Chapter 5

Report
Chapter 7
Random Variables and
Probability Distributions
Random Variable A grocery store
manager
might
be
• A numerical
variable
whose
value
interested in the number of broken
depends
on
the
outcome
of
a
chance
eggs in each carton (dozen of eggs).
experiment
OR
• Associates
a numerical
value
with
An environmental
scientist
might
be each
interestedofin athe
amount
of ozone in an
outcome
chance
experiment
air sample.
• Two types of random variables
– Discrete
Since these values change and are
– Continuous
subject to some uncertainty, these are
examples of random variables.
Two Types of Random Variables:
• Discrete – its set of possible
values
is awe
In this
chapter,
will look
collection of isolated points along
a at
different
number line
This is typically a
“count” of something
distributions of
discrete and
continuous random
variables.
• Continuous - its set of possible values
This is typically a
includes an entire interval
on of
a number
“measure”
something
line
Identify the following variables
as discrete or continuous
1.
The number of broken eggs in each carton
Discrete
2. The amount of ozone in samples of air
Continuous
3. The weight of a pineapple
Continuous
4. The amount of time a customer spends in a
store
Continuous
5. The number of gas pumps in use
Discrete
Probability Distributions for
Discrete Random Variables
Probability distribution is a
model that describes the longrun behavior of a variable.
In a Wolf City (a fictional place), regulations
prohibit This
no more
than afive
dogs or
cats per
is called
discrete
probability
household.
distribution. It can also be displayed
in a number
histogram
with and
the probability
Let x = the
of dogs
cats in a on
What
do
you
notice
about
the
sum
of
the
vertical
axis.
randomly selected household in Wolf City
these probabilities?
Is What
this variable
discrete
or
continuous?
are
the
possible
values
for
x
0
1
2
3
4
5 x?
P(x) .26
Probability
.31 .21 .13 .06 .03
The Department of Animal Control has
collected data over the course of
several years. They have estimated the
long-run probabilities for the values of
x.
Number of Pets
Discrete Probability Distribution
1) Gives the probabilities associated with each
possible x value
2) Each probability is the long-run relative
frequency of occurrence of the
corresponding x-value when the chance
experiment is performed a very large
number of times
3) Usually displayed in a table, but can be
displayed with a histogram or formula
Properties of Discrete Probability
Distributions
1) For every possible x value,
0 < P(x) < 1.
2) For all values of x,
S P(x) = 1.
Dogs and Cats Revisited . . .
Let x =Just
the add
number
dogs or cats
the of
probabilities
forper
0, household
1, and 2
in Wolf City
x
0
P(x) .26
1
2
3
4
5
.31
.21
.13
.06
.03
What does this mean?
What is the probability that a randomly selected
household in Wolf City has at most 2 pets?
P(x < 2) = .26 + .31 + .21 = .78
Dogs and Cats Revisited . . .
Notice that this probability
Let x = the
number
of dogs2!
or cats per household
does
NOT include
in Wolf City
x
0
P(x) .26
1
2
3
4
5
.31
.21
.13
.06
.03
What does this mean?
What is the probability that a randomly selected
household in Wolf City has less than 2 pets?
P(x < 2) = .26 + .31 = .57
Dogs and Cats Revisited . . .
Let x = the number of dogs or cats per household
in Wolf City
When
probabilities
x calculating
0
1
2
3
4for discrete
5
random variables, you MUST pay close
P(x) .26 to.31
.21 certain
.13 .06
.03are
attention
whether
values
included (< or >) orWhat
not included
(< or
>) in the
does this
mean?
calculation.
What is the probability that a randomly selected
household in Wolf City has more than 1 but no
more than 4 pets?
P(1 < x < 4) = .21 + .13 + .06 = .40
Probability Distributions for
Continuous Random Variables
Consider the random variable:
x = the weight (in pounds) of a full-term
newborn child
Suppose that weight is reported to the nearest
pound. The
following
probability
histogram
What
type
of
variable
is
this?
If
weight
is
measured
with
greater
What
is
the
sum
of
the
areas
of all
displays
the
distribution
of
weights.
The
area
of
the
rectangle
and
greater
accuracy,
thecentered
histogram
the
rectangles?
Notice
that
the
rectangles
are
The
shaded
area
represents
the
over
7
pounds
represents
the
This
is
an
example
approaches
ahistogram
smooth
curve.
Nownarrower
suppose
that
and
the
weight
is
reported
begins
to
to
the
probability
6
<
x
<
8.
probability
6.5 appearance.
< xof
< 7.5
a density
curve.
have
a
smoother
nearest 0.1 pound. This would be
the probability
histogram.
Probability Distributions for
Continuous Variables
• 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 of observing a value in a
particular interval is the area under the
curve and above the given 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.
Let x denote the amount of gravel sold (in tons)
during a randomly selected week at a particular
sales facility. Suppose that the density curve
has a height f(x) above the value x, where
2(1  x ) 0  x  1
f (x )  
otherwise
0
The density curve is
shown in the figure:
Density
2
1
Tons
1
Gravel problem continued . . .
What is the probability that at most ½ ton of
gravel is sold during a randomly selected week?
P(x < ½) =
Density
2
1
1 – ½(0.5)(1) = .75
This
area
can
be found
by use
the
OR,The
more
easily,
by
finding
the
probability would be the
formula
for
the
area of a
area
of
the
triangle,
shaded area under the curve and
trapezoid:
1
above the interval
from 0 to 0.5.
A 1 bh
b1  b2 h
A 2
2 that area from 1.
and subtracting
Tons
1
Gravel problem continued . . .
What is the probability that exactly ½ ton of
gravel is sold during a randomly selected week?
P(x = ½) =
2
0
How
do
we
find
the
area
of
a
line
The
probability
would
be
the
area
Since a line segment has NO area,
Density
segment?
the
curve that
and above
0.5.
thenunder
the probability
exactly
½
ton is sold equals 0.
1
Tons
1
Gravel problem continued . . .
What is the probability that less than ½ ton of
gravel is sold during a randomly selected week?
P(x < ½) = P(x < ½)
Density
2
1
= 1 – ½(0.5)(1) = .75
Does theHmmm
probability
change
. . . This
is
whether different
the ½ is included
or not?
than discrete
probability distributions
where it does change
the probability whether
a value is included or
Tons
not!
1
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
f (x )  
0 otherwise
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 more than
5.5 minutes to process the application form?
P(x > 5.5) = .5(.5) = .25
When the density is constant over an interval
the probability
by curve),
calculating
(resultingFind
in a horizontal
density
the
area of the
shaded
region
probabilitythe
distribution
is called
a uniform
(base × height).
distribution.
Density
0.5
4
5
6
Time (in minutes)
Other Density Curves
Some density curves resemble the one
below. Integral calculus is used to find
the area under the 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
This will
be to
useful
laterofinb) –
P(a < x < b) = (cumulative
area
the left
a)
thisarea
chapter!
(cumulative
to the left of
P(a < x < b) = P(x < b) – P(x < a)
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
Mean and Variance for Discrete
Probability Distributions
• Mean is sometimes referred to as the
expected value (denoted E(x)).
μx   xp
• Variance is calculated using
s   x  mx  p
2
2
• Standard deviation is the square root of
the variance.
Dogs and Cats Revisited . . .
Let x = the number of dogs and cats in a
randomly selected household in Wolf City
x
0
1
2
3
4
5
P(x) .26 .31 .21 .13 .06 .03
xP(x) 0 + .31
.31 + .42 + .39 + .24
.24 + .15
What is the mean number of pets per household
in Wolf City?
First multiply
x-value
Next
find the each
sum of
thesetimes
values.
its corresponding probability.
mx = 1.51 pets
Dogs and Cats Revisited . . .
Let x = the number of dogs or cats per household
in Wolf City
x
0
P(x) .26
1
2
3
4
5
.31
.21
.13
.06
.03
What is the standard deviation of the number of
pets per household
in Wolf
City?– take the
This
is
the
variance
First
find the
of each xNext
multiply
bydeviation
the corresponding
square
root
of
this
value.
2
2
fromThen
the
mean.
Then2values.
square
probability.
add
these
sxvalue
= (0-1.51)
(.26)
+ (1-1.51)
(.31)
+
2(.21)deviations.
these
(2-1.51)
+ (3-1.51)2(.13) +
(4-1.51)2(.06) + (5-1.51)2(.03) =
1.7499
sx = 1.323 pets
Mean and Variance for Continuous
Random Variables
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.
• The standard deviation, sx, measures the
extent to which the continuous distribution
spreads out around mx.
A company receives concrete of a certain type
from two different suppliers.
Let
x = compression strength of a randomly selected
Thebatch
firstfrom
supplier
is preferred
to
Supplier
1
second strength
both in terms
of mean
y = the
compression
of a randomly
selected
batch
fromand
Supplier
2
value
variability.
Suppose that
mx = 4650 pounds/inch2 sx = 200 pounds/inch2
my = 4500 pounds/inch2 sy = 275 pounds/inch2
4300
4500
4700
my mx
4900
would
happenhad
to the
mean and
Suppose What
Wolf City
Grocery
a total
standard deviation
if weare
hadthe
to deduct
of 14 employees.
The following
$100 from everyone’s salary because
monthly salaries of all the employees.
of business being bad?
3500
1300
1200
1500
1900
1700
1400
2300
2100
1200
1800
1400
1200
1300
The
and
standard
of thesalaries
Let’smean
graph
boxplots
of deviation
these monthly
monthly
salaries
are to the distributions . . .
to see what
happens
mx = $1700 and sx = $603.56
What
We
see that the distribution
What
happened
just shifts
to theisright
happene
Suppose
business
really100
good, so the manager
to new
the
but the spread
the per month. The
dunits
to the
gives
everyone
a $100israise
standard
same.
means?
mean and standard deviation would be
deviations?
mx = $1800 and sx = $603.56
Wolf City Grocery Continued . . .
mx = $1700 and sx = $603.56
Suppose the manager gives everyone a 20% raise
- the new mean and standard deviation would be
m
= $2040 and sx = $724.27
x
Let’s graph boxplots of these monthly salaries
to see what happens to the distributions . . .
NoticeNotice
that multiplying
that bothbythe mean and standard
a constant stretches the
deviation increased by 1.2.
distribution, thus,
changing the standard
deviation.
Mean and Standard Deviation of
Linear functions
If x is a random variable with mean, mx,
and standard deviation, sx, and a and b
are numerical constants, and the random
variable y is defined by
y  a  bx
and
m y  ma bx  a  bm x
2
2
2 2
sy  sa bx  b sx or sy  b sx
Consider the chance experiment in which a customer
of a propane gas company is randomly selected. Let
x be the number of gallons required to fill a propane
tank. Suppose that the mean and standard deviation
is 318 gallons and 42 gallons, respectively. The
company is considering the pricing model of a service
charge of $50 plus $1.80 per gallon.
Let y be the random variable of the
amount billed.
What is the equation for y?
y = 50 + 1.8x
What are the mean and standard deviation for
the amount billed?
my = 50 + 1.8(318) = $622.40
sy = 1.8(42) = $75.60
Suppose we are going to play a game
called Stat Land! Players spin the two
spinners below and move the sum of the
two numbers.
Find the mean
and
2
1 2
1
3
standard
deviation
for
4 3
6
4
5
these
sums.
Spinner B
Spinner A
Not sure – let’s think
mA = 2.5
mB = 3.5
about
it and return
in
sjust
sB = 1.708
a few minutes!
A = 1.118
are sums
the mean
and
List all the Here
possible
(A + B).
standard
deviation for
Notice
that
the
mean
2 How
3 are
4 theeach
5standard
6
7
spinner.
of
the
sums
is
the
deviations
related?
mA+B = 6
3
4
5
6
7
8
sum of the means!
4
5
5
6
6
7
7
8
8
9
9
10
?
Move
1s
sA+B =2.041
Stat Land Continued . . .
Suppose one variation of the game
had players move the difference
of the spinners
2
1
2
4
3
1
6
?
Move
1s
3
4
5
Find
the
and
weB mean
find the
Spinner
Spinner A How do
standard
deviation
standard
for for
the
mA = 2.5
mBdeviation
= 3.5
these
differences.
sums
or
differences?
sA = 1.118
sB = 1.708
List all the possible differences (B - A).
0
1
2
3
4
5
-1
-2
-3
Notice
that the
mean
0WOW
-1
-2
–
this
is
the
of1 the differences
is
0
-1
same
value asofthe
the
difference
the
2
1
0
standard
deviation of
means!
3
2 sums!
1
the
4
3
2
mB-A= 1
sB-A =2.041
Mean and Standard Deviations
for Linear Combinations
If x1, x2, …, xn are random variables with means
m1, m2, …, mn and variances s12, s22, …, sn2,
respectively,
and is true ONLY if the x’s
This result
y = aare
+ a2x2 + … + anxn
1x1 independent.
then
This result is true regardless of
whether
my  a1the
mx x’s are
a2mindependent.

...

a
m
x
n
xn
2
1
s y  a12s x21  a22s x22  ...  an2s x2n
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 . . .
42 and sx =are
16 all traveling
Since them10
x =passengers
alone,
it
is
reasonable
to
think
that
the
10
The total weight of checked baggage, y, is
baggage weights are unrelated and
therefore
y = x1 + x2independent.
+ … + x10
What isTo
the
standard
deviation
of the total
find
the standard
deviation,
take
weight of the
baggage?
thechecked
square root
of this value.
sx2 = sx12 + sx22 + … + sx102
= 162 + 162 + … + 162
= 2560 pounds
s = 50.596 pounds
Special Distributions
Two Discrete Distributions:
Binomial and Geometric
One Continuous Distribution:
Normal 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
1. There are a fixed number of trials
2. Each trial results in one of two mutually
We use n to denote the fixed
exclusive outcomes.
(success/failure)
number
of trials.
3. Outcomes of different trials are independent
4. The probability that a trial results in success
is the same for all trials
The binomial random variable x is defined as
x = the number of successes observed when a
binomial experiment is performed
Are these binomial distributions?
1) Toss a coin 10 times and count the
number of heads
Yes
2) Deal 10 cards from a shuffled deck
and count the number of red cards
No, probability does not remain constant
3) The number of tickets sold to
children under 12 at a movie theater
in a one hour period
No, no fixed number
Binomial Probability Formula:
Let
n = number of independent trials in a binomial experiment
p = constant probability that any trial results in a success
Where:
n!
x
n x
P (x ) 
p (1  p )
x ! (n  x )!
n  9 can be used

ncalculators
! to find and
Appendix
Table
Technology,
such
as
 n C x 
binomial
probabilities.
x
statistical
software,
x ! (n  x )!will also
 
perform this calculation.
Instead of recording the gender of the next 25
newborns at a particular hospital, let’s record
the gender of the next 5 newborns at this
hospital.
is the
probability of
Is this a What
binomial
experiment?
“success”?
Yes, if the births were not multiple births
(twins, etc).
Define the random variable of interest.
What
will
the largest
value
of the
Will
a
binomial
random
variable
x = the number
of females
born
out
of the next
binomial
random
value
be?
always include the value of 0?
5 births
What are the possible values of x?
x
0
1
2
3
4
5
Newborns Continued . . .
What is the probability that exactly 2 girls will
be born out of the next 5 births?
P (x  2) 5 C2  0.5  0.5  .3125
2
3
What is the probability that less than 2 girls will
be born out of the next 5 births?
P (x  2)  p (0)  p (1)
5 C 0 .5 .5 5 C1 .5 .5
0
 .1875
5
1
4
Newborns Continued . . .
Let’s construct the discrete probability
distribution table for this binomial random
variable:
x
0
1
2
3
p(x)
.03125
.15625
.3125
.3125
4
5
.15625 .03125
is the
multiplying
WhatNotice
is the that
meanthis
number
ofsame
girls as
born
in the
next five births?
n×p
Since
this is a +discrete
mx = 0(.03125)
+ 1(.15625)
2(.3125) +
distribution,
could use:
3(.3125)
+ 4(.15625)we
+ 5(.03125)
=2.5
mx   xp
Formulas for mean and standard deviation
of a binomial distribution
m x  np
s x  np 1  p 
Newborns Continued . . .
How many girls would you expect in the next five
births at a particular hospital?
mx  np  5(.5)  2.5
What is the standard deviation of the number
of girls born in the next five births?
sx  np (1  p )  5(.5)(.5)
 1.118
Remember, in binomial distributions, trials
should be independent.
However, when we sample, we typically sample
without
replacement,
which replacement
would mean that
When
sampling without
if n
the trials
not5%
independent.
. binomial
is atare
most
of N, then. the
distribution
gives
a good
In this case, the
number of
success
observed
to the probability
would notapproximation
be a binomial distribution
but rather
distribution of x.
hypergeometric distribution.
But when
sample size,
n, is small and
The the
calculation
for probabilities
in the
a
population
size, N, is large,
probabilities
hypergeometric
distribution
are even
calculatedmore
usingtedious
binomial
distributions
and
than
the binomial
formula! are VERY close!
hypergeometric distributions
Newborns Revisited . . .
Suppose we were not interested in the
number of females born out of the next
five births, but
which birth would result in the first
female being born?
How is this question different from a
binomial distribution?
Properties of Geometric
Distributions:
• There are two mutually exclusive outcomes
that result in a success or failure
So what are the
• Each trial is independent of the others
possible values of x
• The probability of success is the same for all
trials.
To infinity
How far will this go?
A geometric random variable x is defined as
x = the number of trials UNTIL the FIRST
success is observed ( including the success).
x
1
2
3
4
. . .
Probability Formula for the
Geometric Distribution
Let
p = constant probability that any trial results in a success
x 1
p (x )  (1  p )
Where
x = 1, 2, 3, …
p
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
Is this a geometric distribution?
Yes
Jumper Cables Continued . . .
Let x = the number of students stopped before
finding one with a pair of jumper cables
p = .4
What is the probability that third student
stopped will be the first student to have jumper
cables?
P(x = 3) = (.6)2(.4) = .144
What is the probability that at most three
student are stopped before finding one with
jumper cables? P(x < 3) = P(1) + P(2) + P(3) =
(.6)0(.4) + (.6)1(.4) + (.6)2(.4) = .784
Normal Distributions
• Continuous probability distribution
is this
To overcome the need for How
calculus,
wedone
rely on
• Symmetrical bell-shaped (unimodal)
density
mathematically?
technology or on a table of areas for the
curve defined by m and s
standard normal distribution
• Area under the curve equals 1
• Probability of observing a value in a particular
interval is calculated by finding the area under
the curve
• As s increases, the curve flattens &
spreads out
• As s decreases, the curve gets
taller and thinner
A
6
B
s
s
Do these two normal curves have the same mean?
If so, what is it?
YES
Which normal curve has a standard deviation of 3?
B
Which normal curve has a standard deviation of 1?
A
Notice that the normal curve is curving
downwards from the center (mean) to points
that are one standard deviation on either side
of the mean. At those points, the normal curve
begins to turn upward.
Standard Normal Distribution
• Is a normal distribution with m = 0 and s = 1
• It is customary to use the letter z to
represent a variable whose distribution is
described by the standard normal curve (or z
curve).
Using the Table of Standard
Normal (z) Curve Areas
• For any number z*, from -3.89 to 3.89 and
use theplaces,
table: the Appendix
rounded to twoTodecimal
Table 2 gives the area under the z curve and
to•the
left
ofcorrect
z*.
Find
the
row and column (see
the following
P(z < example)
z*) = P(z < z*)
• The number at the intersection of
Where that row and column is the probability
the letter z is used to represent a random variable
whose distribution is the standard normal
distribution.
Suppose we are interested in the probability
that z* is less than -1.62.
In the table of areas:
P(z < -1.62) = .0526
•Find the row labeled -1.6
•Find the column labeled 0.02
-1.7
-1.6
-1.5
.0446
.0548
.0668
.0436
.0537
.0655
.0427
.0526
.0643
…
…
…
…
…
•Find the intersection of the row and column
…
z*
.00
.01
.02
.0418
.0516
.0618
Suppose we are interested in the probability
that z* is less than 2.31.
P(z < 2.31) = .9896
2.2
2.3
2.4
.9861
.9893
.9918
.9864
.9896
.9920
.9868
.9898
.9922
…
…
.02
…
.01
…
.00
…
…
z*
.9871
.9901
.9925
Suppose we are interested in the probability
that z* is greater than 2.31.
2.2
2.3
2.4
.9861
.9893
.9918
.9864
.9896
.9920
.9868
.9898
.9922
…
…
…
…
…
The Table of Areas gives the area to the
P(z > 2.31) =
LEFT of the z*.
1 - .9896 = .0104
To find the area to the right, subtract
the value in the table from 1
…
z*
.00
.01
.02
.9871
.9901
.9925
Suppose we are interested in the finding the z*
for the smallest 2%.To find z*:
Look for the area .0200 in the body of
P(z < zthe
*) = Table.
.02
Follow the row and column
.0200out
doesn’t
appear
the body
back
to read
the zin-value.
zSince
* = -2.08
*
of the Table, use the zvalue
closest to it.
-2.1
-2.0
-1.9
.0162
.0207
.0262
.0158
.0202
.0256
…
…
.08
…
…
…
…
.07
…
.06
…
…
z*
.0154
.0197
.0250
Suppose we are interested in the finding the z*
for the largest 5%.
Since .9500 is exactly between
.9495
.95
P(z > z*)and
= .05
.9505, we can average the z* for
each of these
z* = 1.645
z*
…
…
…
…
…
Remember the Table of Areas gives the
area to the LEFT of z*.
…
z*
.03
.04
.05
1 – (area to the right of z*)
…
1.5
Then look up this.9382
value in .9398
the body.9406
of
… the.9495
1.6
.9515
table. .9505
…
1.7
.9591
.9599
.9608
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 m
a* 
s
b m
b* 
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 normal distribution with mean
m = 12 minutes and standard deviation s = 2
minutes.
Registration Problem Continued . . .
x = time to register
Standardized this
value.
m = 12 minutes and s = 2 minutes
What is the probability that
willvalue
take up
a in
Lookitthis
randomly selected student less
than
9 minutes to
the
table.
complete registration?
P(x < 9) = .0668
9  12
b* 
 1 . 5
2
9
Registration Problem Continued . . .
x = time to register
Standardized this
value.
m = 12 minutes and s = 2 minutes
What is the probability that
willvalue
take up
a in
Lookitthis
randomly selected student more
than 13
the table
andminutes
to complete registration? subtract from 1.
P(x > 13) = 1 - .6915 = .3085
13  12
a* 
 .5
2
13
Registration Problem Continued . . .
x = time to register
Standardized these
values.
m = 12 minutes and s = 2 minutes
these values
in take
the table
and
What is the Look
probability
that itup
will
a
randomly selected studentsubtract
between 7 and 15
(value
for a*) – (value for b*)
minutes to complete
registration?
P(7 < x < 15) = .9332 - .0062 = .9270
15  12
a* 
 1 .5
2
7  12
b* 
 2.5
2
7
15
Registration Problem Continued . . .
x = time to register
m = 12 minutes and s = 2 minutes
Look
up thedoarea
to off
theproperly, the
Because some
students
not log
Use
the
formula
for
university would
logthe
off
students
automatically
left like
of ato
* in
table.
standardizing
to find
x.
after some time has elapsed.
It is decided
to select
this
time so that only 1% of students will be automatically
logged off while still trying to register.
What time should the automatic log off be set
at?
P(x > a*) = .01
a* = 16.66
.99
x  12
2.33 
2
.01
a*
Ways to Assess Normality
What should
Some
of theifmost
happen
our frequently used statistical
methods
are
valid
only
when
x
,
x
,
…,
x
has
come
1
2
n
data set is
from a population distribution that at least is
normally
approximately normal. One way to see whether an
distributed?
assumption
of population normality is plausible is to
construct a normal probability plot of the data.
A normal probability plot is a scatterplot of (normal
score, observed values) pairs.
Consider a random sample with n = 5.
To find the appropriate normal scores for a
Each
region has
sample
ofthese
size 5, divide the
standard
Why are
an area
equal to
normal
into 5 equal-area
regions.
regionscurve
not the
same width?
0.2.
Consider
a are
random
samplescores
with that
n = 5.
These
the normal
we
Next – find
the median
z-score
for each region.
would
plot our
data against.
Why is the
We
use
technology
(calculators
or
median not in
statistical
software)
to
compute
these
the “middle” of
normal
scores.
each region?
-1.28
-.524
0
1.28
.524
Ways to Assess Normality
Some of the most Such
frequently
used statistical
as curvature
which would
methods Or
areoutliers
valid only
whenskewness
x1, x2, …,inxnthe
hasdata
come
indicate
from a population distribution that at least is
approximately normal. One way to see whether an
assumption of population normality is plausible is to
construct a normal probability plot of the data.
A normal probability plot is a scatterplot of (normal
score, observed values) pairs.
A strong linear pattern in a normal probability plot
suggest that population normality is plausible.
On the other hand, systematic departure from a
straight-line pattern indicates that it is not reasonable
to assume that the population distribution is normal.
Sketch
a scatterplot
byprobability
pairing
theis plot.
The
Let’s
following
construct
data
a normal
represent
eggplot
weights
(in
Since
the
normal
probability
smallest
normal
score
the
grams)
Since approximately
for
the
avalues
sample
ofof
the
10
normal
eggs.
scores
linear,
it with
is plausible
smallest
observation
from
data
set
depend
the
sample size
nthe
, the
normal
thaton
the
distribution
of egg
weights
is
&
so
on
approximately
normal.
scores when n = 10 are below:
53.04
53.50
52.53
53.00
53.07
53.5
52.86
52.66
53.23
53.26
53.16
53.0
-1.539 -1.001 -0.656 -0.376 -0.123
0.123 0.376 0.656 1.001 1.539
52.5
-1.5
-1.0 -0.5
0.5
1.0
1.5
Using the Correlation Coefficient
to Assess Normality
•The correlation coefficient, r, can be calculated for
the n (normal score, observed value) pairs.
•If r is too much smaller than 1, then normality of the
Since
underlying distribution is questionable.
r > to
critical
, Normality
Values to Which r Can be Compared
Check rfor
How
smaller
iseggs
“too
then
it is
plausible
that
the60
sample
Consider
from
the
weight
data:
n
5 these
10 points
15
20
25
30
40 of 50
75
of much
egg weights
came
from
a
smaller
than
1”?
(-1.539, 52.53)
(-1.001, 52.66)
(-.656,52.86)
(-.376,53.00)
Critical
(-.123, 53.04)
.832 .880 (.123,53.07)
911distribution
.929 .941 (.376,53.16)
.949
.960
.966(.656,53.23)
.971 .976
that
was
approximately
r
(1.001,53.26)
(1.539,53.50)
normal.
Calculate the correlation coefficient for these points.
r = .986
Transforming Data to Achieve
Normality
• When the data is not normal, it is common to
use a transformation of the data.
• For data that shows strong positive skewness
(long upper tail), a logarithmic transformation
usually applied.
• Square root, cube root, and other
transformations can also be applied to the
data to determine which transformation best
normalizes the data.
Consider the data set in Table 7.4 (page 463)
about plasma and urinary AGT levels.
A histogram of the
urinary AGT levels
is strongly
positively skewed.
A logarithmic
transformation is
applied to the data.
The histogram of
the log urinary
AGT levels is more
symmetrical.
Using the Normal Distribution to
Suppose thisabar
is centered
at x = 6.
Approximate
Discrete
Distribution
The bar actually begins at 5.5 and ends
at 6.5.
endpoints
will be used
Suppose
theTheses
probability
distribution
of a in
Often,
a probability
histogram
can be
well
calculations.
discrete
random
variable
x is displayed
in the
approximated by a normal curve. If so, it
histogram below.
is customary
to sayofthat
x has an
The
probability
a
particular
This
is
called
a
continuity
correction.
approximately
normal
distribution.
value is the area
of the
rectangle
centered at that value.
6
Normal Approximation to a
Binomial Distribution
Let x be a random variable based on n trials and
success probability p, so that:
m  np
s  np (1  p )
If n and p are such that:
np > 10 and n (1 – p) > 10
then x has an approximately normal distribution.
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 are randomly selected and
that the number of these births that occurred
prior to 34 weeks, x, is to be determined.
Since both are
greater than 10, the
np = 1000(.02)
= 20
> 10
distribution
of x can
Can
the
distribution
of x be
be
by
by approximated
a normal
n(1 – p) = 1000(.98)approximated
= 980 > 10
distribution?a normal
distribution
Find the mean and standard deviation for the
approximated normal distribution.
m  np  1000(.02)  20
s  np (1  p )  1000 (.02)(.98)  4.427
Premature Babies Continued . . .
m = 20 and s = 4.427
What is the probability
that
the number of
Look up
these
babies in the sample
ofin1000
values
the born
tableprior to 34
weeks will be between
10 and 25
and subtract
the(inclusive)?
To find the shaded
probabilities.
standardize
= .8836
P(10 < x < 25) = .8925 - .0089 area,
the endpoints.
a* 
9.5  20
 2.37
4.427
b* 
25 .5  20
 1.24
4.427

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