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Chapter 6
Random Variables
and Probability
Distributions
Created by Kathy Fritz
Consider the chance experiment of randomly selecting a
customer who is leaving a store.
What are possible
One numerical variable of interest to the store manager
values
for purchased
x?
might be the number
of items
by the customer.
Let’s use the letter x to denote this variable.
In this example, the values of x are isolated points.
One possible value of y is 3.0 minutes and another 4.0
minutes, but any other number between 3.0 and 4.0 is
also a possibility.
Another
of interest
mightand
be ythe
= number
Until variable
a customer
is selected
numberofof items
minutes spent
in a checkout
line. of x is uncertain.
counted,
the value
The possible y values form an entire interval on the number
line.
Random Variables
Random Variable
In this chapter,
we will look
at different
A random variable
is a numerical
variable
whose
distributions
of discrete
continuous
value depends
on the outcome
of aand
chance
random variables.
experiment.
Thisvariable
is typically
a “count” a numerical value
A random
associates
of something.
with each outcome
of a chance experiment.
• A random variable is discrete if its possible values
This is typically a “measure”
are isolated points along the number line.
of something
• A random variable is continuous if its possible values
are all points in some interval.
Identify the following variables as
discrete or continuous
1.
The number of items purchased by each
customer
Discrete
2. The amount of time spent in the checkout
line by each customer
Continuous
3. The weight of a pineapple
Continuous
4. The number of gas pumps in use
Discrete
Probability Distributions for
Discrete Random Variables
Properties
In Wolf City (a fictional place), regulations prohibit more
than five dogs or cats per household.
Let x = the number of dogs or cats per household in Wolf
City
X
0
1
2
3
4
5
Is this variable discrete or continuous?
What are the possible values for x?
Although you know what the possible values for
x are, it would also be useful to know how this
variable would behave if it were observed for
many houses.
A discrete probability distribution
provides this information.
Discrete Probability Distribution
The probability distribution of a discrete
random variable x gives the probability
associated with each possible x value.
Each probability is the long-run proportion of
the time that the corresponding x value will
occur.
If one possible
value
of x isa2,probability
it is common to write
Common
ways to
display
p(2)aindiscrete
place of P(x
= 2). variable are
distribution for
random
a table, probability histogram, or formula.
Properties of Discrete Probability
Distributions
1) For every possible x value,
0 < P(x) < 1.
2) The sum of P(x) over all values of x is
equal to one.
SP(x) = 1.
Suppose that each of four randomly selected customers
purchasing a refrigerator at an appliance store chooses
either an energy-efficient model (E) or one from a less
expensive group of models (G) that do not have an energyefficient rating.
Assume that these customers make their choices
independently of one another and that 40% of all
customers select an energy-efficient model.
What are the possible values for x?
Consider the next four customers. Let:
x = the number of energy efficient refrigerators
purchased by the four customers
x
0
1
2
3
4
Refrigerators continued . . .
x = the number of energy efficient refrigerators
purchased by the four customers
P(0) = P(GGGG) = 0.6(0.6)(0.6)(0.6) = 0.1296
P(1) = P(EGGG) + P(GEGG) + P(GGEG) + P(GGGE)
= 0.0864 + 0.0864 + 0.0864 + 0.0864 = 0.3456
Similarly,
P(2) = 0.3459
P(3) = 0.1536
P(4) = 0.0256
The probability distribution of x is summarized in the
following table:
x
0
1
2
3
4
P(x)
0.1296 0.3456 0.3456 0.1536 0.0256
Refrigerators continued . . .
x
P(x)
0
0.1296
1
2
0.3456 0.3456
3
0.1536
4
0.0256
The probability distribution can be used to determine
probabilities of various events involving x.
This means that in the long run, a group of four
refrigerator purchasers will include at least two
who select energy-efficient models about 52.48%
of the time.
Refrigerators continued . . .
x
P(x)
0
0.1296
1
2
0.3456 0.3456
3
0.1536
Does this include the x value of 2?
In discrete probability distributions, pay
close attention to whether the value in the
probability statement is included (≤ or ≥) or
the value is not included (< or >).
4
0.0256
Refrigerators continued . . .
x
P(x)
0
0.1296
1
2
0.3456 0.3456
3
0.1536
4
0.0256
A probability histogram is a graphical representation of a
discrete probability distribution.
The graph has a rectangle centered above each possible
value of x. The area of each rectangle is proportional to
the probability of the corresponding value.
Probability Distributions for
Continuous Random Variables
Properties
Consider the random variable:
x = the weight (in pounds) of a
full-term newborn child
Suppose that weight is reported to the nearest
What type
of variablehistogram
is this?
pound. The following
probability
What
is distribution
the sum of theof
areas
displays
the
weights.
This is an example of a
of all the rectangles?
Notice
theisrectangles
narrower
Ifthat
weight
measured are
with
greater
and
density
curve.
Now
suppose
that
weightcentered
is reported
to
the
The area
of
the
rectangle
over
7
pounds
the
histogram
begins
to
have
a
smoother
greater accuracy, the histogram approaches
The
shaded
areaThis
represents
the
probability
nearest
0.1
pound.
would
be
probability
represents
the
probability
6.5
<
x < 7.5
appearance.
a smooth curve. the
6 < x < 8.
histogram.
Probability Distributions for Continuous
Variables
A probability distribution for a continuous random
variable x is specified by a curve called a density
curve.
The function that describes this curve is denoted
by f(x) and is called the density function.
The probability that x falls in any particular
interval is the area under the density curve and
above the interval.
Properties of continuous probability
distributions
1. f(x) > 0
(the curve cannot dip below
the horizontal axis)
2. The total area under the density curve
equals one.
Suppose x is a continuous random variable defined as the
amount of time (in minutes) taken by a clerk to process a
certain type of application form.
Suppose x has a probability distribution with density
function:
.5 4  x  6
When
constant
over an
interval
fdensity
(xheight
)  is of
Why the
is the
this density
curve
0.5?
(resulting in a horizontal
density curve), the
0 otherwise
probability distribution is called a uniform
distribution.
The following is the graph of f(x), the density curve:
Density
0.5
4
5
6
Time (in minutes)
Application Problem Continued . . .
What is the probability that it takes at least 5.5 minutes
to process the application form?
P(x ≥ 5.5) = (6 - 5.5)(.5) = .25
Find the probability by calculating the area of
the shaded region (base × height).
Density
0.5
4
5
6
Time (in minutes)
Application Problem Continued . . .
What is the probability that it takes exactly 5.5 minutes
to process the application form?
P(x = 5.5) = 0
x = 5.5 is represented by a
line segment.
What is the area of this
line segment?
Density
0.5
4
5
6
Time (in minutes)
Application Problem Continued . . .
What is the probability that it takes more than 5.5
minutes to process the application form?
P(x > 5.5) = (6 - 5.5)(.5) = .25
In continuous probability distributions,
P(x > a) and P(x ≥ a) are equal!
Density
0.5
4
5
6
Time (in minutes)
Two hundred packages shipped using the Priority Mail rate
for packages less than 2 pounds were weighed, resulting in
a sample of 200 observations of the variable
x = package weight (in pounds)
from the population of all Priority Mail packages under 2
pounds.
A histogram (using the density scale, where
height = (relative frequency)/(interval width)) of 200
weights is shown below.
The shape of this
histogram suggests
1.0
that a reasonable
model for the
0.5
distribution of x
might be a
triangular
distribution.
1
2
Two hundred packages shipped using the Priority Mail rate
for packages less than 2 pounds were weighed, resulting in
a sample of 200 observations of the variable
x = package weight (in pounds)
from the population of all Priority Mail packages under 2
The easiest way to find the area of the shaded
pounds.
region is toof
find
– the areaweigh
of x over
≤ 1.5. 1.5 pounds?
What proportion
the1 packages
1.0
h = 0.75
0.5
1
2
b = 1.5
Students at a university use an online registration system
to register for courses. The variable
x = length of time (in minutes) required for a student to register
was recorded for a large number of students using the
system. The resulting values were used to construct a
How(below).
can you find
probability histogram
the area under this A smooth curve has
been superimposed
smooth
curve?
The general form
on the histogram
of the histogram
and is a reasonable
can be described
model for the
as bell shaped and
probability
symmetric.
distribution of x.
Some density curves resemble the one below.
Integral calculus is used to find the area under
these curves.
Don’t worry – we will use tables (with the
values already calculated). We can also use
calculators or statistical software to find
the area.
The probability that a continuous random variable
x lies between a lower limit a and an upper limit b
is
P(a < x < b) = (cumulative area to the left of b) –
(cumulative area to the left of a)
P(a < x < b) = P(x < b) – P(x < a)
=
-
Mean and Standard Deviation of
a Random Variable
Of Discrete Random Variables
Of Continuous Random Variables
Means and Standard Deviations of
Probability Distributions
The mean value of a random variable x, denoted
by mx, describes where the probability
distribution of x is centered.
The standard deviation of a random variable x,
denoted by sx, describes variability in the
probability distribution.
When the value of sxThe
is small,
larger the value of sx the
observed values of x will
more
tend
variability
to
there will be in
be close to the mean value.
observed x values.
How do the means and standard deviations of these
three density curves compare?
These two density curves have the same mean but
different standard deviations.
What happens to the appearance of the density
curve as the standard deviation increases?
Mean Value for a Discrete Random Variable
The mean value of a discrete random variable x,
denoted by mx , is computed by first multiplying each
possible x value by the probability of observing that
value and then adding the resulting quantities.
Symbolically,
all possible x values
The term expected value is sometimes used in place of
mean value and E(x) is another way to denote mx .
Individuals applying for a certain license are allowed up to
four attempts to pass the licensing exam. Consider the
random variable
x = the number of attempts made by a randomly selected applicant
The probability distribution of x is as follows:
x
1
2
3
4
p(x)
0.10
0.20
0.30
0.40
Standard Deviation for a Discrete Random
Variable
all possible x values
x
1
2
3
4
p(x)
0.10
0.20
0.30
0.40
Mean and Standard Deviation When x is
Continuous
For continuous probability distributions, mx and sx can
be defined and computed using methods from calculus.
The mean value mx locates the center of the
continuous distribution and gives the approximate
long-run average of observed x values.
The standard deviation, sx, measures the extent to
which the continuous distribution (density curve)
spreads out around mx and indicates the amount of
variability that can be expected in observed x values.
A company can purchase concrete of a certain type
from two different suppliers.
Let
x = compression strength of a randomly selected
batch from Supplier 1
y = compression
strength
a randomly
Which supplier
shouldofthe
company selected
The
density
curves
look similar
to these
batch the
from
Supplier
2
purchase
concrete
from?
Explain.
below.
Suppose that
mx = 4650 pounds/inch2
sx = 200 pounds/inch2
my = 4500 pounds/inch2
sy = 275 pounds/inch2
4300
4500
my
4700
mx
4900
Consider the experiment in which a customer of a propane
gas company is randomly selected. Suppose that the mean
and standard deviation of the random variable
x = number of gallons required to fill a propane tank
is 318 gallons and 42 gallons, respectively.
The company is considering two different pricing models.
Model 1: $3 per gallon
Model 2: service charge of $50 + $2.80 per gallon
The company is interested in the variable y = amount billed
For each of the two models, y can be expressed as a
function of the random variable x :
ymodel 1 = 3x
ymodel 2 = 50 + 2.8x
Mean and Standard Deviation of Linear
Functions
Revisit the propane gas company . . .
mean billingsamount
for Model 1 is a bit
m = The
318 gallons
= 42 gallons
higher than for Model 2, as is the
The company is considering
twoindifferent
pricing models.
variability
billing amounts.
Model 1: $3 per gallon
Model 2 results in slightly more consistency
Model 2: from
service
$50amount
+ $2.80charged.
per gallon
billcharge
to bill of
in the
Let’s consider a different type of problem . . .
Suppose that you have three tasks that you plan to do on
the way home.
x1 = time required to return book
x2 = time required to deposit check
x3 = time required to buy printer paper
Return library
book
You can define a new variable, y, to
represent the total amount of time
to complete these tasks
y = x1 + x2 + x3
Deposit paycheck
Purchase
printer paper
Linear Combinations
If x1, x2, …, xn are random variables and a1, a2, …, an are
numerical constants, the random variable y defined as
y = a1x1 + a2x2 + … + anxn
is a linear combination of the xi’s.
Let’s see how to compute the mean, variance,
and standard deviation of a linear combination.
Mean and Standard Deviations for Linear
Combinations
If x1, x2, …, xn are random variables with means m1, m2, …, mn
2, s 2result
2 true regardless of whether
is
and variances s1This
2 , …, sn , respectively, and
the x ’s are independent.
y = a1x1 + ai2x2 + … + anxn
then
This result is true ONLY if the xi’s are
independent.
A commuter airline flies small planes between San Luis
Obispo and San Francisco. For small planes the baggage
weight is a concern.
Suppose it is known that the variable x = weight (in pounds)
of baggage checked by a randomly selected passenger has a
mean and standard deviation of 42 and 16, respectively.
Consider a flight on which 10 passengers, all traveling alone,
are flying.
The total weight of checked baggage, y, is
y = x1 + x2 + … + x10
Airline Problem Continued . . .
mx = 42 and sx = 16
The total weight of checked baggage, y, is
y = x1 + x2 + … + x10
What is the mean total weight of the checked baggage?
mx = m1 + m2 + … + m10
= 42 + 42 + … + 42
= 420 pounds
Airline Problem Continued . . .
mx = 42 and sx = 16
The total weight of checked baggage, y, is
y = x1 + x2 + … + x10
What is the standard deviation of the total weight of the
checked baggage?
Since the 10 passengers are all traveling alone, it
is reasonable to think that the 10 baggage
weights are unrelated and therefore
independent.
Binomial and Geometric
Distributions
Properties of Binomial Distributions
Mean of Binomial Distributions
Standard Deviation of Binomial Distributions
Properties of Geometric Distributions
Suppose we decide to record the gender of the
next 25 newborns at a particular hospital.
These questions can be
answered using a binomial
distribution.
Properties of a Binomial Experiment
A binomial experiment consists of a sequence of trials
with the following conditions:
1. There are a fixed number of trials
2. Each trial results in one of only two possible outcomes,
We use n to denote the fixed number
labeled success (S) and failure (F).
of trials.
3. Outcomes of different trials are independent
Theterm
probability
of x is called
The
successdistribution
does not necessarily
mean
4. The probability of success is the same for each trial.
the binomial
distribution.
something
positive.probability
For example,
if the random
variable is the number of defective items
produced,random
then being
“defective”
is a success.
The binomial
variable
x is defined
as
x = the number of successes observed when a
binomial experiment is performed
Binomial Probability Formula:
Let
n = number of independent trials in a binomial experiment
p = constant
probability
that
any particular
trial results
in
Notice
that the
probability
distribution
is
a success
specified by a formula rather than a table
or probability histogram.
Then
p( x )  P( x successes among the n trials )
n!

p x (1  p)n  x
x  0,1, 2, . . ., n
x!(n  x )!
. . . can be written as nCx
Sixty percent of all computers sold by a large
computer retailer are laptops and 40% are
desktop models. The type of computer
purchased by each of the next 12 customers
will be recorded.
Define the random variable of interest as
x = the number of laptops among these 12
The binomial random variable x counts the number of
laptops purchased. The purchase of a laptop is
considered a success and is denoted by S. The
probability distribution of x is given by
12!
p (x ) 
(0.6)x (0.4)12 x
x ! (12  x )!
x  0, 1, 2, . . ., 12
What is the probability that exactly four of the next 12
computers sold are laptops?
p ( 4)  P ( x  4)
12!
4
8
0.6 0.4

4!8!
 0.042
If many groups of 12 purchases are examined, about
4.2% of them include exactly four laptops.
What is the probability that between four and seven
(inclusive) are laptops?
P(4  x  7)  p(4)  p(5)  p(6)  p(7)
12!
12!
4
8
7
5
0.6 0.4  . . .  0.6 0.4

4!8!
7!5!
 0.042  0.101  0.177  0.227
calculations can become very tedious.
These
0.547
We will examine how to use Appendix Table 9
to perform these calculations.
What is the probability that between four and seven
(exclusive) are laptops?
P( 4  x  7)  p(5)  p(6)
12!
12!
5
7
6
6
0.6 0.4   0.6 0.4 

5!7!
6!6!
 0.101  0.177
Notice that the probability depends on
whether
< or ≤ appears. This is typical of
 0.278
discrete random variables.
Using Appendix Table 9 to Compute
Binomial Probabilities
To find p(x) for any particular value of x,
1. Locate the part of the table corresponding to the
value of n (5, 10, 15, 20, or 25).
2. Move down to the row labeled with the value of x.
3. Go across to the column headed by the specified
value of p.
The desired probability is at the intersection
of the designated x row and p column.
Sampling Without Replacement
One of the properties of a binomial probability
distribution is . . .
“Outcomes of different trials are independent”
If we sample with replacement (that is, we return the
element
the population
next
Ifto
(n/N)
≤ 0.05, i.e., before
no morethe
than
5%selection),
then the
outcomes
are independent.
of the
population
is sampled, then the
The calculations
even more
binomial distribution
gives aare
good
for
thewithout
hypergeometric
However,approximation
sampling tedious
is usually
done
returning
to the
probability
distribution
forpopulation
the
(without replacement)
the element
to the
distribution
of x. than
binomial
distribution.
before the next selection.
Therefore,
the outcomes
are dependent and the observed number of successes
have a hypergeometric distribution.
Formulas for mean and standard
deviation of a binomial distribution
mx  np
sx  np 1  p 
Let’s revisit the computer example:
Sixty percent of all computers sold by a large
computer retailer are laptops and 40% are desktop
models. The type of computer purchased by each of the
next 12 customers will be recorded.
Computers Revisited . . .
Suppose we were NOT interested in the number
of laptops purchased by the next 12 customers,
but which of the next customers would be the
first one to purchase a laptop.
How is this question different
from a binomial distribution?
Properties of a Geometric Experiment
Suppose an experiment consists of a sequence of trials
with the following conditions:
1. The trials are independent.
2. Each trial can result in one of two possible outcomes,
success (S) or failure (F).
3. The probability of success is the same for all trials.
A geometric random variable is defined as
x = number of trials until the first success is
observed
(including
the success
How do
these properties
differtrial)
from
those of a binomial probability
The probability distribution
distribution?of x is called the
geometric probability distribution.
Suppose that 40% of students who drive to campus at your
school or university carry jumper cables.
Your car has a dead battery and you don’t have jumper
cables, so you decide to stop students as they are headed
to the parking lot and ask them whether they have a pair
of jumper cables.
Let:
x = the number of students stopped before finding one
with a pair of jumper cables
This is an example of a
geometric random variable.
Geometric Probability Distribution
If x is a geometric random variable with probability of
success = p for each trial, then
x 1
p (x )  (1  p )
Where
x = 1, 2, 3, …
p
Jumper Cables Continued . . .
Let:
x = the number of students stopped before finding one
with a pair of jumper cables
Recall that p = .4
What is the probability that third student stopped will be
the first student to have jumper cables?
p(3) = (0.6)2(0.4) = 0.144
What is the probability that three or fewer students are
stopped before finding one with jumper cables?
P(x < 3) = p(1) + p(2) + p(3) =
(0.6)0(0.4) + (0.6)1(0.4) + (0.6)2(0.4) = 0.784
Normal Distributions
Standard Normal Curve
Using a Table to Calculate Probabilities
Other Normal Curves
Normal Distributions . . .
Normal Distributions . . .
Standard Normal Distribution . . .
is customary
to usethe
thestandard
letter z normal
to represent
A It
table
of areas under
curve isa
The standard
normal
distribution
theby
normal
variable
distribution
is described
the
used towhose
calculate
probabilities
ofisevents.
standard
distribution
with normal curve (or z curve).
m = 0 and s = 1
Thus, the z curve is comprised of z values instead of
x values.
Using the Table of Standard
Normal Curve Areas
For any number z*, from -3.89 to 3.89 and rounded
Todecimal
find thisplaces,
probability
using the table,
to two
the Appendix
Tablelocate
2 gives
the following:
(area under z curve to the left of z*) = P(z < z*) = P(z < z*)
• The row labeled with the sign of z* and the
digit to either side of the decimal point
(for example, -1.7 or 0.5)
Where
• The column identified with the second digit
to the right
of the
pointa in
z*
the letter
z is used
todecimal
represent
random
• The number
at the intersection
this row
variable
whose distribution
is the of
standard
anddistribution.
column is the desired probability.
normal
Suppose we are interested in the probability that z is
less than 1.42.
P(z < 1.42)
P(z < 1.42) = 0.9222
Find the intersection of the row 1.4 and column .02.
.00
.01
.02
.03
…
…
…
…
…
…
z*
…
1.42
1.3
.9032
.9049
.9066
.9082
1.4
.9192
.9207
.9222
.9236
1.5
.9332
.9345
.9357
.9370
…
…
…
Suppose we are interested in the probability that z* is
less than 0.58.
0.5
0.6
…
…
…
.6808
.08
.6844
.09
…
…
0.4
.07
…
…
…
z*
…
P(z < 0.58) = 0.7190
P(z < 0.58)
.6879
.7157
.7190
.7224
.7486
.7517
.7549
Find the following probability:
P(-1.76 < z < 0.58) =
P(z < 0.58) - P(z < -1.76)
.7190 - .0392
= 0.6798
P(z < -1.76)
Suppose we are interested in the probability that z* is
greater than 2.31.
The Table of Areas gives the area to the LEFT
of the z*.
P(z > 2.31) =
z*
.00
.01
.02
…
…
…
…
…
…
To find the area to the right, subtract the
1 - .9896 = 0.0104
value in the table from 1
2.2
.9861
.9864
.9868
.9871
2.3
.9893
.9896
.9898
.9901
2.4
.9918
.9920
.9922
.9925
Suppose we are interested in the finding the z* for the
smallest 2%.
To find z*:
P(z < z*) = .02
Since
.0200
doesn’t
in the
the body
body of
of the
the
Look
for the
areaappear
.0200 in
Table,
usethe
therow
value
it. out to
Follow
andclosest
columntoback
z*Table.
= -2.05
read thez*z-value.
-2.1
-2.0
-1.9
…
…
.05
…
…
…
…
.04
…
.03
…
…
z*
.0162
.0158
.0154
.0207
.0202
.0197
.0262
.0256
.0250
Finding Probabilities for Other
Normal Curves
To find the probabilities for other normal curves,
standardize the relevant values and then use the table of
z areas.
If x is a random variable whose behavior is described by a
normal distribution with mean m and standard deviation s ,
then
P(x < b) = P(z < b*)
P(x > a) = P(z > a*)
P(a < x < b) = P(a* < z < b*)
Where z is a variable whose distribution is standard
normal and
a* 
a m
s
b* 
b m
s
Data on the length of time to complete registration
for classes using an on-line registration system
suggest that the distribution of the variable
x = time to register
for students at a particular university can well be
approximated by a Standardizes
normal
9. up with
Look distribution
this value
in themean
m = 12 minutes and standard deviation
table. s = 2 minutes.
What is the probability that it will take a randomly
selected student less than 9 minutes to complete
registration?
9  12
b* 
 1.5
2
P(x < 9) = 0.0668
9
Registration Problem Continued . . .
x = time to register
m = 12 minutes and s = 2 minutes
What is the probability that it will take a randomly
selected student more than 13 minutes to complete
registration?
13  12
a* 
 .5
2
13
P(x > 13) = 1 - .6915 = 0.3085
Registration Problem Continued . . .
x = time to register
m = 12 minutes and s = 2 minutes
What is the probability that it will take a randomly
selected student between 7 and 15 minutes to complete
registration?
15  12
a* 
 1.5
2
7  12
b* 
 2.5
2
7
P(7 < x < 15) = .9332 - .0062 = 0.9270
15
Ways to Assess Normality
Normal Probability Plot
Using Correlation Coefficient
Normal Probability Plot
A normal probability plot is a scatterplot of
(normal One
score,
values)anpairs.
wayobserved
to see whether
assumption of
population normality is plausible is to construct
a normal
probability
plot
ofprobability
Normal
scores
are z-scores
from
the data. plot
A strong
linear
pattern
in a normal
Or outliers
standard normal distribution.
suggests
that
the population
distribution is
Such as
curvature
which would
approximately
normal.in the data
indicate skewness
On the other hand, systematic departure from a
straight-line pattern (such as curvature in the plot)
suggests that the population distribution is not
normal.
What are normal scores?
Consider a random sample with n = 5.
To find the appropriate normal scores for a sample of
size 5, divide the standard normal curve into 5 equalWhy are these
area regions
regions.not the
same width?
Each region has an
area equal to 0.2.
What are normal scores?
Next – find the median z-score for each region.
We use technology (calculators or statistical
software) to compute these normal scores.
These are the normal scores that we would plot our
data against.
Why is the median
not in the “middle”
of each region?
-1.28
-.524
0
1.28
.524
The
following
dataa represent
egg weights
(in
Let’s
construct
normal probability
plot.
Sincefor
theanormal
probability
plot is approximately
grams)
sample
of 10 eggs.
linear, it is plausible that the distribution of egg
is approximately
normal. depend
Since the weights
values of
the normal scores
on the sample size n, the normal scores when
53.552.53
53.04
53.50
53.00 normal
53.07
nSketch
= 10 are
below:
a scatterplot by pairing the smallest
with the smallest
observation
from the53.16
52.86score52.66
53.23
53.26
data set and so on.
53.0
-1.539 -1.001 -0.656 -0.376 -0.123
0.123 0.376 52.5
0.656 1.001 1.539
-1.5
-1.0 -0.5
0.5
1.0
1.5
Using the Correlation Coefficient to
Assess Normality
How r,
smaller
“too muchfor the n
The correlation coefficient,
can beiscalculated
smaller”
(normal score, observed value)
pairs. than 1?
If r is too much smaller than 1, then normality of the
underlying distribution is questionable.
Since
r > critical r,
Values to Which r Can
be it
Compared
to Check
for sample
Normality
then
is plausible
thatofthe
of
Consider these points
from
the weight
eggs
data:
n
5
10 (-1.001,
15 egg20
25 came
30 from
40 a50
60
75
weights
distribution
(-1.539,
52.53)
52.66)
(-.656,52.86)
(-.376,53.00)
(-.123, 53.04)
(.123,53.07)
(.376,53.16)
(.656,53.23)
that was approximately
normal.
Critical
.832 .880 (1.539,53.50)
911 .929 .941 .949 .960 .966 .971 .976
(1.001,53.26)
r
Calculate the correlation coefficient for
these points.
r = .986
Using the Normal Distribution to
Approximate a Discrete
Distribution (optional)
Suppose the probability distribution of a
discrete random variable x is displayed in the
Often,below.
a probability
canatbe
Suppose
this
rectanglehistogram
is centered
x =well
6. The
histogram
approximated
by abegins
normalatcurve.
Ifends
so, itatis6.5.
rectangle
actually
5.5 and
customary
to say that
has
an approximately
These endpoints
willxbe
used
in calculations.
normal distribution.
The probability of a particular value is
This is called a continuity correction.
the area of the rectangle centered at
that value.
6
Normal Approximation to a Binomial
Distribution
When either np < 10 or n (1 - p) < 10,
the binomial distribution is too skewed for the
normal approximation to give accurate
probability estimates.
Premature babies are born before 37 weeks, and those
born before 34 weeks are most at risk. A study reported
that 2% of births in the United States occur before 34
weeks.
Suppose that 1000 births will be randomly selected and
that the value of
x = number of births that occur prior to 34 weeks
is to be determined. Because
np = 1000(.02) = 20 ≥ 10
n(1 – p) = 1000(.98) = 980 ≥ 10
Premature Babies Continued . . .
m = 20 and s = 4.427
What is the probability that the number of babies in
the sample of 1000 born prior to 34 weeks will be
up these values in
between 10 and Look
25 (inclusive)?
the table and subtract
To find the shaded
the
probabilities.
P(10 < x < 25) = 0.8925 - 0.0089
= 0.8836
area, standardize
the
Since 10 is
included
in the
Similarly,
the
endpoint
of the
endpoints.
probability,
of
rectanglethe
for endpoint
25 is 25.5.
the rectangle for 10 is 9.5. 25.5  20
9.5  20
a* 
4.427
 2.37
b* 
4.427
 1.24