# 4 questions for introduction to statistic by using minitab 19 graphing

STATISTICS 1000Q
Winter Intersession 2020
Assignment 2
Professor Suman Majumdar
After you complete the assignment, save it under the filename
yourlastname2.
General Instructions
Answer the questions in the fields provided for and submit the resulting document through
HuskyCT.
Question Number
1a
1b
1c
1d
2a
2b
2c
3
4
Point Allotted
2
2
2
2
2
2
2
3
3
Point Scored
QUESTION 1
The shape of the graph of a binomial distribution depends on the value of both n and p. To
see how the shape changes for a fixed value of n, you will let p vary and graph each
probability distribution. Let X be a binomial random variable with n = 10.
a. For p = 0.11 obtain a bar chart of the binomial probability distribution.
b. For p = 0.50 obtain a bar chart of the binomial probability distribution.
c. For p = 0.89 obtain a bar chart of the binomial probability distribution.
d. Describe the effect of changing p.
QUESTION 2
Now see what happens when you hold p constant and vary n. Let X be a binomial random
variable with p = 0.25.
a. Obtain a bar chart of the binomial probability distribution for n = 5.
b. Obtain a bar chart of the binomial probability distribution for n = 50.
c. Describe the effect of changing n.
QUESTION 3
The location of a Normal distribution is determined by its mean , where as its shape is
determined by the standard deviation . To see the effect of changing , you are going to
graph two Normal probability density functions, one with  = 100 and another with  =
105, both having  = 10. Recall that for each distribution the first value should be 3 = 30
below the mean, and the last value should be 3 above the mean. When MINITAB creates
the X values for you, for both distributions set the data IN STEPS OF 1. Overlay the two
density functions on the same graph (in MINITAB), and paste in the box below.
QUESTION 4
Now you want to see graphically the effect of changing the standard deviation of a Normal
distribution. Let  = 100 for both distributions, but let  = 10 for one and  = 16 for the
other distribution. Recall that for each distribution the first value should be 3 below the
mean of 100, and the last value should be 3 above the mean of 100. When MINITAB
creates the X values for you, for both distributions this time set the data IN STEPS OF 2.
Overlay the two density functions on the same graph (in MINITAB), and paste in the box
below.
Chapter 4
Random
Variables and
Probability
Distributions
A L WA YS L E A R N I N G
Slide – 1
Content
1. Two Types of Random Variables
2. Probability Distributions for Discrete
Random Variables
3. The Binomial Distribution
4. Probability Distributions for Continuous
Random Variables
5. The Uniform Distribution
6. The Normal Distribution
A L WA YS L E A R N I N G
Slide – 2
Learning Objectives
1. Develop the notion of a random variable
2. Learn that numerical data are observed
values of either discrete or continuous
random variables
3. Present some simple discrete and continuous
random variables
4. Study two important families of random
variables and their probability models in
detail: the binomial and the normal families
A L WA YS L E A R N I N G
Slide – 3
Thinking Challenge
You’re taking a 33 question
multiple choice test. Each
question has 4 choices.
Clueless on 1 question, you
decide to guess. What’s the
chance you’ll get it right?
If you guessed on all 33
questions, what would be your
A L WA YS L E A R N I N G
Slide – 4
4.1
Two Types of Random
Variables
A L WA YS L E A R N I N G
Slide – 5
Discrete Random Variable
Random variables that can assume a
countable (finite or infinite) number of values
are called discrete.
A L WA YS L E A R N I N G
Slide – 6
Discrete Random Variable
Examples
Experiment
Random
Variable
Possible
Values
Make 100 Sales Calls
# Sales
0, 1, 2, …, 100
# Defective
0, 1, 2, …, 70
# Correct
0, 1, 2, …, 33
Count Cars at Toll
Between 11:00 & 1:00
# Cars
Arriving
0, 1, 2, …, ∞
A L WA YS L E A R N I N G
Slide – 7
Continuous
Random Variable
If the number of possible values of the
random variable is uncountable, then the
random variable is called continuous.
A L WA YS L E A R N I N G
Slide – 8
Continuous Random Variable
Examples
Experiment
Random
Variable
Possible
Values
Weigh 100 People
Weight
45.1, 78, …
Measure Part Life
Hours
900, 875.9, …
Amount spent on food
\$ amount
54.12, 42, …
Measure Time
Between Arrivals
Inter-Arrival 0, 1.3, 2.78, …
Time
A L WA YS L E A R N I N G
Slide – 9
4.2
Probability Distributions for
Discrete Random Variables
A L WA YS L E A R N I N G
Slide – 10
Discrete
Probability Distribution
The probability distribution of a discrete
random variable is a graph, table, or
formula that specifies the probability
associated with each possible value of the
random variable.
A L WA YS L E A R N I N G
Slide – 11
Requirements for the
Probability Distribution of a
Discrete Random Variable X
1. p(x) ≥ 0 for all values of x
2.  p(x) = 1
where the summation of p(x) is over all
possible values of X.
A L WA YS L E A R N I N G
Slide – 12
Discrete Probability
Distribution Example
Experiment: Toss 2 coins. Count number of
tails.
Probability Distribution
Values, x Probabilities, p(x)
0
1/4 = 0.25
1
2/4 = 0.50
2
1/4 = 0.25
A L WA YS L E A R N I N G
Slide – 13
Visualizing Discrete
Probability Distributions
Listing
Table
{ (0, .25), (1, .50), (2, .25) }
# Tails
f(x)
Count
p(x)
0
1
2
1
2
1
.25
.50
.25
Graph
p(x)
.50
.25
.00
Formula
x
0
1
A L WA YS L E A R N I N G
2
p (x ) =
n!
px(1 – p)n – x
x!(n – x)!
Slide – 14
Characterization of the
Probability Distribution of a
Discrete Random Variable X
If we have a two column table with the
columns labeled x and p(x), and the entries
in the p(x) column satisfy the two properties
1. p(x) ≥ 0 for all values of x
2.  p(x) = 1,
then, the table in question represents the
probability distribution of a discrete random
variable X.
A L WA YS L E A R N I N G
Slide – 15
Modifying Discrete Probability
Distribution Example
x_
_p(x)_
0
0.16
1
0.48
2
0.36
There exists a discrete random variable X
such that the table above represents the
distribution of X.
A L WA YS L E A R N I N G
Slide – 16
Summary Measures
1. Expected Value (Mean of probability
distribution)
Weighted average of all possible values
 = E(X) = x p(x)
2. Variance
Weighted average of squared deviation
2 = E[(X 2(x 2 p(x)
3.
Standard Deviation
●   2
A L WA YS L E A R N I N G
Slide – 17
Summary Measures
Calculation Table
x
p(x)
Total
x p(x)
x p(x)
A L WA YS L E A R N I N G
x–
(x – 2
(x – 2p(x)
(x 2 p(x)
Slide – 18
Variance – Alternative Calculation
A L WA YS L E A R N I N G
Slide – 19
Variance – Alternative Calculation
A L WA YS L E A R N I N G
Slide – 20
Variance – Alternative Calculation
A L WA YS L E A R N I N G
Slide – 21
Thinking Challenge
You toss 2 coins. You’re
interested in the number
of tails. What are the
expected value,
variance, and standard
deviation of this random
variable, number of tails?
A L WA YS L E A R N I N G
Slide – 22
Expected Value & Variance
Solution*
x
p(x)
x p(x)
x–
0
.25
0
–1.00
1.00
.25
1
.50
.50
0
0
0
2
.25
.50
1.00
1.00
.25
 = 1.0
(x –  2 (x –  2p(x)
2 .50
 .71
A L WA YS L E A R N I N G
Slide – 23
Examples of Discrete Random Variables
The Experiment of Rolling a Die – Part 1
The Experiment of Rolling a Die – Part 2
The Minimum Problem
A L WA YS L E A R N I N G
Slide – 24
Probability Rules for Discrete
Random Variables
Let x be a discrete random variable with probability
distribution p(x), mean µ, and standard deviation .
Then, depending on the shape of p(x), the
following probability statements can be made:
Chebyshev’s Rule
Empirical Rule
P x    x  µ   
0
 .68
P x  2  x  µ  2 
 34
 .95
P x  3  x  µ  3 
 89
 1.00
A L WA YS L E A R N I N G
Slide – 25
Suggested Exercises
Work out the following exercises from the
Textbook :
4.12, 4.14, 4.15, 4.26, 4.32, 4.35, 4.39.
These exercises will not be collected or graded,
but let me know as questions arise.
A L WA YS L E A R N I N G
Slide – 26
4.3
The Binomial Distribution
A L WA YS L E A R N I N G
Slide – 27
Binomial Distribution
Number of ‘successes’ in a sample of n
observations (trials)
Number of reds in 15 spins of roulette wheel
Number of defective items in a batch of 5 items
Number correct on a 33 question exam
Number of customers who purchase out of 100
customers who enter store (each customer is
equally likely to purchase)
A L WA YS L E A R N I N G
Slide – 28
Binomial Probability
Characteristics of a Binomial Experiment
1. The experiment consists of n identical trials.
2. There are only two possible outcomes on each trial.
We will denote one outcome by S (for success) and
the other by F (for failure).
3. The probability of S remains the same from trial to trial.
This probability is denoted by p, and the probability of
F is denoted by q. Note that q = 1 – p.
4. The trials are independent.
5. The binomial random variable x is the number of S’s in
n trials.
A L WA YS L E A R N I N G
Slide – 29
Binomial Probability
Distribution
 n  x n x
n!
x
n x
p( x)    p q 
p (1  p )
x ! (n  x)!
 x
p(x) = Probability of x ‘Successes’
p = Probability of a ‘Success’ on a single trial
q = 1–p
n = Number of trials
x = Number of ‘Successes’ in n trials
(x = 0, 1, 2, …, n)
n – x = Number of failures in n trials
A L WA YS L E A R N I N G
Slide – 30
Binomial Probability
Distribution Example
Experiment: Toss 1 coin 5 times in a row. Note
number of tails. What’s the probability of 3
n!
tails?
x
n x
p( x) 
p (1  p )
x !(n  x)!
5!
p (3) 
.53 (1  .5)53
3!(5  3)!
 .3125
A L WA YS L E A R N I N G
Slide – 31
Binomial Probability Table
(Portion)
n=5
p
k
.01

0.50

.99
0
.951

.031

.000
1
.999

.188

.000
2
1.000

.500

.000
3
1.000

.812

.001
4
1.000

.969

.049
Cumulative Probabilities
p(x ≤ 3) – p(x ≤ 2) = .812 – .500 = .312
A L WA YS L E A R N I N G
Slide – 32
Binomial Distribution
Characteristics
n = 5 p = 0.1
Mean
  E(x)  np
P(X)
1.0
.5
.0
X
Standard Deviation
  npq
0
2
3
4
5
n = 5 p = 0.5
.6
.4
.2
.0
P(X)
X
0
A L WA YS L E A R N I N G
1
1
2
3
4
5
Slide – 33
Binomial Distribution
Thinking Challenge
You’re a telemarketer selling
service contracts for Macy’s.
You’ve sold 20 in your last 100
calls (p = .20). If you call 12
people tonight, what’s the
probability of
A. No sales?
B. Exactly 2 sales?
C. At most 2 sales?
D. At least 2 sales?
A L WA YS L E A R N I N G
Slide – 34
Binomial Distribution Solution*
n = 12, p = .20
A. p(0) = .0687
B. p(2) = .2835
C. p(at most 2) = p(0) + p(1) + p(2)
= .0687 + .2062 + .2835
= .5584
D. p(at least 2) = p(2) + p(3)…+ p(12)
= 1 – [p(0) + p(1)]
= 1 – .0687 – .2062
= .7251
A L WA YS L E A R N I N G
Slide – 35
Binomial probability calculations
using Minitab
Graphing a Binomial Distribution
Binomial Distribution Word Problems
Simulating from a Binomial Distribution
A L WA YS L E A R N I N G
Slide – 36
Suggested Exercises
Work out the following exercises from the
Textbook :
4.43, 4.47, 4.50, 4.52, 4.56, 4.59.
These exercises will not be collected or graded,
but let me know as questions arise.
A L WA YS L E A R N I N G
Slide – 37
4.5
Probability Distributions for
Continuous Random Variables
A L WA YS L E A R N I N G
Slide – 38
Continuous Probability
Distribution
To describe the probability distribution for a discrete random
variable, we specified the possible values of the variable along
with the corresponding probabilities.
That approach will not work for a continuous random variable for
two related reasons. First, the number of possible values for a
continuous random variable is uncountable. Second,
if X is a continuous random variable,
then, for any number b,
𝑃 𝑋 = 𝑏 = 0.
A L WA YS L E A R N I N G
Slide – 39
Continuous Probability
Distribution
To describe the probability distribution for a continuous random
variable X, we have to specify 𝑃 𝑎 < 𝑋 < 𝑏 for every pair of numbers a and b such that 𝑎 < 𝑏. Since, for a continuous random variable X, 𝑃 𝑋 = 𝑏 = 0 for any number b, we conclude 𝑃 𝑎

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