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Chapter 6
Random Variablesand Probability
DistributionsCreated by Kathy Fritz
Consider the chance experiment of randomly selecting a customer who is leaving a store.
One numerical variable of interest to the store manager might be the number of items purchased by the customer.
Let’s use the letter x to denote this variable. In this example, the values of x are isolated points.
Another variable of interest might be y = number of minutes spent in a checkout line.The possible y values form an entire interval on the number line.
What are possible values for x?
Until a customer is selected and the number of items counted, the value of x is uncertain.
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.
Random Variables
Random Variable
A random variable is a numerical variable whose value depends on the outcome of a chance experiment.
A random variable associates a numerical value with each outcome of a chance experiment.
• A random variable is discrete if its possible values are isolated points along the number line.
• A random variable is continuous if its possible values are all points in some interval.
This is typically a “count” of something.
This is typically a “measure” of something
In this chapter, we will look at different distributions of discrete and continuous
random variables.
Identify the following variables as discrete or continuous
1. The number of items purchased by each customer
2. The amount of time spent in the checkout line by each customer
3. The weight of a pineapple
4. The number of gas pumps in use
Discrete
Discrete
Continuous
Continuous
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 5Is 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.
Common ways to display a probability distribution for a discrete random variable are a table, probability histogram, or formula.
If one possible value of x is 2, it is common to write p(2) in place of P(x = 2).
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 energy-efficient rating.
Assume that these customers make their choices independently of one another and that 40% of all customers select an energy-efficient model.
Consider the next four customers. Let: x = the number of energy efficient refrigerators purchased by the four customers
What are the possible values for x?
x 10 42 3
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) = 0.1296P(1) = P(EGGG) + P(GEGG) + P(GGEG) + P(GGGE) = 0.0864 + 0.0864 + 0.0864 + 0.0864 = 0.3456
Similarly, P(2) = 0.3459P(3) = 0.1536P(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 0 1 2 3 4
P(x) 0.1296 0.3456 0.3456 0.1536 0.0256
The probability distribution can be used to determine probabilities of various events involving x.
For example, the probability that at least two of the four customers choose energy-efficient models is
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 0 1 2 3 4
P(x) 0.1296 0.3456 0.3456 0.1536 0.0256
What is the probability that more than two of the four customers choose energy-efficient models?
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
>).
Refrigerators continued . . .
x 0 1 2 3 4
P(x) 0.1296 0.3456 0.3456 0.1536 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 childSuppose that weight is reported to the nearest pound. The following probability histogram displays the distribution of weights.
Now suppose that weight is reported to the nearest 0.1 pound. This would be the probability histogram.
What type of variable is this?What is the sum of the areas
of all the rectangles?If weight is measured with greater and
greater accuracy, the histogram approaches a smooth curve.
The area of the rectangle centered over 7 pounds represents the probability 6.5 < x <
7.5
Notice that the rectangles are narrower and the histogram begins to have a
smoother appearance.The shaded area represents the
probability 6 < x < 8.
This is an example of a density curve.
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:
The following is the graph of f(x), the density curve:
otherwise0
645.)(
xxf
0.5
4 5 6
Time (in minutes)
Densi
ty
When the density is constant over an interval (resulting in a horizontal density curve), the probability distribution is called a uniform
distribution.
Why is the height of this density curve 0.5?
Application Problem Continued . . .
What is the probability that it takes at least 5.5 minutes to process the application form?
0.5
4 5 6
Time (in minutes)
Densi
tyP(x ≥ 5.5) =(6 - 5.5)(.5) = .25
Find the probability by calculating the area of the shaded region (base × height).
Application Problem Continued . . .
What is the probability that it takes exactly 5.5 minutes to process the application form?
0.5
4 5 6
Time (in minutes)
Densi
tyP(x = 5.5) =0 x = 5.5 is represented by
a line segment.
What is the area of this line segment?
Application Problem Continued . . .
What is the probability that it takes more than 5.5 minutes to process the application form?
0.5
4 5 6
Time (in minutes)
Densi
tyP(x > 5.5) =(6 - 5.5)(.5) = .25
In continuous probability distributions, P(x > a) and P(x ≥ a) are equal!
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 that a reasonable model
for the distribution of x
might be a triangular
distribution.1 2
0.5
1.0
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.What proportion of the packages weigh over 1.5 pounds?
1 2
0.5
1.0
𝑃 (𝑥>1.5 )=1−𝑃 (𝑥≤1.5)
The easiest way to find the area of the shaded region is to find 1 – the area of x ≤
1.5.The area of a triangle is
b = 1.5
h = 0.75
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 probability histogram (below).
The general form of the histogram can be described
as bell shaped and symmetric.
A smooth curve has been
superimposed on the histogram
and is a reasonable model for the probability distribution of x.
How can you find the area under
this smooth curve?
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 VariablesOf 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 sx is small, observed values of x will tend to be close to the mean value.
The larger the value of sx the more variability there will be in
observed x values.
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?
How do the means and standard deviations of these three density curves compare?
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,
The term expected value is sometimes used in place of mean value and E(x) is another way to denote mx .
𝜇𝑥=∑ 𝑥 ∙𝑝(𝑥 )all possible x values
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
Then x has mean value
(1)(0.10)+(2)(0.20)+(3)(0.30)+(4)(0.40) 3.00
The variance of a discrete random variable x, denoted by , is computed by first subtracting the mean from each possible x value to obtain the deviations, then squaring each deviation and multiplying the result by the probability of the corresponding x value, and finally adding these quantities. Symbolically
The standard deviation of x, denoted by sx, is the square root of the variance.
Standard Deviation for a Discrete Random Variable
𝜎 𝑥2=∑ (𝑥−𝜇 )2𝑝 (𝑥)all possible x values
Revisit the license example . . .
x = the number of attempts made by a randomly selected applicant
The probability distribution of x is as follows:
Then x has variance
(1-3)2(0.10) + (2-3)2(0.20) + (3-3)2(0.30) + (4-3)2(0.40) 1.00
The standard deviation of x is
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 of a randomly selected batch from Supplier 2
Suppose that mx = 4650 pounds/inch2 sx = 200 pounds/inch2
my = 4500 pounds/inch2 sy = 275 pounds/inch2
Which supplier should the company purchase the concrete from? Explain.
45004300 4700 4900my mx
The density curves look similar to these below.
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 = 3xymodel 2 = 50 + 2.8x
Mean and Standard Deviation of Linear FunctionsIf x is a random variable with mean, mx, and variance, sx
2, and a and b are numerical constants, then the random variable y defined by
is called a linear function of the random variable x.The mean of is
The variance of is
from which it follows that the standard deviation of y is
Revisit the propane gas company . . .
m = 318 gallons s = 42 gallons
The company is considering two different pricing models.
Model 1: $3 per gallon
Model 2: service charge of $50 + $2.80 per gallon For Model 1:
For Model 2:
117.60
The mean billing amount for Model 1 is a bit higher than for Model 2, as is the
variability in billing amounts.
Model 2 results in slightly more consistency from bill to bill in the
amount charged.
Let’s consider a different type of problem . . .
Suppose that you have three tasks that you plan to do on the way home.
Return library book
Deposit paycheck
Purchase printer paper
x1 = time required to return book
x2 = time required to deposit check
x3 = time required to buy printer paperYou can define a new variable, y, to represent the total amount of time to complete these tasks
y = x1 + x2 + x3
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 and variances s1
2, s22, …, sn
2, respectively, and
y = a1x1 + a2x2 + … + anxn
then
This result is true regardless of whether the xi’s are independent.
1.
2. When x1, x2, …, xn are independent random variables,
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
Where:x1 = weight of the first passenger’s luggagex2 = weight of the first passenger’s luggage x10 = weight of the first passenger’s luggage
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?
= 162 + 162 + … + 162
= 2560 pounds
s = 50.596 pounds
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 DistributionsMean 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.
What is the chance that at
least 15 are female?What is the chance that between
10 and 15 are female?
Out of the 25 newborns,
how many can we expect
to be female? These questions can be answered using a binomial
distribution.
A binomial experiment consists of a sequence of trials with the following conditions:
1. There are a fixed number of trials2. Each trial results in one of only two possible
outcomes, labeled success (S) and failure (F).3. Outcomes of different trials are independent4. The probability of success is the same for each trial.
The binomial random variable x is defined as x = the number of successes observed when a
binomial experiment is performed
Properties of a Binomial Experiment
We use n to denote the fixed number of trials.
The probability distribution of x is called the binomial probability distribution.
The term success does not necessarily mean something positive. For example, if the
random variable is the number of defective items produced, then being “defective” is a
success.
Binomial Probability Formula:
Let n = number of independent trials in a binomial experimentp = constant probability that any particular trial results in a success
Then
nxppxnx
n
trialsntheamongsuccessesxPxp
xnx .,..,2,1,0)1()!(!
!
)()(
Notice that the probability distribution is specified by a formula rather than a
table or probability histogram.
. . . can be written as nCx
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.,..,2,1,0)4.0()6.0()!12(!
!12)( 12
x
xxxp xx
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.
What is the probability that exactly four of the next 12 computers sold are laptops?
If many groups of 12 purchases are examined, about 4.2% of them include exactly four laptops.
042.0
4.06.0!8!4
!12
)4()4(
84
xPp
What is the probability that between four and seven (inclusive) are laptops?
547.0
227.0177.0101.0042.0
4.06.0!5!7
!12...4.06.0
!8!4
!12
)7()6()5()4()74(
5784
ppppxP
These calculations can become very tedious. 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?
278.0
177.0101.0
4.06.0!6!6
!124.06.0
!7!5
!12
)6()5()74(
6675
ppxP
Notice that the probability depends on whether < or ≤ appears. This is typical
of discrete random variables.
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.
Using Appendix Table 9 to Compute Binomial Probabilities
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 to the population before the next selection), then the outcomes are independent.
However, sampling is usually done without returning (without replacement) the element to the population before the next selection. Therefore, the outcomes are dependent and the observed number of successes have a hypergeometric distribution.
The calculations are even more tedious for the
hypergeometric distribution than for the binomial
distribution.
If (n/N) ≤ 0.05, i.e., no more than 5% of the population is sampled,
then the binomial distribution gives a good approximation to the probability distribution of x.
Formulas for mean and standard deviation of a binomial distribution
pnp
np
x
x
1
Define the random variable of interest as
x = the number of laptops among these 12
Compute the mean and standard deviation for the binomial distribution of x.
laptops
laptops
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?
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 trial)
The probability distribution of x is called the geometric probability distribution.
Properties of a Geometric Experiment
How do these properties differ from those of a binomial 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
Where x = 1, 2, 3, …
ppxp x 1)1()(
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?
What is the probability that three or fewer students are stopped before finding one with jumper cables?
p(3) =
(0.6)2(0.4) = 0.144
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 CurveUsing a Table to Calculate Probabilities
Other Normal Curves
Normal Distributions . . .
are continuous
distributions
are bell shaped and continuousapproximate the
distributions of
many different
variablesare used in inferential procedures
are distinguished from
one another by their
mean m and standard
deviation s
have an area under
the curve equal to 1
Normal Distributions . . .The value of m is the
number on the
measurement axis lying
directly below the top of
the bell.
The value of s can be approximated from a picture of the curve.
It is the distance to
either side of m at
which a normal curve
changes from turning
downward to turning
upward.
Standard Normal Distribution . . .
The standard normal distribution is the 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).
Thus, the z curve is comprised of z values instead of x values.
A table of areas under the standard normal curve is used to calculate probabilities of
events.
Using the Table of Standard Normal Curve Areas
For any number z*, from -3.89 to 3.89 and rounded to two decimal places, the Appendix Table 2 gives
(area under z curve to the left of z*) = P(z < z*) = P(z < z*)
Where the letter z is used to represent a random variable whose distribution is the standard normal distribution.
To find this probability using the table, locate the following:
• 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)
• The column identified with the second digit to the right of the decimal point in z*
• The number at the intersection of this row and column is the desired probability.
Suppose we are interested in the probability that z is less than 1.42.
P(z < 1.42) =
z* .00 .01 .02 .03
1.3 .9032 .9049 .9066 .9082
1.4 .9192 .9207 .9222 .9236
1.5 .9332 .9345 .9357 .9370
0.9222
P(z < 1.42)
1.42
Find the intersection of the row 1.4 and column .02.
………
… … … … …
……
Suppose we are interested in the probability that z* is less than 0.58.
P(z < 0.58) =
z* .07 .08 .09
0.4 .6808 .6844 .6879
0.5 .7157 .7190 .7224
0.6 .7486 .7517 .7549
0.7190
… … … … …
…
………
P(z < 0.58)
Find the following probability:
P(-1.76 < z < 0.58) =
P(z < 0.58)
.7190
- P(z < -1.76)
P(z < -1.76)
- .0392
= 0.6798
Suppose we are interested in the probability that z* is greater than 2.31.
P(z > 2.31) =
z* .00 .01 .02
2.2 .9861 .9864 .9868 .9871
2.3 .9893 .9896 .9898 .9901
2.4 .9918 .9920 .9922 .9925
… … … … …
…
1 - .9896 = 0.0104
The Table of Areas gives the area to the LEFT of the z*.
To find the area to the right, subtract the value in the table from 1
Suppose we are interested in the finding the z* for the smallest 2%.
P(z < z*) = .02
z* .03 .04 .05
-2.1 .0162 .0158 .0154
-2.0 .0207 .0202 .0197
-1.9 .0262 .0256 .0250
… … … … …
…
z* = -2.05z*
To find z*:
Look for the area .0200 in the body of the Table. Follow the row and column back out
to read the z-value.
………
Since .0200 doesn’t appear in the body of the Table, use the value closest to it.
Finding Probabilities for Other Normal CurvesTo 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
b
b*
aa*
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.
What is the probability that it will take a randomly selected student less than 9 minutes to complete registration?
9
5.12129
*
b
P(x < 9) = 0.0668
Look this value up in the table.
Standardizes 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?
P(x > 13) =
5.2
1213*
a
1 - .6915 = 0.3085
13
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?
P(7 < x < 15) =
5.12
1215*
a
.9332 - .0062 = 0.9270
5.22127
*
b 7 15
Ways to Assess Normality
Normal Probability PlotUsing Correlation Coefficient
Normal Probability Plot
A normal probability plot is a scatterplot of (normal score, observed values) pairs.
A strong linear pattern in a normal probability plot suggests that the population distribution is approximately normal.
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.
One way to see whether an assumption of population normality is plausible is to
construct a normal probability plot of the data.
Normal scores are z-scores from the standard normal distribution.Or outliers
Such as curvature which would indicate skewness in the data
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 equal-area regions.Why are these
regions not the same width?
Each region has an area equal to
0.2.
What are normal scores?
1.28
.524-.524
Next – find the median z-score for each region.
-1.28 0
Why is the median not in the “middle”
of each region?
These are the normal scores that we would plot our data against.
We use technology (calculators or statistical software) to compute these normal scores.
What are normal scores?
Let’s construct a normal probability plot.
Since the values of the normal scores depend on the sample size n, the normal scores when n = 10 are below:
-1.539 -1.001 -0.656 -0.376 -0.123 0.123 0.376 0.656 1.001 1.539
The following data represent egg weights (in grams) for a sample of 10 eggs.
53.04 53.50 52.53 53.00 53.07 52.86 52.66 53.23 53.26 53.16
Sketch a scatterplot by pairing the smallest normal score with the smallest observation
from the data set and so on.
-1.5 -1.0 -0.5 0.5 1.0 1.5
52.5
53.0
53.5
Since the normal probability plot is approximately linear, it is plausible that the distribution of egg weights is approximately
normal.
Using the Correlation Coefficient to Assess NormalityThe 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 underlying distribution is questionable.
Consider these points from the weight of eggs data:(-1.539, 52.53) (-1.001, 52.66) (-.656,52.86) (-.376,53.00) (-.123, 53.04) (.123,53.07) (.376,53.16)
(.656,53.23) (1.001,53.26) (1.539,53.50)
Calculate the correlation coefficient for these points.
How smaller is “too much smaller” than 1?
Values to Which r Can be Compared to Check for Normality
n 5 10 15 20 25 30 40 50 60 75
Critical r .832 .880 911 .929 .941 .949 .960 .966 .971 .976
Since r > critical r,
then it is plausible that the sample of egg weights came from a
distribution that was approximately normal.
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 histogram below.
The probability of a particular value is the area of the rectangle centered at
that value.
Often, a probability histogram can be well approximated by a normal curve. If so, it is
customary to say that x has an approximately normal distribution.
6
Suppose this rectangle is centered at x = 6. The rectangle actually begins at 5.5 and
ends at 6.5. These endpoints will be used in calculations.
This is called a continuity correction.
In general, if possible x values are consecutive whole numbers, then P(a ≤ x ≤ b) (including a and b) will be approximately the normal curve
area between limits and .
However, P(a < x < b) (excluding a and b) will be approximately the area between limits
and .
Normal Approximation to a Binomial Distribution
Suppose x is a binomial random variable based on n trials and success probability p, so that:
If n and p are such that:
np ≥ 10 and n (1 – p) ≥ 10
then the distribution of x is approximately normal.
Combining this result with the continuity correction implies that
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
The distribution of x is approximately normal with
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 between 10 and 25 (inclusive)?
P(10 < x < 25) =0.8925 - 0.0089 = 0.8836
24.1427.4
205.25*
b
To find the shaded area, standardize the
endpoints.
37.2427.4
205.9*
a
Look up these values in the table and
subtract the probabilities.Since 10 is included in the
probability, the endpoint of the rectangle for 10 is 9.5.
Similarly, the endpoint of the rectangle for 25 is 25.5.
