### Chap5-8x

```Random Variables and
Probability Distributions
Chapters 5-8
McGraw-Hill/Irwin
Learning Objectives
Chapter 5
1. Define probability.
2. Define Random Variable.
3. Understand discrete random variable and continuous random
variable.
Chapter 6
Identify the characteristics of a probability distribution.
Chapter 7
1. List the characteristics of the normal probability distribution.
2. Convert a normal distribution to the standard normal distribution.
3. Find the probability for a normally distributed random variable.
4. Find the Z-value.
5-2
Learning Objectives
Chapter 8
1 Explain why a sample is often the only feasible way to learn
4 Describe the sampling distribution of the sample mean.
5 Understand the central limit theorem.
6 Define the standard error of the mean.
7 Apply the central limit theorem to find probabilities.
8-3
A Survey of
Probability Concepts
Chapter 5
McGraw-Hill/Irwin
Random Variables (r.v)
RANDOM VARIABLE: A quantity resulting from an experiment
that, by chance, can assume different values. Random variable is
usually denoted by X.
Discrete r.v.: A random variable that can assume only certain clearly
separated values.
1. The outcome when throwing a die of six sides
2. The number of students in a class.
3. The number of cars entering a carwash in a hour.
Continuous r.v.: can assume an infinite number of values within a
given range.
1. The weight of each student in this class.
2. The temperature outside as you are reading this book.
3. The amount of money earned by each of the more than 750 players
currently on Major League Baseball team rosters.
Probability
PROBABILITY: A value between zero and one, inclusive,
describing the relative possibility (chance or likelihood) an
event will occur. A probability is the relative possibility that a
random variable assumes certain values.
Notations:
For discrete r.v.: P(X = x) or P(x)
For continuous r.v.: P(X > x)
Discrete
Probability
Distributions
Chapter 6
McGraw-Hill/Irwin
Probability Distribution
PROBABILITY DISTRIBUTION A listing of all the outcomes of
an experiment and the probability associated with each outcome.
Experiment:
Outcome
P(X=x)
Toss a die of six sides.
The outcome of the
experiment is a r.v,
denoted by X. The
possible results are: 1,
2, 3, 4, 5, and 6.
1
1/6
2
1/6
3
1/6
4
1/6
5
1/6
What is the probability
distribution for each of
the outcome?
6
1/6
Total
1
Characteristics of a Probability Distribution
1.The probability of a particular outcome is between
0 and 1 inclusive.
2. The outcomes are mutually exclusive events.
3. The list is exhaustive. So the sum of the probabilities
of the various events is equal to 1.
4. For continuous distributions, P(X = x) = 0
Continuous
Probability
Distributions
Chapter 7
McGraw-Hill/Irwin
Normal Probability Distribution
1.
2.
3.
4.
It is bell-shaped and symmetrical about the mean.
The area under the curve denotes probability. The total area
under the curve is 1.00.
The location of a normal distribution curve is determined by
the mean , the shape of the curve is determined by the
standard deviation,σ.
For a r.v. that follows normal distribution with mean  and s.d.
σ, we denote X~(, σ).
mean 
The Family of Normal Distribution
The Standard Normal Probability
Distribution




The standard normal distribution is a normal
distribution with a mean of 0 and a standard
deviation of 1. It is also called the z distribution.
A r.v. that follows standard normal distribution is
denoted by Z.
Converting a normally distributed r.v., X to a
standard normally distributed r.v., Z: subtract X by
its mean and divide by its s.d.
Z score
Probabilities: Areas Under Curve
P(Z > 1.80)= P(Z < -1.80) = .0359;
P(Z < −1.52) = .0643;
=NORMSDIST(-1.8)
=NORMSDIST(-1.52)
P(−1.30 < Z < 2.10) = P(Z<2.1) – P(Z< - 1.3) = .8853;
=NORMSDIST(2.1) - NORMSDIST(-1.3)
Excel function: =NORMSDIST(z)
Normal distribution table
Normal distribution table
The Normal Distribution – Example
The weekly incomes of shift
foremen in the glass industry
distribution with a mean of
\$1,000 and a standard
deviation of \$100.
What is the probability of
selecting a shift foreman in
the glass industry whose
income is between \$840 and
\$1,200?
the Z score is
the Z score is
P(840 < X <1,200)
= P (-1.6 < Z < 2.0 )
=P (Z < 2.0) – P(Z < -1.6)
= .9224
Excel function:
=NORMDIST(mean, s.d, x, true)
=NORMDIST(1000, 100, 1200, true) - NORMDIST(1000, 100, 840,
true)
Finding Values of Z For Given
Areas
Given an area () under the curve, what is the
corresponding value of z (z ) on the horizontal axis
such that the area to its right is ?
P(Z > z) =
Example: Z0.05=?

Excel function
=NORMSINV(1-)
=NORMSINV(1-0.05)

8.18
Z0.05=1.645
Finding Values of Z For Given
Areas
0.025
-z0.025
Example: - Z0.025=?
Excel function
=NORMSINV()
=NORMSINV(0.025)
- Z0.025= -1.96
Other commonly used
z-values
z0.10 = 1.28
-z0.10 = -1.28
z0.01 = 2.326
-z0.01= -2.326
Finding z values Using the
Normal Table
Example: z0.025
To find the z value in the normal table, we look for .025 (the
subscript of Z) inside the table. Here we can find an exact match
for .025. Thus the corresponding z value is 1.96.
Example: z0.05



When we look for .05 (the subscript of Z) inside the table, we
cannot find an exact match for .05. Instead we find two values
that are close to .05: .495 and .505.
In this class only: if we cannot find an exact match for the
subscript of Z, we will take the average of the two corresponding
Z values, 1.64 and 1.65.
z0.05=(1.64+1.65)/2=1.645

z values found this way can be slightly different from the z
values found using Excel.
Using Z in Finding X Given Area - Example
Layton Tire and Rubber
Company wishes to set a
minimum mileage guarantee on
its new MX100 tire. Tests
reveal the mean mileage is
67,900 with a standard
deviation of 2,050 miles and
that the distribution of miles
follows the normal probability
distribution. Layton wants to set
the minimum guaranteed
mileage so that no more than 4
percent of the tires will have to
be replaced.
What minimum guaranteed
mileage should Layton
announce?
21
Using Z in Finding X Given Area - Example
Solve X using the formula :
z
x-


x  67,900
2,050
The value of z is found using the 4% informatio n
- z 0 .04  - 1.75.
- 1.75 
Excel function
=NORMINV(, mean, s. .)
x - 67,900
, then solving for x
2,050
- 1.75(2,050 )  x - 67,900
=NORMINV(0.04, 67900,
x  67,900 - 1.75(2,050 )
2050)
x  64,312
The Empirical Rule



the area under the
normal curve is within
one standard
deviation of the
mean.
within two standard
deviations of the
mean.
Practically all is
within three standard
deviations of the
mean.
7-23
Sampling Methods and
the Central Limit Theorem
Chapter 8
McGraw-Hill/Irwin
Sampling Method—Simple
Random Sample
Simple Random Sample: A sample selected so that each item
or person in the population has the same chance of being
included.
EXAMPLE:
A population consists of 845 employees of Nitra Industries. A
sample of 52 employees is to be selected from that population.
The name of each employee is written on a small slip of paper and
deposited all of the slips in a box. After they have been thoroughly
mixed, the first selection is made by drawing a slip out of the box
without looking at it. This process is repeated until the sample of
52 employees is chosen.
8-25
Sampling Distribution of the
Sample Mean
The sampling distribution of the sample mean is a
probability distribution consisting of all possible
sample means of a given sample size selected from a
population.
If all samples of a particular
size are selected from any population, the sampling
distribution of the sample mean is approximately a
normal distribution. This approximation improves with
larger samples.
CENTRAL LIMIT THEOREM
8-26
Central Limit Theorem

8-27
The Quality Assurance Department for Cola, Inc., maintains records
regarding the amount of cola in its Jumbo bottle. The actual amount
of cola in each bottle is critical, but varies a small amount from one
bottle to the next. Cola, Inc., does not wish to underfill the bottles.
On the other hand, it cannot overfill each bottle. Its records indicate
that the amount of cola follows the normal probability distribution.
The mean amount per bottle is 31.2 ounces and the population
standard deviation is 0.4 ounces.
At 8 A.M. today the quality technician randomly selected 16 bottles
from the filling line. The mean amount of cola contained in the
bottles is 31.38 ounces.
Is this an unlikely result? Is it likely the process is putting too much
soda in the bottles? To put it another way, does the population
represented by the sample of 16 bottles really has a population
mean of only 31.2 ounces?
8-28
Using the Sampling Distribution of the Sample Mean
(Sigma Known) - Example

X   31.38  31.20
z

 1.80
 n
\$0.4 16
8-29
Using the Sampling Distribution of the Sample Mean
(Sigma Known) - Example
8-30
Using the Sampling Distribution of the Sample Mean
(Sigma Known) - Example
What do we conclude?
With less than a 4 percent
chance, It is unlikely for us to
observe such a sample of 16
observations from a normal
population (which has a mean
of 31.2 ounces and a
population standard deviation
of 0.4 ounces) that the sample
mean is as high as 31.38
ounces.
We conclude the process is putting too much cola in the
bottles.
8-31
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