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Chapter 4
Discrete Probability
Distributions
§ 4.1
Probability
Distributions
Random Variables
A random variable x represents a numerical value
associated with each outcome of a probability distribution.
A random variable is discrete if it has a finite or countable
number of possible outcomes that can be listed.
x
0
2
4
6
8
10
A random variable is continuous if it has an uncountable
number or possible outcomes, represented by the intervals
on a number line.
x
0
2
4
6
8
10
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Random Variables
Example:
Decide if the random variable x is discrete or continuous.
a.) The distance your car travels on a tank of gas
The distance your car travels is a continuous
random variable because it is a measurement that
cannot be counted. (All measurements are
continuous random variables.)
b.) The number of students in a statistics class
The number of students is a discrete random
variable because it can be counted.
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Discrete Probability Distributions
A discrete probability distribution lists each possible value
the random variable can assume, together with its
probability. A probability distribution must satisfy the
following conditions.
In Words
In Symbols
1. The probability of each value of
the discrete random variable is
between 0 and 1, inclusive.
0  P (x)  1
2. The sum of all the probabilities
is 1.
ΣP (x) = 1
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Constructing a Discrete Probability Distribution
Guidelines
Let x be a discrete random variable with possible
outcomes x1, x2, … , xn.
1. Make a frequency distribution for the possible
outcomes.
2. Find the sum of the frequencies.
3. Find the probability of each possible outcome by
dividing its frequency by the sum of the frequencies.
4. Check that each probability is between 0 and 1 and
that the sum is 1.
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Constructing a Discrete Probability Distribution
Example:
The spinner below is divided into two sections. The
probability of landing on the 1 is 0.25. The probability of
landing on the 2 is 0.75. Let x be the number the spinner
lands on. Construct a probability distribution for the
random variable x.
1
2
x
P ( x)
1
0.25
2
0.75
Each probability is
between 0 and 1.
The sum of the probabilities is 1.
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Constructing a Discrete Probability Distribution
Example:
The spinner below is spun two times. The probability of
landing on the 1 is 0.25. The probability of landing on the 2
is 0.75. Let x be the sum of the two spins. Construct a
probability distribution for the random variable x.
The possible sums are 2, 3, and 4.
P (sum of 2) = 0.25  0.25 = 0.0625
1
2
Spin a 1 on “and”
the first spin.
Spin a 1 on the
second spin.
Continued.
Larson & Farber, Elementary Statistics: Picturing the World, 3e
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Constructing a Discrete Probability Distribution
Example continued:
P (sum of 3) = 0.25  0.75 = 0.1875
1
Spin a 1 on “and”
the first spin.
2
Spin a 2 on the
second spin.
“or”
Sum of
P ( x)
spins, x
2
0.0625
0.375
3
4
P (sum of 3) = 0.75  0.25 = 0.1875
Spin a 2 on “and”
the first spin.
Spin a 1 on the
second spin.
0.1875 + 0.1875
Larson & Farber, Elementary Statistics: Picturing the World, 3e
Continued.
9
Constructing a Discrete Probability Distribution
Example continued:
1
P (sum of 4) = 0.75  0.75 = 0.5625
2
Spin a 2 on “and”
the first spin.
Sum of
P ( x)
spins, x
2
0.0625
3
0.375
0.5625
4
Spin a 2 on the
second spin.
Each probability is between
0 and 1, and the sum of the
probabilities is 1.
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Graphing a Discrete Probability Distribution
Example:
Graph the following probability distribution using a histogram.
P(x)
Sum of
Sum of Two Spins
P ( x)
spins, x
0.6
0.0625
0.375
0.5625
0.5
Probability
2
3
4
0.4
0.3
0.2
0.1
x
0
2
3
4
Sum
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Mean
The mean of a discrete random variable is given by
μ = ΣxP(x).
Each value of x is multiplied by its corresponding
probability and the products are added.
Example:
Find the mean of the probability distribution for the sum of
the two spins.
xP (x)
x
P ( x)
2
0.0625
2(0.0625) = 0.125
ΣxP(x) = 3.5
3
0.375
3(0.375) = 1.125
4
0.5625
4(0.5625) = 2.25
The mean for the
two spins is 3.5.
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Variance
The variance of a discrete random variable is given by
2 = Σ(x – μ)2P (x).
Example:
Find the variance of the probability distribution for the sum
of the two spins. The mean is 3.5.
x – μ (x – μ)2 P (x)(x – μ)2
x
P ( x)
2
0.0625
–1.5
2.25
 0.141
3
0.375
–0.5
0.25
 0.094
4
0.5625
0.5
0.25
 0.141
ΣP(x)(x – 3.5)2
 0.376
The variance for the
two spins is
approximately 0.376
Larson & Farber, Elementary Statistics: Picturing the World, 3e
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Standard Deviation
The standard deviation of a discrete random variable is
given by
σ = σ 2.
Example:
Find the standard deviation of the probability distribution
for the sum of the two spins. The variance is 0.376.
x – μ (x – μ)2 P (x)(x – μ)2
x
P ( x)
2
0.0625
–1.5
2.25
0.141
3
0.375
–0.5
0.25
0.094
4
0.5625
0.5
0.25
0.141
σ  σ2
 0.376  0.613
Most of the sums
differ from the
mean by no more
than 0.6 points.
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Expected Value
The expected value of a discrete random variable is equal to
the mean of the random variable.
Expected Value = E(x) = μ = ΣxP(x).
Example:
At a raffle, 500 tickets are sold for $1 each for two prizes of
$100 and $50. What is the expected value of your gain?
Your gain for the $100 prize is $100 – $1 = $99.
Your gain for the $50 prize is $50 – $1 = $49.
Write a probability distribution for the possible gains
(or outcomes).
Continued.
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Expected Value
Example continued:
At a raffle, 500 tickets are sold for $1 each for two prizes of
$100 and $50. What is the expected value of your gain?
Gain, x
P ( x)
$99
1
500
$49
1
500
–$1
498
500
Winning
no prize
E(x) = ΣxP(x).
 $99 
1
1
498
 $49 
 ($1) 
500
500
500
 $0.70
Because the expected value is
negative, you can expect to lose
$0.70 for each ticket you buy.
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Page 179-183, 5-6, 12-24 even; 2731 odd, 33-35 odd
5 True
6 False, it is equal to the mean
12 Continuous
14 Discrete
16 Continuous
18 Discrete
20a 0.45 b 0.80
22 0.15
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Page 179-183, 5-6, 12-24 even; 2731 odd, 33-35 odd
24 No the sum is not equal to 1
26 Yes
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27
X
P(X)
0
0.686
1
0.195
2
0.077
3
0.022
4
0.013
5
0.006
Mean = 0.497
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29
X
P(X)
0
0.432
1
0.403
2
0.137
3
0.029
Mean = 0.764
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31
X
P(X)
0
0.031
1
0.063
2
0.151
3
0.297
4
0.219
5
0.156
6
0.083
Mean = 3.410
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homework
33a 5.3
35a 2.1
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homework
Page 179-183
28 - 40 even
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Page 179-183 28
X
P(X)
0
0.7275
1
0.1308
2
0.0761
3
0.0292
4
0.0214
5
0.0150
b≈0.5
c≈1.1
d≈1.0
E a household has
on average 0.5 cats
with a standard
deviation of 1
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Page 179-183 30
X
P(X)
0
0.152
1
0.292
2
0.249
3
0.178
4
0.102
5
0.026
b≈1.9
c≈1.7
d≈1.3
E The average
number of accidents
per student is 1.9
with a standard
deviation of 1.3
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Page 179-183 32
X
P(X)
0
0.059
1
0.122
2
0.163
3
0.178
4
0.213
5
0.128
6
0.084
7
0.053
b≈3.4
c≈3.4
d≈1.8
E The average
number of school
related
extracurricular
activities per student
is 3.4 with a
standard deviation of
1.8
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Page 179-183, 34, 36
34 a ≈1.9 b≈2.6 c≈1.6 d≈1.9
e The average number of ATM
transactions made in one day is 1.9
with a standard deviation of 1.6
36 a ≈1.7 b ≈ 1.0 c≈1.0 d≈1.7
e The average car occupancy is 1.7
with a standard deviation of 1.0
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Page 179-183, 38 & 40
38 a ≈1.6 b ≈ 1.9 c≈1.4 d≈1.6
e The average car occupancy is 1.6
with a standard deviation of 1.4
40 A 0.432 B 0.568 C 0.568
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Page 179-183, 38 & 40
38 a ≈1.6 b ≈ 1.9 c≈1.4 d≈1.6
e The average car occupancy is 1.6
with a standard deviation of 1.4
40 A 0.432 B 0.568 C 0.568
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Page 179-183, 34
34 a ≈1.9 b≈2.6 c≈1.6 d≈1.9
e The average number of ATM
transactions made in one day is 1.9
with a standard deviation of 1.6
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Homework
Page 179-183, 27, 29, 37-43 odd, 44
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44
X
-4
21 (gift certificate)
446 (tent)
3146 (boat)
P(X)
4983/5000=0.9966
15/5000=0.003
1/5000=0.0002
1/5000=0.0002
The expected value of the raffle for the
person buying the ticket is -3.205
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Homework
Page 179-180 1-4, 7-10 and 11-21 odd
And read section 4.1
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Homework
Page 179-183, 25, 26, 31, 42
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Homework - worksheet
1 μ ≈ 1.5
σ2 ≈ 0.75
σ ≈ 0.87
2 μ ≈ 0.60 σ2 ≈ 0.51
σ ≈ 0.72
3μ≈2
σ2 ≈ 1
σ≈1
4 Not a probability distribution sum of
probabilities not equal to 1
5 μ ≈ 0.80 σ2 ≈ 0.48
σ ≈ 0.69
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Homework
Page 179-183, 5-6, 12-24 even; 27-31
odd parts a and b only, 33-35 odd part
a only.
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