STA 2023 Elementary Statistics
Lecture Notes
Chapter 4 – Discrete Probability Distributions
Professor Achenbach
Random Variables
A random variable is a variable whose value depends on the outcome of a probability
experiment. As in algebra, random variables are represented by letters.
Examples:




T = the number of tails when a coin is flipped 3 times.
s = the sum of the values showing when two dice are rolled.
h = the height of a woman chosen at random from a group.
V = the liquid volume of soda in a can marked 12 oz.
There are two basic types of random variables:
Discrete Random Variables – have a finite or countable number of possible values.
Continuous Random Variables – can take on any value in some interval.
Examples:


The variables T and s from above are discrete random variables
The variables h and V from above are continuous random variables.
Probability Distributions of Discrete Random Variables
A probability distribution for a discrete random variable x is a discrete is a list of each
possible value for x together with the probability that when the experiment is run, x will
have that value. This probability is denoted by P ( x ) .
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Examples:

As above, let T be the random variable that represents the number of tails
obtained when a coin is flipped three times. Then T has 4 possible values: 0, 1, 2,
and 3. The probability distribution for T is given in the following table:
T
0
1/8
P(T )

1
3/8
2
3/8
3
1/8
A statistics class of 25 students is given a 5 point quiz. 3 students scored 0, 1
scored 1, 4 scored 2, 8 scored 3, 6 scored 4, and 3 students scored 5. If a student
is chosen at random, and the random variable s is the student’s quiz score then the
probability distribution of s is:
s
P( s)
0
0.12
1
0.04
2
0.16
3
0.32
Note: For any discrete random variable x:
0  P( x)  1 and
4
0.24
5
0.12
 P( x)  1
Finding Probabilities from a Probability Distribution
Since a random variable can only take on one value at a time, the events of a variable
assuming two different values are always mutually exclusive. The probability of the
variable taking on any number of different values can thus be found by simply adding the
appropriate probabilities.
Exercises:

Find the probability that a student scored 3 or more on the quiz from the previous
example.

Find the probability that a student did not get a perfect score.

Find the probability of getting 2 or more tails when a coin is flipped 3 times.

Find the probability of getting at least one tail.

Find the missing probability in the following distribution:
X
P( X )
-3
0.21
0
0.15
5
2
13
0.33
Mean or Expected Value
The mean or expected value of a random variable x is the average value that we should
expect for x over many trials of the experiment.
Notation: The mean or expected value of a random variable x will be represented by
 ( x ) or E ( x)
We can calculate the mean theoretically by using the formula:
E( x)   ( x)   xP( x)
Examples:

The expected value of the random variable T from above is:
3 6 3 12 3
 1  3  3  1
E (T )   T  P(T )   0    1    2     3    0     
8 8 8 8 2
 8  8  8  8
Thus if 3 coins are flipped a large number of times, we should expect the average
number of tails (per 3 flips) to be about 1.5.

The mean of the random variable s from above is:
E ( s)   s  P( s )  (0)(.12)  (1)(.04)  (2)(.16)  (3)(.32)  (4)(.24)  (5)(.12) 
0  .04  .32  .96  .96  .60  2.88
Note that this is actually the class average on the quiz as well.
Exercise:
Suppose an instant lottery ticket is purchased for $2. The possible prizes are $0, $2, $20,
$200, and $1000. Let Z be the random variable representing the amount won on the
ticket, and suppose Z has the following distribution:
Z
P( Z )
0
2
.2
20
.05
200
.001
1000
.0001
Determine P (0) . Determine E ( Z ) and interpret its meaning. How much should you
expect to gain or lose on average per ticket?
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Variance and Standard Deviation
Often, we are also interested in how much the values of a random variable differ from
trial to trial. To measure this, we can define the variance and standard deviation for a
random variable.
For a random variable x, the variance of x, denoted by  2 ( x) can be calculated by the
formula:
 2 ( x)   ( x   )2 P( x)
The standard deviation of x, denoted by  ( x ) is just the square root of  2 ( x) .
 ( x) 
 ( x   ) P( x)
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As before, standard deviation estimates the average difference between a value of x and
the average.
Calculating Variance and Standard Deviation
The calculation of standard deviation for a random variable is similar to the calculation of
weighted standard deviation we made for data in frequency tables. (In fact, P ( x ) can
just be thought of as the relative frequency of x.)
The calculation can be made using the following steps:
1. Calculate  ( x ) .
2. Subtract the mean from each of the possible values of x. Recall that these are
called the deviations of the x values.
3. Square each of the deviations calculated in Step 2.
4. Multiply each squared deviation calculated in Step 3 by the corresponding
probability P ( x ) .
5. Sum the results of Step 4. This is  2 ( x) .
6. Take the square root of the result of Step 5 to obtain  ( x ) .
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Exercises:


Calculate the standard deviation of the random variable T from above.
Calculate the standard deviation of the random variable Z from above.
Binomial Random Variables
A discrete random variable x is said to have a binomial distribution if x satisfies the
following conditions:

An experiment is repeated for a fixed number of trials n.

All trials of the experiment are independent from one another.

All possible outcomes for each trial of the experiment can be divided into two
complementary events one S called “success” and one F called “failure”.

The probability of success P( S ) has a constant value of p for every trial and the
probability of failure P ( F ) has a constant value of q for every trial.
Note: q  1  p

The random variable x counts the number of trials on which S (success) occurred.
Examples:



Consider the experiment of flipping a coin 5 times. If we let the event of getting
tails on a flip be considered “success”, and if the random variable T represents the
1
number of tails obtained, then T will be binomially distributed with n  5 , p  ,
2
1
and q  .
2
A student takes a 10 question multiple-choice quiz and guesses each answer. For
each question, there are 4 possible answers, only one of which is correct. If we
consider “success” to be getting a question right and consider the 10 questions as
10 independent trials, then the random variable X representing the number of
1
3
correct answers will be binomially distributed with n  10 , p  , and q  .
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4
Fourteen percent of flights from Tampa International Airport are delayed. If 20
flights are chosen at random, then we can consider each flight to be an
independent trial. If we define a successful trial to be that a flight takes off on
time, then the random variable z representing the number of on-time flights will
be binomially distributed with n  20 , p  .86 , and q  .14 .
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Calculating Probabilities for a Binomial Random Variable
If X is a binomial random variable with n trials, probability of success p, and probability
of failure q, then by the Fundamental Counting Principle, the probability of any outcome
in which there are x successes (and therefore n  x failures) is:
( p  p  ...  p)  (q  q  ...  q)  p x q n  x
x successes
n x failures
To count the number of outcomes with x successes and n  x failures, we observe that the
x successes could occur on any x of the n trials. The number of ways of choosing x trials
out of n is n C x , so the probability of x successes becomes:
P( x)  n Cx p x q n  x
Examples:

As in the previous examples, let T be the random variable representing the
number of tails when a coin is flipped 3 times. Using the formula above, we can
calculate the probability of exactly 2 tails as:
2
1
1
3
1 1
P(2)  3 C2      3    .375
2
8
2 2

3
Let the random variable X represent the number correct answers on the multiplechoice quiz described above. Then the probability of a student guessing 3
answers correctly is:
3
7
1 2187
1 3
P(3)  10 C3      120  
 .25
64 16384
4 4
while the probability of guessing seven answers correctly is:
7
3
1
27
1 3
P(7)  10 C7      120 

 .003
16384 64
4 4
Exercises:
Let z be the random variable defined above as the number of on-time flights out of 20 at
Tampa International Airport.



Find the probability that 15 out of the 20 flights depart on-time.
Find the probability that only 6 out of the 20 flights depart on-time.
Find the probability that all 20 flights are on time.
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Properties of Binomial Distributions
In many cases, we are interested in the mean and standard deviation of a binomial
random variable. If x is a binomial random variable with n trials, probability of success p
and probability of failure q, then the mean and standard deviation of x can be calculated
by the following:
E ( x)   ( x)  np
and
 ( x)  npq
Example:

For T the number of tails when a coin is flipped 3 times:
1 3
E (T )    3   
2 2
and
 1  1 
3
 (T )  3    
 .866
4
 2  2 
Exercises:

Compute  ( X ) and  ( X ) for the random variable X from the multiple-choice
test example.

Compute  ( z ) and  ( z ) for the random variable z from the example using
flights from Tampa International.
Note: A binomial distribution is symmetric if p  q , left skewed if p  q and right
skewed if p  q .
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Using the TI-83 to Calculate Mean and Standard Deviation
The mean and standard deviation for a discrete random variable can be calculated in one
step using the TI-83.
Step 1: Enter the Probability Distribution
Enter the list of possible x values in L1 and the set of corresponding P ( x ) values
in L2 as was done for frequency tables in Chapter 2.
Step 2: Calculate
Choose 1-Var Stats from the CALC menu as when finding the mean of a data set
and when it appears on the screen choose L1 comma L2:
1-Var Stats L1, L2
Press [ENTER] to calculate.
Step 3: Read the Calculated Values
x
is the mean of the random variable
x
is the standard deviation of the random variable
Exercise:

Use your calculator to recalculate the mean and standard deviation of the random
variable Z from the lottery example above.
Using the TI-83 to Calculate Binomial Probabilities
If x is a binomial random variable with number of trials n, probability of success p, then
P ( x ) can be calculated on the TI-83 as follows:

Press [2nd][VARS] to choose DISTR.

From the menu use the down arrow to select 0: binompdf( and press [ENTER].
binompdf( should appear on the screen.

Enter the number of trials n, probability of success p, and desired number of
successes x separated by commas so that the screen reads: binompdf(n,p,x)

To calculate press [ENTER] again.
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Examples:


binompdf(10, .25, 3) yields .2502822876
binompdf(10, .25, 7) yields .0030899048
Creating Binomial Probability Distributions with the TI-83
Omitting the last entry x in the parenthesis when using the binompdf function will yield
a list of all binomial probabilities starting with x  0 and ending with x  n .
Example:

Entering binompdf(3, .5) yields: {.125, .375, .375, .125} the probability
distribution of the binomial experiment of flipping three coins.
Exercise:

Use the TI-83 to construct a probability distribution for the random variable X
defined above as the number of correct answers on a multiple choice quiz.
Cumulative Binomial Probabilities with the TI-83
The binomcdf function can also be accessed from the DISTR menu, and
binomcdf(n, p, x) gives the probability of at most x successes.
Example:

The probability of at most 12 on-time flights (out of 20) from Tampa can be
found on the TI-83 as follows:
binomcdf (20, 0.86,12)  .0038413869
Exercise:

Use your calculator to calculate the probability that at least 15 flights (out of 20)
from Tampa are on time.
Note: As with binompdf, omitting the last entry x in using the binomcdf function will
produce the list of cumulative probabilities starting with x  0 and ending with x  n .
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