Expert answer:Statistic R Studio

Solved by verified expert:Files attached.
stat_assign..docx

psbe4e_lectureslides_ch05.pptx

psbe4e_lectureslides_ch04.pptx

Unformatted Attachment Preview

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=<10)? d) What is the probability that more than 25 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
p = p
p =
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<10.  Let μ = np. The Poisson probability will approximately equal the binomial probability.   Suppose for the distribution B(1000, 0.001) we want P(X ≤ 2). Then let μ = 1000 × 0.001 = 1. The results from Excel show how accurate the Poisson approximation is in this situation. Distribution for Counts and Proportions Toward Statistical Inference PSBE Chapter 5.3 © 2016 W. H. Freeman and Company Objectives (PSBE Chapter 5.3) Toward Statistical Inference Sampling distributions Sampling variability Bias Population versus sample  Population: The entire group of individuals in which we are interested.  Sample: The part of the population we actually examine and for which we do have data. Population Sample   A parameter is a number describing a characteristic of the population. In practice we do not know its value.  A statistic is a number describing a characteristic of a sample. The value can change from one sample to another. We often use a statistic to estimate an unknown parameter. Sampling variability Each time we take a random sample from a population, we are likely to get a different set of individuals and calculate a different statistic. This is called sampling variability. The good news is that, if we take lots of random samples of the same size from a given population, the variation from sample to sample—the sampling distribution—will follow a predictable pattern. Sampling variability The variability of a statistic is described by the spread of its sampling distribution. This spread depends on the sampling design and the sample size n. If we increase the sample size, the statistic from the larger sample is less variable. Bias  Bias concerns the center of the sampling distribution.  If the mean of the sampling distribution of a statistic is equal to the true value of the parameter being estimated, the statistic is an unbiased estimator of the parameter.  When taking an SRS, pො is an unbiased estimator of the population proportion p. Bias and variability When trying to locate the population parameter, bias and variability are two things that can cause our estimates to be less than ideal. Think of the true population parameter as being the bull’seye of a target. Bias looks like hits to the target that are centered away from the bull’s-eye as in (a) and (c). Variability relates to how spread out the hits may be. Larger variability is shown in (b) and (c). Managing bias and variability  To reduce bias, use random sampling. The values of a statistic computed from an SRS neither consistently overestimate or underestimate the value of a population parameter.  To reduce the variability of a statistic from an SRS, use a larger sample. You can make the variability as small as you want by taking a large enough sample. 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 Probability: The Study of Randomness Randomness and Probability Models PSBE Chapters 4.1 and 4.2 © 2016 W.H. Freeman and Company Objectives (PSBE Chapters 4.1 and 4.2) Randomness and Probability Models Randomness and probability Probability rules Assigning probabilities: finite number of outcomes Assigning probabilities: equally likely outcomes Independence and the multiplication rule Randomness and probability A phenomenon is random if individual outcomes are uncertain, but there is nonetheless a regular distribution of outcomes in a large number of repetitions. The probability of any outcome of a random phenomenon can be defined as the proportion of times the outcome would occur in a very long series of repetitions. Coin toss The result of any single coin toss is random. But the result over many tosses is predictable, as long as the trials are independent (i.e., the outcome of a new coin flip is not influenced by the result of the previous flip). The probability of heads is 0.5 = the proportion of times you get heads in many repeated trials. First series of tosses Second series Independence Two events are independent if the outcome of one trial does not influence or change the outcome of any other trial. When are trials not independent? Imagine that these coins were spread out so that half were heads up and half were tails up. Close your eyes and pick one. The probability of it being heads is 0.5. However, if you don’t put it back in the pile, the probability of picking up another coin and having it be heads is now less than 0.5. The trials are independent only when you put the coin back each time. This is called sampling with replacement. Probability models Probability models describe mathematically the outcome of random processes and consist of two parts: 1) S = Sample Space: This is a set, or list, of all possible outcomes of a random process. An event is a subset of the sample space. 2) A probability for each possible event in the sample space S. Example: Probability Model for a Coin Toss: S = {Head, Tail} Probability of heads = 0.5 Probability of tails = 0.5 Sample spaces It’s the question that determines the sample space. A. A basketball player shoots three free throws. What are the possible sequences of hits (H) and misses (M)? H H - HHH M - HHM H M M… H - HMH M - HMM … B. A basketball player shoots three free throws. What is the number of baskets made? S = { 0, 1, 2, 3 } S = { HHH, HHM, HMH, HMM, MHH, MHM, MMH, MMM } Note: 8 elements, 23 Probability rules 1) Probabilities range from 0 (no chance of the event) to 1 (the event has to happen). For any event A, 0 ≤ P(A) ≤ 1 Coin Toss Example: S = {Head, Tail} Probability of heads = 0.5 Probability of tails = 0.5 Probability of getting a Head = 0.5 We write this as: P(Head) = 0.5 P(neither Head nor Tail) = 0 P(getting either a Head or a Tail) = 1 2) Because some outcome must occur on every trial, the sum of the probabilities Coin toss: S = {Head, Tail} for all possible outcomes of the sample P(head) + P(tail) = 0.5 + 0.5 =1 space must be exactly 1.  P(sample space) = 1 P(sample space) = 1 Probability rules (cont. ) Venn diagrams: A and B disjoint 3) Two events A and B are disjoint if they have no outcomes in common and can never happen together. The probability that A or B occurs is then the sum of their individual probabilities. P(A or B) = P(A) + P(B) This is the addition rule for disjoint events. A and B not disjoint Example: If you flip two coins, and the first flip does not affect the second flip: S = {HH, HT, TH, TT}. The probability of each of these events is 1/4, or 0.25. The probability that you obtain “only heads or only tails” is: P(HH or TT) = P(HH) + P(TT) = 0.25 + 0.25 = 0.50 Probability rules (cont.) Coin Toss Example: S = {Head, Tail} Probability of heads = 0.5 Probability of tails = 0.5 4) The complement of any event A is the event that A does not occur, written as Ac. The complement rule states that the probability of an event not occurring is 1 minus the probability that it does occur. P(not A) = P(Ac) = 1 − P(A) Tailc = not Tail = Head P(Tailc) = 1 − P(Tail) = 0.5 Venn diagram: Sample space made up of an event A and its complementary Ac, i.e., everything that is not A. Assigning probabilities: finite number of outcomes Finite sample spaces deal with discrete data — data that can only take on a limited number of values. These values are often integers or whole numbers. Throwing a die: S = {1, 2, 3, 4, 5, 6} The individual outcomes of a random phenomenon are always disjoint.  The probability of any event is the sum of the probabilities of the outcomes making up the event (addition rule). M&M’s candies If you draw an M&M’s candy at random from a bag, the candy will have one of six colors. The probability of drawing each color depends on the proportions manufactured, as described here: Color Probability Brown Red Yellow Green Orange Blue 0.3 0.2 0.2 0.1 0.1 ? What is the probability that an M&M’s candy chosen at random is blue? S = {brown, red, yellow, green, orange, blue} P(S) = P(brown) + P(red) + P(yellow) + P(green) + P(orange) + P(blue) = 1 P(blue) = 1 – [P(brown) + P(red) + P(yellow) + P(green) + P(orange)] = 1 – [0.3 + 0.2 + 0.2 + 0.1 + 0.1] = 0.1 What is the probability that a random M&M’s candy is red, yellow, or orange? P(red or yellow or orange) = P(red) + P(yellow) + P(orange) = 0.2 + 0.2 + 0.1 = 0.5 Probabilities: equally likely outcomes We can assign probabilities either:  empirically  from our knowledge of numerous similar past events  Mendel discovered the probabilities of inheritance of a given trait from experiments on peas without knowing about genes or DNA. or theoretically  from ou ... Purchase answer to see full attachment

How it works

1. Paste your instructions in the instructions box. You can also attach an instructions file
2. Select the writer category, deadline, education level and review the instructions 
3. Make a payment for the order to be assignment to a writer
4.  Download the paper after the writer uploads it 

Will the writer plagiarize my essay?

You will get a plagiarism-free paper and you can get an originality report upon request.

Is this service safe?

All the personal information is confidential and we have 100% safe payment methods. We also guarantee good grades