Chapter 4 & 5 – Continuous & Discrete Distributions Assignment #3
Your answers must be clear, concise (one or two sentences) and supported by attached output from R
(include code and outputs).
1. A technical support call center measures the amount of time that each customer waits on hold. After
analyzing the data it has been determined that the wait times follow a Uniform distribution with a
minimum wait time of 0 minutes and a maximum wait time of 22 minutes.
a) Sketch the distribution, calculate the mean and include this on your sketch.
b) What is the probability that a customer waits on hold for less than 5 minutes?
c) What is the probability that a customer waits on hold for more than 10 minutes?
d) What is the probability that a customer waits on hold for between 5 and 15 minutes?
e) The call center wants to publish a service standard on their website i.e. “90% of calls will be answered
within ……… minutes.” Calculate what the service standard should be.
f) Simulate the wait time of 100 callers by generating random values from your Uniform distribution.
How many of the 100 callers waited for longer than the service standard computed in part e?
2. The final grades for a business statistics class follow a Normal distribution with a mean of 78.3 and a
standard deviation of 12.1.
a) Sketch the distribution, include the mean on your sketch.
b) What is the probability that a student scored less than 65?
c) What is the probability that a student scored more than 90?
d) What is the probability that a student scored between 70 and 90?
e) The professor wishes to award an ‘A’ letter grade to the top 25% of students. Calculate what the
cutoff-grade for an ‘A’ should be.
f) Simulate the grades of 50 students by generating random values from your Normal distribution. How
many of the 50 students scored above the ‘A’ cut-off computed in part e?
3. Suppose the number of customers using a bank’s ATM during the lunch period between 12noon and
2pm can be modelled with a Poisson distribution with a mean of 18.
a) Calculate the standard deviation of the distribution.
b) What is the probability that exactly 18 customers use the ATM between 12noon and 2pm?
c) What is the probability that 15 or less customers use the ATM between 12noon and 2pm (i.e. X=25)?
e) Simulate 100 lunch periods by generating random values from your Poisson distribution. In how many
of these lunch periods were there more than 25 customers?
The Practice of Statistics for
Business and Economics
Fourth Edition
David S. Moore
George P. McCabe
Layth C. Alwan
Bruce A. Craig
© 2016 W.H. Freeman and Company
Distribution for Counts and
Proportions
The Binomial Distributions
PSBE Chapter 5.1
© 2016 W. H. Freeman and Company
Objectives (PSBE Chapter 5.1)
The Binomial Distributions
▪
The Binomial setting
▪
The Binomial distribution
▪
Binomial probabilities
▪
Binomial mean and standard deviation
▪
Sample proportions
▪
The Normal approximation for counts and proportions
Binomial setting
Binomial distributions are models for some categorical variables,
typically representing the number of successes in a series of n trials.
The observations must meet these requirements:
The total number of observations n is fixed in advance.
The outcomes of all n observations are independent.
Each observation falls into just one of two categories: success or failure.
All n observations have the same probability of “success,” p.
We record the next 50 vehicles sold at a dealership. Each vehicle sold is
either an SUV (success) or not (failure).
Binomial distribution
The distribution of the count X of successes in the binomial setting is the
binomial distribution with parameters n and p: B(n,p).
The parameter n is the total number of observations.
The parameter p is the probability of success on any one observation.
The count of successes X can be any whole number between 0 and n.
A coin is flipped 10 times. Each outcome is either a head or a tail.
The variable X is the number of heads among those 10 flips, our count
of “successes.”
On each flip, the probability of success, “head,” is 0.5. The number X of
heads among 10 flips has the binomial distribution B(n = 10, p = 0.5).
Applications for binomial distributions
Binomial distributions describe the possible number of times that a
particular event will occur in a sequence of observations.
They are used when we want to know about the occurrence of an
event, not its magnitude.
In a clinical trial, a patient’s condition may improve or not. We study the
number of patients who improved, not how much better they feel.
Was a sales transaction considered pleasant? The binomial distribution
describes the number of pleasant transactions, not how pleasant they
are.
In quality control we assess the number of defective items in a lot of
goods, irrespective of the type of defect.
Binomial coefficient
The number of ways of arranging k successes in a series of n
observations (with constant probability p of success) is the number of
possible combinations (unordered sequences).
This can be calculated with the binomial coefficient:
n!
n
k k!(n k )!
Where k = 0, 1, 2, ..., or n.
n
, “n choose k” uses the factorial notation “!”.
k
The factorial n! for any strictly positive whole number n is:
n! = n× (n – 1) × (n – 2) × … × 3× 2× 1.
The binomial coefficient
For example, 5! = 5 × 4 × 3 × 2 × 1 = 120. Note 0! = 1.
Calculations for binomial probabilities
The binomial coefficient counts the number of ways in which k
successes can be arranged among n observations.
The binomial probability P(X = k) is this count multiplied by the
probability of any specific arrangement of the k successes:
P( X k ) n p k (1 p) n k
k
The probability that a binomial random variable takes any range of values is
the sum of each probability for getting exactly that many successes in n
observations.
P(X ≤ 2) = P(X = 0) + P(X = 1) + P(X = 2)
Finding binomial probabilities: tables
You can also look up the probabilities for some values of n and p in
Table C in the back of the book.
The entries in the table are the probabilities P(X = k) of individual
outcomes.
The values of p that appear in Table C are all 0.5 or smaller. When
the probability of a success is greater than 0.5, restate the problem
in terms of the number of failures.
Customer satisfaction
Each consumer has probability of 0.25 of preferring your product over a
competitor’s product. If we question five consumers, what is the
probability that exactly two of them prefer your product?
P(X= 2)=
P(X= 2)=
n!
5!
pk(1 p)n-k =
×0.252×0.753
2!
×
3!
k!(n k)!
5×4
× 0.252×0.753
2×1
P(X= 2)= 10 ×0.0625 ×0.421875 = 0.2637
Binomial mean and standard deviation
0.3
a)
distribution for a count X are defined by
P(X=x)
The center and spread of the binomial
0.25
0.2
0.15
0.1
0.05
the mean μ and standard deviation σ:
0
0
np(1 p)
2
0.3
3
4
5
6
7
8
9
10
8
9
10
8
9
10
Number of successes
0.25
b)
P(X=x)
np
1
0.2
0.15
0.1
0.05
0
Effect of changing p when n is fixed.
a) n = 10, p = 0.25
0
1
2
3
4
5
6
7
Number of successes
0.3
b) n = 10, p = 0.5
c) n = 10, p = 0.75
For small samples, binomial distributions
are skewed when p is different from 0.5.
c)
P(X=x)
0.25
0.2
0.15
0.1
0.05
0
0
1
2
3
4
5
6
7
Number of successes
Sample proportions
Let X be the count of successes in n samples. Then
X
pො = .
n
In an SRS of size n drawn from a large population with population
proportion p of successes, the sample proportion pො has mean and
standard deviation
𝜇p = p
𝜎p =
p 1−p
.
n
Use when the population is at least 20 times as large as the sample.
Normal approximation
If n is large, the binomial distribution can be approximated by the
Normal distribution.
X is approximately N np,
pො is approximately N p,
np(1−p)
p(1−p)
n
The Normal approximation to the binomial distribution can be used
when by np ≥ 10 and n(1-p) ≥ 10.
Normal approximation
If a quality control engineer determines that 8% of switches fail to meet
specifications, what is the probability that no more than 10 of 150
switches will be nonconforming?
µ= np = (150)(0.08) = 12
σ=
np(1 p)=
150(0.08)(0.92)= 3.3226
So Xis approximately N(12, 3.3226); Note: np≥10 and n(1 p) ≥10.
P(X ≤ 10) = P
X – 12 10 – 12
≤
3.3226 3.3226
= P(Z≤ -0.60) = 0.2743
The continuity correction
When using the Normal approximation to the binomial distribution, a
continuity correction is needed since the binomial distribution puts
probability only on whole numbers. To estimate P(X ≤ k) for a binomial
distribution, find P(X ≤ k + 0.5) in the Normal distribution.
P(X ≤ 10) = P(X ≤ 10.5) in the
binomial distribution as shown in
the probability histogram.
Looking at the Normal curve,
P(X ≤ 10.5) will give a better
approximation of the binomial
probability than P(X ≤ 10).
Distribution for Counts and
Proportions
The Poisson Distributions
PSBE Chapter 5.2
© 2016 W. H. Freeman and Company
Objectives (PSBE Chapter 5.2)
The Poisson Distributions
▪
The Poisson setting
▪
The Poisson distribution
▪
Approximations to the Poisson
▪
Assessing Poisson assumptions with data
The Poisson setting
A count X of successes has a Poisson distribution in the following
Poisson setting:
➢
The number of successes that occur in two non-overlapping units
unit of measure are independent.
➢
The probability that a success will occur in a unit of measure is
the same for all units of equal size and is proportional to the size
of the unit.
➢
The probability that more than one event occurs in a unit of
measure is negligible for very small-sized units. (Events occur
one at a time.)
Poisson distribution
Poisson distribution
Let 0.9 be the average number of Wi-Fi interruptions on your home
network per day. The count of X flaws per day is modeled by the
Poisson distribution with μ= 0.9.
The probability of two or less flaws per square yard is P(X ≤ 2).
P(X ≤ 2) = P(X = 0) + P(X = 1) + P(X = 2)
e−0.9(0.9)0 e−0.9(0.9)1 e−0.9(0.9)2
=
+
+
0!
1!
2!
= 0.4066 + 0.3659 + 0.1647
= 0.9372
Poisson approximations
Using the Normal to approximate the Poisson:
Use N(μ, μ) to approximate Poisson probabilities when μ is
large.
Americans send an average of approximately 616 text messages
per week. What is the probability that you would send more than
650 text messages over a week?
P(X> 650)
=P
X−616 650−616
>
616
616
= P(Z>1.37)
= 1 - 0.9147 = 0.0853
The actual Poisson probability is 0.083176, so this approximation
is quite accurate.
Poisson approximations
Using the Poisson approximation to the binomial distribution:
Use when n is large and p is small such that np 0.8) = 1 – P(0.5 < X < 0.8) = 0.7
Continuous random variable and population distribution
% individuals with X
such that x1 < X < x2
The shaded area under a density
curve shows the proportion, or %,
of individuals in a population with
values of X between x1 and x2.
Because the probability of drawing
one individual at random
depends on the frequency of this
type of individual in the population,
the probability is also the shaded
area under the curve.
Normal probability distributions
The probability distribution of many random variables is a normal
distribution, N(μ, σ).
Example: Probability
distribution of women’s
heights.
Here, because we chose a
woman randomly, her height,
X, is a random variable.
To calculate probabilities with the normal distribution, we will
standardize the random variable (z score) and use Table A.
What is the probability, if we pick one woman at random, that her height will be
some value X? For instance, between 68 and 70 inches P(68 < X < 70)?
Because the woman is selected at random, X is a random variable.
z
(x )
N(µ, ) =
N(64.5, 2.5)
As before, we calculate the zscores for 68 and 70.
For x1 = 68",
z
(68 64.5)
1.4
2.5
For x2 = 70",
z
(70 64.5)
2.2
2.5
0.9192
0.9861
The area under the curve for the interval [68" to 70"] is 0.9861 − 0.9192 = 0.0669.
Thus, the probability that a randomly chosen woman falls into this range is 6.69%.
P(68 < X < 70) = 6.69%
Probability: The Study of
Randomness
Means and Variances of Random Variables
PSBE Chapter 4.5
© 2016 W.H. Freeman and Company
Objectives (PSBE Chapter 4.5)
Means and Variances of Random Variables
▪ Mean of a random variable
▪ Law of large numbers
▪ Rule for means
▪ Variance of a random variable
▪ Rule for variances
Mean of a random variable
The mean of a set of observations is their arithmetic average.
The mean µ of a random variable X is a weighted average of the
possible values of X, reflecting the fact that all outcomes might not be
equally likely.
A basketball player shoots three free throws. The random variable X is the
number of baskets successfully made (“H”).
MMM
HMM
MHM
MMH
HHM
HMH
MHH
HHH
Value of X
0
1
2
3
Probability
1/8
3/8
3/8
1/8
The mean of a random variable X is also called expected value of X.
We would be interested in the µX, the expected number of baskets
successfully made in three shots.
Mean of a discrete random variable
For a discrete random variable X with
probability distribution
the mean µ of X is found by multiplying each possible value of X by its
probability, and then adding the products.
A basketball player shoots three free throws. The random variable X is the
number of baskets successfully made.
Value of X
0
1
2
3
Probability
1/8
3/8
3/8
1/8
The mean µ of X is
µX = (0*1/8) + (1*3/8) + (2*3/8) + (3*1/8)
= 12/8 = 3/2 = 1.5
Law of large numbers
As the number of randomly
drawn independent
observations in a sample
x
increases, the mean of the
sample gets closer and
closer to the population
mean .
This is the law of large numbers. It is valid for any population.
Note: We often intuitively expect predictability over a few random observations,
but this is wrong. The law of large numbers only applies to really large numbers.
Rules for means
Rule 1: If X is a random variable and a and b are fixed numbers, then
µa+bX = a + bµX
Rule 2: If X and Y are random variables, then
µX+Y = µX + µY
Rule 3: If X and Y are random variables, then
µX-Y = µX - µY
Investment
You invest 20% of your funds in Treasury bills and 80% in an “index fund” that
represents all U.S. common stocks. Your rate of return over time is proportional
to that of the T-bills (X) and of the index fund (Y), such that R = 0.2X + 0.8Y.
Based on annual returns between 1950 and 2003:
Annual return on T-bills µX = 5.0% σX = 2.9%
Annual return on stocks µY = 13.2% σY = 17.6%
Correlation between X and Y ρ = −0.11
µR = 0.2µX + 0.8µY = (0.2*5) + (0.8*13.2) = 11.56%
This portfolio has a smaller mean return than an all-stock portfolio.
Variance of a random variable
The variance and the standard deviation are the measures of spread
that accompany the choice of the mean to measure center.
The variance σ2X of a random variable is a weighted average of the
squared deviations (X − µX)2 of the variable X from its mean µX. Each
outcome is weighted by its probability in order to take into account
outcomes that are not equally likely.
The larger the variance of X, the more scattered the values of X on
average. The positive square root of the variance gives the standard
deviation σ of X.
Variance of a discrete random variable
For a discrete random variable X with probability distribution
and mean µX, the variance σ2 of X is
A basketball player shoots three free throws. The random variable X is the
number of baskets successfully made.
µX = 1.5.
Value of X
0
1
2
3
Probability
1/8
3/8
3/8
1/8
The variance σ2 of X is
σ2 = 1/8*(0−1.5)2 + 3/8*(1−1.5)2 + 3/8*(2−1.5)2 + 1/8*(3−1.5)2
= 2*(1/8*9/4) + 2*(3/8*1/4) = 24/32 = 3/4 = 0.75
Rules for variances
Rule 1: If X is a random variable and a and b are fixed numbers, then
σ2a+bX = b2σ2X
Rule 2: If X and Y are independent random variables, then
σ2X+Y = σ2X + σ2Y
σ2X-Y = σ2X + σ2Y
Rule 3” If X and Y are NOT independent but have correlation ρ, then
σ2X+Y = σ2X + σ2Y + 2ρσXσY
σ2X-Y = σ2X + σ2Y - 2ρσXσY
Investment
You invest 20% of your funds in Treasury bills and 80% in an “index fund” that
represents all U.S. common stocks. Your rate of return over time is proportional
to that of the T-bills (X) and of the index fund (Y), such that R = 0.2X + 0.8Y.
Based on annual returns between 1950 and 2003:
Annual return on T-bills µX = 5.0% σX = 2.9%
Annual return on stocks µY = 13.2% σY = 17.6%
Correlation between X and Y ρ = −0.11
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