Chapter 11
Elementary Statistics
Continuous Random Variables
&
Probability Distributions
Chapter 11
Elementary Statistics
What is a Random Variable?
It is a quantity whose values are real numbers and are determined
by the number of desired outcomes of an experiment.
Is there any special Random Variable?
We can categorize random variables in two groups:
◮
Discrete random variable
◮
Continuous random variable
Chapter 11
Elementary Statistics
What are Discrete Random Variables?
It is a numerical value associated with the desired outcomes and
has either a finite number of values or infinitely many values but
countable such as whole numbers 0, 1, 2, 3, · · · .
What are Continuous Random Variables?
It has infinitely numerical values associated with any interval on
the number line system without any gaps or breaks.
Chapter 11
Elementary Statistics
What is a Probability Distribution?
It is a description and often given in the form of a graph, formula,
or table that provides the probability for all possible random
variables of the desired outcomes.
Are there any Requirements?
Let x be any random variable and P(x) be the probability of the
random variable x, then
X
◮
P(x) = 1
◮
0 ≤ P(x) ≤ 1
Chapter 11
Elementary Statistics
What is Discrete Probability Distributions?
It is a probability distribution for a discrete random variable x with
probability P(x) such that
X
◮
P(x) = 1, and
◮
0 ≤ P(x) ≤ 1.
Here are some examples of continuous probability distribution:
◮
Discrete Probability Distribution
◮
Binomial Probability Distribution
Chapter 11
Elementary Statistics
What is Continuous Probability Distributions?
It is a probability distribution for a continuous random variable x
with probability P(x) such that
X
◮
P(x) = 1,
◮
◮
0 ≤ P(x) ≤ 1, and
P(x = a) = 0.
Here are some examples of continuous probability distribution:
◮
Uniform Probability Distribution
◮
Standard Normal Probability Distribution
◮
Normal Probability Distribution
Chapter 11
Elementary Statistics
What is Uniform Probability Distributions?
It is a probability distribution for a continuous random variable x
that can assumes all values on the interval [a, b] such that
◮
All values are evenly spread over the interval [a, b],
◮
The graph of the distribution has a rectangular shape, and
1
as displayed below.
P(c < x < d) = (d − c) ·
b−a
1
−
b−a
◮
a
c
d
b
Chapter 11
Elementary Statistics
Example:
The amount of time, in minutes, that a person must wait for a bus
is uniformly distributed between 0 and 12 minutes, inclusive.
◮
Draw and label the uniform probability distribution,
◮
Find the probability that a randomly selected person has to
wait for a bus is between 7.5 to 10 minutes?
◮
Find k = P90 , and explain what this number represents.
Solution:
We have a uniform probability distribution with a = 0, b = 12, and
1
1
a rectangular graph with the width of
=
b−a
12
Chapter 11
Elementary Statistics
Continued Solution:
◮
Draw and label the uniform probability distribution,
1
−
12
0
◮
12
Find the probability that a randomly selected person has to
wait for a bus is between 7.5 to 10 minutes?
We need to redraw our uniform probability distribution, label,
and shade a region that represents the probability for wait
time from 7.5 to 10 minutes.
Chapter 11
Elementary Statistics
Continued Solution:
1
−
12
0
7.5
Now to the find the probability,
10 12
P(7.5 < x < 10) = (10 − 7.5) ·
= 2.5 ·
=
1
12
5
≈ 0.208
24
1
12
Chapter 11
Elementary Statistics
Continued Solution:
◮
Find k = P90 , and explain what this number represents.
The gray area below is 0.9 since it represents k = P90 ,
1
−
12
0.9
0
k
1
P(x < k) = (k − 0) ·
12
1
⇒ k = 10.8
0.9 = k ·
12
12
So 90% of the customers have a wait time that is below 10.8
minutes.
Chapter 11
Elementary Statistics
Finding Mean, Variance, and Standard Deviation of a
Uniform Probability Distribution:
Given a continuous random variable x that can assumes all values
on the interval [a, b] with a uniform probability distribution, then
◮
◮
◮
b+a
,
2
(b − a)2
, and
Variance ⇒ σ 2 =
12
√
Standard Deviation ⇒ σ = σ 2 .
Mean ⇒ µ =
Chapter 11
Elementary Statistics
Example:
A continuous random variable x that assumes all values from
x = 5 and x = 30 with uniform probability distribution.
◮
Draw and label the uniform probability distribution,
◮
Find the mean of the distribution.
◮
Find the variance of the distribution.
◮
Find the standard deviation of the distribution.
Solution:
We have a uniform probability distribution with a = 5, b = 30, and
1
1
a rectangular graph with the width of
=
.
b−a
25
Chapter 11
Elementary Statistics
Continued Solution:
A continuous random variable x that assumes all values from
x = 5 and x = 30 has a uniform probability distribution.
◮
Draw and label the uniform probability distribution,
1
−
25
5
◮
Find the mean of the distribution.
µ =
=
b+a
2
30 + 5
⇒ µ = 17.5
2
30
Chapter 11
Elementary Statistics
Continued Solution:
◮
Find the variance of the distribution.
σ2 =
=
◮
(b − a)2
12
625
(30 − 5)2
⇒ σ2 =
12
12
Find the standard deviation of the distribution.
√
σ = rσ 2
625
=
⇒ σ ≈ 7.1217
12
Chapter 11
Elementary Statistics
Example:
The amount of coffee dispensed by a certain machine into a cup is
a continuous random variable x that assumes all values from
x = 11 and x = 25 ounces with uniform probability distribution.
◮
Draw and label the uniform probability distribution,
◮
Find the probability that a cup filled by this machine will
contain at least 14.5 ounces.
◮
Find the probability that a cup filled by this machine will
contain at most 20 ounces.
◮
Find the probability that a cup filled by this machine will
contain more than 16.5 ounces but less than 18 ounces.
◮
Find k such that P(x > k) = 0.3. Explain what this number
represents.
Chapter 11
Elementary Statistics
Solution:
In this example, we have a uniform probability distribution with
a = 11, b = 25, and a rectangular graph with the width of
1
1
=
b−a
14
◮ Draw and label the uniform probability distribution,
1
−
14
11
◮
25
Find the probability that a cup filled by this machine will
contain at least 14.5 ounces.
1
= 0.75
P(x ≥ 14.5) = P(x > 14.5) = (25 − 14.5) ·
14
Chapter 11
Elementary Statistics
Continued Solution:
Find the probability that a cup filled by this machine will
contain at most 20 ounces.
1
≈ 0.643
P(x ≤ 20) = P(x < 20) = (20 − 11) ·
14
◮ Find the probability that a cup filled by this machine will
contain more than 16.5 ounces but less than 18 ounces.
1
≈ 0.107
P(16.5 < x < 18) = (18 − 16.5) ·
14
◮ Find k such that P(x > k) = 0.3. Explain what this number
represents.
1
= 0.3 ⇒ k = 20.8
(25 − k) ·
14
The probability that the machine will fill a cup more than 20.8
ounces in 0.3.
◮
Chapter 11
Elementary Statistics
What is Standard Normal Distributions?
It is a probability distribution for a continuous random variable z
that can assumes any values such that
◮ The graph of the distribution is symmetric, and bell-shaped,
◮ The total are under the curve is equal to 1,
◮ Mean, mode, and median are all equal,
◮ µ = 0 and σ = 1, and
◮ P(a < z < b) is the area under the curve on the interval
(a, b).
P(a < z < b)
a
|
µ=0
σ=1
b
Chapter 11
Elementary Statistics
Chapter 11
Elementary Statistics
Example:
Find P(−1.5 < z < 1).
Solution:
We start by drawing the normal curve, then shade and label
accordingly.
P(−1.5 < z < 1)
−1.5
|
µ=0
σ=1
1
Now we can use the normal distribution chart or different
technology to compute the area which is the probability that we
are looking for. In this case, P(−1.5 < z < 1) = 0.7745.
Chapter 11
Elementary Statistics
Example:
Find P(z > 1.5).
Solution:
We start by drawing the normal curve, then shade and label
accordingly.
P(z > 1.5)
|
µ = 0 1.5
σ=1
Now we can use the normal distribution chart or different
technology to compute the area which is the probability that we
are looking for. In this case, P(z > 1.5) = 0.0668.
Chapter 11
Elementary Statistics
Example:
Use any method that you prefer to find the following and very your
results with the drawing given below.
1
P(0 < z < 1)
2
P(−2 < z < 2)
3
P(z > 1)
4
P(z < −2)
Chapter 11
Elementary Statistics
How do we find Z Score from Known Area?
It is a probability distribution for a continuous random variable z
that can assumes any values such that
◮ Draw a bell-shaped curve,
◮ Clearly identify and shade the region that represents the
known area,
◮ Working with the cumulative area from the left, use the
normal distribution chart or different technology to find the
corresponding Z score.
Needed Area
Z
|
µ=0
σ=1
Chapter 11
Elementary Statistics
Example:
Use any method that you prefer to find k such that
P(z > k) = 0.85.
Solution:
Needed Area
0.15
Known Area
0.85
k
|
µ=0
σ=1
Using any method, we get k = −1.036, that is
P(z > −1.036) = 0.85 and P(z < −1.036) = 0.15. We can also
conclude that k = P15 = −1.036.
Chapter 11
Elementary Statistics
What is Zα ?
It is a notation that describes a z score with an area α to its right.
How do we find Zα ?
We use the fact that Zα = Z1−α .
◮ Draw a bell-shaped curve, clearly identify, and shade the
regions that represent the area for α, and 1 − α,
◮ Working with the cumulative area from the left, use the
normal distribution chart or different technology to find the
corresponding Z score.
1−α
α
|
µ=0
σ=1
Zα
Chapter 11
Elementary Statistics
Example:
Use any method that you prefer to find Z0.025 .
Solution:
Given Z0.025 , we have α = 0.025, and 1 − α = 0.975, so we draw
and label our normal curve.
0.025
0.975
|
µ=0
σ=1
Z0.025
Using any method, we get Z0.025 = 1.960.
Chapter 11
Elementary Statistics
What is Normal Distributions?
It is a probability distribution for a continuous random variable x
that can assumes any values such that
◮ The graph of the distribution is symmetric, and bell-shaped,
◮ The total are under the curve is equal to 1,
◮ Mean, mode, and median are all equal,
◮ µ and σ are given, and
◮ P(a < x < b) is the area under the curve on the interval
(a, b).
P(a < x < b)
a
|
µ
σ
b
Chapter 11
Elementary Statistics
Example:
Consider a normal distribution with the mean of 115 and standard
deviation of 10. Find P(100 < x < 130).
Solution:
We start by drawing the normal curve, then shade and label
accordingly.
P(100 < x < 130)
100
|
µ = 115
130
σ = 10
Now we can use different technology to compute the area which is
the probability that we are looking for. In this case,
P(100 < x < 130) = 0.8664.
Chapter 11
Elementary Statistics
Example:
Consider a normal distribution with the mean of 75 and standard
deviation of 7.5. Find P(x > 90).
Solution:
We start by drawing the normal curve, then shade and label
accordingly.
P(x > 90)
|
µ = 75
90
σ = 7.5
Now we can use different technology to compute the area which is
the probability that we are looking for. In this case,
P(x > 90) = 0.0228.
Chapter 11
Elementary Statistics
Example:
Consider a normal distribution with the mean of 6375 and standard
deviation of 200. Find P(x < 6000).
Solution:
We start by drawing the normal curve, then shade and label
accordingly.
P(x < 6000)
6000
|
µ = 6375
σ = 200
Now we can use different technology to compute the area which is
the probability that we are looking for. In this case,
P(x < 6000) = 0.0304.
Chapter 11
Elementary Statistics
Example:
Consider a normal distribution with the mean of 125 and standard
deviation of 12. Find x = P80 .
Solution:
0.8
0.2
|
µ = 125
P80
σ = 12
Using any method, we get x = P80 = 135.1, that is
P(x < 135.1) = 0.85 and P(x > 135.1) = 0.2.
Chapter 11
Elementary Statistics
Normal Probability Distributions & TI
Normal Distribution
P(a < x < b)
P(x < a)
P(x > a)
Pk
Zα
TI Command
normalcdf(a, b, µ, σ)
normalcdf(−E 99, a, µ, σ)
normalcdf(a, E 99, µ, σ)
invNorm(k%, µ, σ)
invNorm(1 − α, 0, 1)
Chapter 11
Elementary Statistics