
–
2
=6/3=2
= 2/3
What are the mean & variance of the sample mean X ?
x
p(x)
1.0
1/9
1/9
1.5
2/9
3/9
2.0
3/9
6/9
2.5
3.0
2/9
1/9
xp(x)
E( X )  18 / 9  2
5/9
3/9
What are the mean & variance of the sample mean X ?
2
x
p(x)
1.0
1/9
1/9
1.5
2/9
3/9
2.0
3/9
6/9
2.5
3.0
2/9
1/9
xp(x)
E( X )  18 / 9  2
2
x
x p(x)
1.00
2.25
4.00
6.25
9.00
0.111
0.500
1.333
1.389
1.000
2
E ( X )  4.333
2
5/9 2
2
V ( X )  E ( X )  [ E ( X )]
3/9
2
 4.333  2
 0.333 or 1/3
In our chip example, we found that
E( X )  2
E( X )  2
2
V (X ) 
3
1
V (X ) 
3
The expected value (or mean) of the original population
and the sample mean are the same.
However, the variance of the sample mean is smaller
than the variance of the original population.
In general,
E(X )  
V (X )  
2
E(X )  
V (X ) 

2
n
The mean of the original population and the mean of the
sample are the same.
However, as long as the sample size is more than one,
the variance of the sample mean is smaller than the
variance of the original population.
Intuition: Suppose that each person in the class
randomly sampled 50 men from a population of
men whose average height is 70 inches. Each
student then calculated his/her sample mean.
If you averaged together all the sample means, you’d expect
to get something very close to 70 inches.
The expected value or mean of the original population and
the expected value of the sample mean are the same.
Why does the sample mean have a smaller
variance than the original population?
Suppose 5% of the population is taller than 78 inches (6’6”).
Then there’s a 0.05 probability of selecting at random an
individual with a height over 6’6” inches.
It is much less likely that you will select at random 50 men
whose average is over 6’6”.
You get a mixture of tall guys & short guys, so your average
tends to be pretty close to the average for the population.
So the distribution of the sample mean clusters more tightly
about the average than does the distribution of the original
population.
That is, the variance of the sample mean is smaller than the
variance of the original population.
Central Limit Theorem
As the sample size n increases, the distribution of
the sample mean X of a random sample from a
population (not necessarily normal) with mean 
and variance 2 approaches normal with mean 
and variance 2/n.
Example: Suppose the grades of a large class
have a mean of 72 and a standard deviation of 9.
a. What is the probability that the average grade of a
random sample of 25 students will be above 77?
b. If the population is normal, what is the probability that
an individual student drawn at random will have a
grade over 77?
Notice that in part b, we need to assume normality but we
didn’t in part a.
This is because the central limit theorem assures us of an
approximately normal distribution for the sample mean
of a reasonably large sample, but not for the
distribution of a single observation.
To be assured of that, we need to know that the distribution
of the original population was normal.
What is the probability that the average grade of a
random sample of 25 students will be above 77?
Pr(X  77)


 X   77  72 

 Pr

9
 



n
25


 PrZ  2.78)
= 1 - 0.9973
0.9973
0
2.78
Z
= 0.0027
If the population is normal, what is the probability
that an individual student drawn at random
will have a grade over 77?
 X   77  72 


Pr(X  77)  Pr
9 
 
 PrZ  0.56)
= 1 - 0.7123
0.7123
0
0.56
Z
= 0.2877
Notice
The probability of selecting an individual at
random who has a grade five points above
the class mean is much greater than the
probability of randomly selecting a sample
of 25 students who have an average that
is five points above the class mean.
(0.2877 versus 0.0027)
Problem
Suppose we are sampling (without replacement) and we
sample the entire population.
Then our sample mean will always be the same as the
population mean. No spread. Zero variance.
If we don’t sample the entire population but do sample a
large part of it, our variance will not be zero but it will be
very small.
Thus, as the sample size n approaches the population size
N, the variance of the sample mean approaches zero.
Our formula for the variance of the sample mean was 2/n.
2/n approaches zero as n approaches infinity, not as n
approaches N.
So we need to make some adjustment when we sample a
large part of our population.
Finite Population Correction Factor
When we sample a substantial part of
our population (more than 5%), we need
to multiply the standard deviation of our
sample mean by this factor:
N n
N 1

So the standard deviation of the
sample mean which was:
becomes:
n

n
N n
N 1
The standard deviation of the sample
mean adjusted for a large sample

n
N n
N 1
Notice that if you sample just one observation
(n=1), the square root part becomes 1, and the

formula becomes the old formula:
n
If you sample the entire population (n=N), the
formula has a value of 0.
So the adjustment works the way it should.
So our standardiz ed X is now
X

n
Nn
N 1
Example: What is the probability that the mean of a
sample of 36 observations, from a population of 300, will be
less than 14, if the population mean & population standard
deviation are 15.30 & 4.10 respectively?
If we sample more than 5% of our population, we need to use the finite
population correction factor, and here we are sampling 12%.
Pr(X  14.00)



X 
14.00  15.30
 Pr

 
N n
4.10 300  36

36 300  1
 n N 1







 PrZ  2.02)
= 0.0217
-2.02
0
Z
Up until now, as long as our sample
was not too large, we have used this
standardization formula:
Z
X 

n
However, what if we don’t know
what the population standard
deviation  is?
The next best thing is the
sample standard deviation:
But when we use s instead of , the
result is not a normally distributed
variable.
It has what is called a t or
student’s t distribution.
n
s
(X  X )
i 1
i
n 1
X 
t
s
n
2
Our t distribution looks very similar to the Z distribution.
Both are symmetric, bell-shaped, & have a mean
of zero.
But while the Z has a variance of 1, the t has a
variance of (n-1)/(n-3), which is greater than one.
So the t is wider than the Z.
So, the numbers are different & we need to use a
different table.
The t distribution is tabulated based on
the number of “degrees of freedom.”
n
The number of degrees of freedom, is
denoted by dof, df, or the Greek letter nu:
In this context, the number of degrees of freedom
is n-1, where n is the number of observations.
For each number of degrees of freedom, there is a
different t distribution.
The degrees of freedom are often indicated as a
subscript on the t.
For example, t15 is a t distribution with 15 degrees
of freedom.
When the sample size is large, the Z and t
distributions are virtually indistinguishable.
When we have at least 100 observations, we
will be able to use values from the Z table for
our t.
Some people use the Z to approximate the t
when there are 30 or more observations.
When we’re working with s, I prefer to use the t
until 100 observations.
When you take Intermediate Statistics (EC252),
make sure you know what your instructor uses.
The t distribution is set up differently from
the Z table.
The table in your text book gives probabilities that
the t is greater than a specified positive number,
that is, Pr(t ≥ a).
Some tables are set up differently, so you need to
notice how a table is computed when you use it.
Example:
For 6 observations or 5 df,
Pr(t5 > 3.3649) = 0.01
0
a
d.f. = 1
2
3
4
5
6
100
∞
0.10
0.05
0.025
t5
0.01
3.3649
3.3649
0.005
Example: What is the probability that the sample
mean is less than 800, if the population mean is
843.3419 and the sample standard deviation is
105, based on a sample of 25.
Pr(X  800)


X   800  843.3419 
 Pr 


105
 s

n
25


-2.0639
0
t24
 Pr( t24  2.0639 )
 0.025
Using Excel to solve t distribution problems
On an Excel spreadsheet, you can get the t
distribution as follows:
Either click the fx (insert function) button OR
click the formulas tab, and then click insert function
select statistical as the category of function, scroll
down to the t.dist or t.inv function, and click on it
fill in the information in the dialog box .
Using Excel to solve t distribution problems
We’re going to look at Excel’s two-tailed t functions.
There are also some other t functions on Excel
that you can explore.
t.dist.2t - you provide the number x on the
horizontal axis and the degrees of freedom, and
it then gives you the area or probability in the two
tails. (Note that the number you provide must be
positive.)
t.inv.2t - you provide the probability or area that
you want in the two tails and the degrees of
freedom, and it then gives you the number on the
horizontal axis.
Example: Suppose that you wanted to use
Excel to find the probability that a t with 24
degrees of freedom is less than -2.0639.
Note that the area to the left of
-2.0639 is the same as the area
to the right of +2.0639.
Select t.dist.2t . Specify
2.0639 as x, 24 for degrees of
freedom.
-2.0639
0
2.0639
t24
Excel will provide you the
probability value of 0.05. Since
you only want the left tail, the
number you want is half of 0.05
or 0.025 .
Example: Suppose instead you wanted to
use Excel to find out for what value of k
Pr(t24 < -k or t24 > k) is equal to 0.05.
Select t.inv.2t . Specify 0.05
for probability, and 24 for
degrees of freedom.
-k
0
k
t24
Excel will provide you the value
on the horizontal axis of
2.0639.
Normal Approximations
Recall that in the section on discrete distributions,
we discussed the binomial distribution, which we
used when we were considering the number of
successes (x) on n independent trials with a
probability of success on any given trial  .
We said that the binomial distribution is
approximately symmetric if n ≥ 5 and n(1) ≥ 5 .
We also said that
the mean of the binomial distribution is  = n , and
the variance is 2 = n(1),
so the standard deviation is   n (1- ) .
It is sometimes easier to use the normal
distribution to approximate a binomial probability
than to calculate the precise binomial probability.
The normal distribution provides a reasonable
approximation for the binomial if n ≥ 5 and n(1) ≥ 5.
However, because the binomial is discrete and the normal
is continuous, we make an adjustment which we call the
“continuity correction”.
Recall that for discrete probability distributions, the
probability that the variable takes on a particular value x
is the height of a spike at that point. However, for a
continuous probability distribution, the probability is the
area under the probability density curve.
To adjust for that difference, we include an extra half a unit
on the ends of our intervals when we calculate normal
probabilities as approximations for binomial probabilities.
Examples
(1) To estimate binomial probability Pr(X = 20),
calculate normal probability Pr(19.5 ≤ X ≤ 20.5).
(2) To estimate binomial probability Pr(3 ≤ X ≤ 7),
calculate normal probability Pr(2.5 ≤ X ≤ 7.5).
(3) To estimate binomial probability Pr(X ≥ 10),
calculate normal probability Pr(X ≥ 9.5).
(4) To estimate binomial probability Pr(X < 15),
calculate normal probability Pr(X < 15.5).
Example: Suppose that the probability that any
individual in a given population supports a particular
proposition is 0.20. Consider a group of 500 people.
Determine the probability that at least 75 people
support the proposition.
In this problem, n = 500 and   0.20.
So n = 100 ≥ 5 and n(1) = 400 ≥ 5 , and this should
be an excellent approximation.
The mean is n = 100 and the standard deviation is
 
n (1- ) = 500(0.20)(0.80)  80  8.9443 .
For the continuity correction, estimate the binomial
probability Pr(X ≥ 75) using the normal probability
Pr(X ≥ 74.5).
We need to standardize the normal by subtracting off the
mean and dividing by the standard deviation.
So we have this:
Pr(X ≥ 74.5) = Pr[(X- )/ ≥ (74.5 – 100)/8.9443]
= Pr(Z ≥ -2.85)
0.9978
-2.85
0
Z
= 1- Pr(Z < -2.85)
= 1 - 0.0022 = 0.9978
So it’s extremely likely that at least 75 people in
a sample of 500 will support the proposition.