Week 7 Quiz Notes

Week 7 Quiz Notes
Math 221
Quiz Review for Weeks 5 and 6
1. Find the area under the standard normal curve between z = 1.6 and z = 2.6. First we look for the
area of both by doing "2nd ,Vars, NORMALCDF" and inputting "–1000, "Z," 0, 1 then find the
difference between both.
2. A business wants to estimate the true mean annual income of its customers. It randomly samples
220 of its customers. The mean annual income was $61,400 with a standard deviation of $2,200.
Find a 95% confidence interval for the true mean annual income of the business' customers.
First we find E by doing Zc(standard deviation/square root of number of trials.) Now we add and
subtract that number from the mean income to find both endpoints. The Zc of 95% is 1.96 so we
would do ... Show more content on Helpwriting.net ...
11. What z–scores would be used to create an 89% confidence interval?
12. A soccer ball manufacturer wants to estimate the mean circumference of mini–soccer balls
within .12 inch. Assume that the population of circumferences is normally distributed.
Determine the minimum sample size required to construct a 95% confidence interval for the
population mean. Assume the population's standard deviation is .2 inches.
13. Find the area under the standard normal curve to the left of z = –1.25.
14. True or False. In the standard normal distribution the standard deviation is always exactly 0.
15. Compute the population mean margin of error for a 90% confidence interval when sigma is 7
and the sample size is 36.
16. The area under a normal curve with mu = 35 and sigma = 7 is 0, 1, or 2?
17. If John gets an 90 on a physics test where the mean is 85 and the standard deviation is 3, where
does he stand in relation to his classmates? (he is in the top 5%, he is in the top 10%, he is in the
bottom 5%, or bottom 1%)
18. Find P(12 < x < 23) when mu = 19 and sigma = 6. Write your steps in probability notation.
19. In a normal distribution with mu = 34 and sigma = 5 what number corresponds to z = –4?
20. Let's assume you have taken 1000 samples of size 64 each from a normally distributed
population. Calculate the standard deviation of the sample means if the population's variance is 49.
21. Interpret a 93% confidence interval of (7.46, 12.84) for a population mean.
22. The
... Get more on HelpWriting.net ...

Standard deviation abstract
Standard Deviation Abstract
QRB/501
Standard Deviation Abstract
Standard Deviations Are Not Perverse
Purpose:
The purpose of this article is to illustrate how using statistical data, such as standard deviation, can
help a cattleman choose the best lot of calf's at auction. The statistical data used in these decision
making processes can also help the cattleman with future analysis of the lots purchased and existing
stock.
Research Question:
How can understanding the standard deviation of weights in a lot of calf's be used to determine
which lot should be purchased?
Hypothesis:
The hypothesis proposed in this article is that in using the group average and standard deviation a
cattleman can choose a lot of calf's ... Show more content on Helpwriting.net ...
For example a customer can complain about the service received and the employee can try their best
to make the situation better this is where double deviation plays. There is no right way to make the
situation better, it just has to be handled and keep it moving with the service to make the business
move forward.
The research questions(s):
What is the best way to handle a complaint or something that went wrong with the question?
The hypothesis:
To come together with the business and come up with different ways to handle different situations
that may come up with handling the customers. Brainstorm about different ways to solve the
questions if a customer is dissatisfied with their service. Make sure that the customers leaves
satisfied and to make it a possibly that they will come back.
The main finding of the study:
If the customer can be recovered a second time, after the same unfavorable service experience, it
represents a double service recovery, whereas a failure of this outcome or process constitutes a triple
deviation. The concepts of double service recovery and triple deviation thus extend the well‐known
concepts of service recovery and double deviation. Because customers seldom are pleased with
service recoveries (Rust et al., 1992), double deviations are frequent, often caused by the same
factors that initiated the first failure.
Relative deviation metrics and the problem of strategy replication,

Purpose:
Funds

Normal Distribution and Standard Deviation
Student Exploration: Sight vs. Sound Reactions
Vocabulary: histogram, mean, normal distribution, range, standard deviation, stimulus
Prior Knowledge Questions (Do these BEFORE using the Gizmo.)
Most professional baseball pitchers can throw a fastball over 145 km/h (90 mph). This gives the
batter less than half a second to read the pitch, decide whether to swing, and then try to hit the ball.
No wonder hitting a baseball is considered one of the hardest things to do in sports!
1. What are some things in your life you must react to quickly? You need to react quickly when you
are in danger, and you need to get away. You also need to react quickly when you are in a car so you
don't get hurt
2. In general, do you think ... Show more content on Helpwriting.net ...
Explain. I don't think that mine has normal distribution. Mine has a wide range of numbers, but I
don't think that it is distributed like a normal graph should be distributed
7. Experiment: On the GRAPH tab, click Clear data. This time, try to vary your response times
randomly as you take the test. Click Start and take the test.
A. What is the mean of this new set of values? 0.344
B. What is the range of this set? 0.26
C. What is the standard deviation of this set? 0.0872
8. Analyze: How does the standard deviation relate to the consistency and range of a data set?
The standard deviation is the mean multiplied by the range of the numbers.
Activity B:
Visual and auditory stimuli
Get the Gizmo ready:
Click Clear data.

Introduction: We perceive stimuli through nerve cells in our eyes, ears, nose, tongue, and skin.
When a nerve cell is stimulated, it sends an electrical signal to the brain. After the signal is
processed by the brain, other signals are sent to our muscles as we react to the stimulus.
Question: Do we react more quickly to visual or auditory stimuli?
1. Form hypothesis: Do you think you will react more quickly to sights or sounds? Explain why.
I will react quicker to sounds because my brain has a slower reaction time for sight because of a
concussion I got when I was younger.
2. Gather data: Select the TABLE tab. Use the Gizmo to run

Ilab Week 6 Math 221 Essay
Elementary Statistics iLab Week 6
Statistical Concepts: * Data Simulation * Discrete Probability Distribution * Confidence Intervals
Calculations for a set of variables
Mean Median
3.2 3.5
4.5 5.0
3.7 4.0
3.7 3.0
3.1 3.5
3.6 3.5
3.1 3.0
3.6 3.0
3.8 4.0
2.6 2.0
4.3 4.0
3.5 3.5
3.3 3.5
4.1 4.5
4.2 5.0
2.9 2.5
3.5 4.0
3.7 3.5
3.5 3.0
3.3 4.0
Calculating Descriptive Statistics
Descriptive Statistics: Mean, Median
Variable N N* Mean SE Mean StDev Minimum Q1 Median Q3 Maximum
Mean 20 0 3.560 0.106 0.476 2.600 3.225 3.550 3.775 4.500
Median 20 0 3.600 0.169 0.754 2.000 3.000 3.500 4.000 5.000

Calculating Confidence Intervals for ... Show more content on Helpwriting.net ...
The mean for the column "mean" is 3.56. It is very close to the parameter of interest but is not equal
to it. You can calculate a confidence interval for the mean of the mean column, but a specific
confidence interval would need to be provided. In that case, the confidence interval would be
centered on 3.56, not 3.5. |
4. Give the mean for the median column of the Worksheet. Is this estimate centered about the
parameter of interest (the parameter of interest is the answer for the mean in question 2)
The mean for the median column is 3.6, which is close to the mean in question 2 but not as close as
the answer in question 3. |
5. Give the standard deviation for the mean and median column. Compare these and be sure to
identify which has the least variability?
Standard Deviation of Mean= 0.4762Standard Deviation of Median= 0.7539The standard deviation
of the Mean is smaller, which means all of the data points will tend to be very close to the Mean.
The Median with a larger Standard Deviation will tend to have data points spread out over a large
range of values. Since the Mean has the smaller value of the Standard Deviations, it has the least
variability. |
6. Based on questions 3, 4, and 5 is the mean or median a better estimate for the parameter of
interest? Explain your reasoning.
The Mean seems to be the better estimate as

Business Report for Corvette: Mean, Standard Deviation and...
Business Report for Corvette
Executive Summary: This report talks about a Company is called Corvette, which sells luxury sports
cars in twelve months from now. There is a table shows the order of five customers and in which
currencies. Using those data I will find out mean, standard deviation and some probability for
analysis. In addition, these is a case involves an offer was given to Corvette by HSBC for estimating
whether it is risk or not. Furthermore, we also thought about what role banks plays in the case, and
analyzed the bank itself. Banks expected profit is also mentioned. Finally I used the data about my
birthday graphed the two models.
Introduction: Reading through the whole report, you can ... Show more content on Helpwriting.net
...
Actually, the company can invest the money or operate own company whatever. The key is that
company can get money quickly, maybe it can solve problem of liquidity of money in the company.
On the contrary, payment in twelve months' time does not have this usage.
In my view, bank would prefer the payment could be dealt as late as possible, because bank can hold
that money to investment and reduce the risk of liquidity. On the other side, company would prefer
the payment could be dealt as soon as possible, as the company can hold that money to operate
itself, such like paying salary, buying assets and so on.
Part 3: Risk of bank taking
Liquidity risk
Liquidity risk refers to the probability that a bank may not have sufficient funds with which to meet
the demands for cash (or to make transfers on the part of customers). In addition, a bank is insolvent
(has negative net worth) only if its total assets are insufficient to meet total liabilities.
(bankofengland)
Furthermore, there is the danger that a liquidity crisis at one bank can cause a 'systemic' failure of
the banking system through a process of 'contagion'.
Payment risk
Payment risk refers to the possibility that a bank may be asked to transfer funds which it does not
receive. The risk arises from the fact that instructions to a bank to make payment can be

Motion Picture Industry Essay
Motion Picture Industry
Data is important and a main–stream form of communication throughout the world; however,
without being able to represent the data in a format that is easily understandable, then the data will
not be as useful. Anderson, Sweeney, Williams, Camm, & Cochran (2015) define data as "the facts
and figures collected, analyzed, and summarized for presentation and interpretation" (p. 5). It is
important for statisticians, mangers, and employees to collect and present statistical data in accurate
and easy to read format. When presenting data to managers for instance, the easiest form of
communication, yet effectively provide details as to the importance of the data to the applicable
situation should be used. Scientists use data to measure scientific results; such as, comparing the
amount of carbon dioxide in the atmosphere over the last 650,000 years (NASA, 2015) – See
Appendix A. In business, data is found in all parts of industry; such as, accounting, finance,
marketing, production, economics, and information systems (Anderson et al, 2015, p. 3–5). In the
case of this study, the results will be shown as a managerial report from an accumulation of data
showing the financial success of 100 motion pictures on the years 20XX in the following categories:
Rank, Movie Title, Opening Gross, Total Gross, Theaters (number of) and Weeks (weeks shown).
The managerial report will show the descriptive statistics (mean, median, range and standard
deviation see data

Standard Deviation
I 'll be honest. Standard deviation is a more difficult concept than the others we 've covered. And
unless you are writing for a specialized, professional audience, you 'll probably never use the words
"standard deviation" in a story. But that doesn 't mean you should ignore this concept.
The standard deviation is kind of the "mean of the mean," and often can help you find the story
behind the data. To understand this concept, it can help to learn about what statisticians call normal
distribution of data.
A normal distribution of data means that most of the examples in a set of data are close to the
"average," while relatively few examples tend to one extreme or the other.
Let 's say you are writing a story about nutrition. You ... Show more content on Helpwriting.net ...
In this way, looking at the standard deviation can help point you in the right direction when asking
why information is the way it is.
The standard deviation can also help you evaluate the worth of all those so–called "studies" that
seem to be released to the press everyday. A large standard deviation in a study that claims to show a
relationship between eating Twinkies and shooting politicians, for example, might tip you off that
the study 's claims aren 't all that trustworthy.
Of course, you 'll want to seek the advice of a trained statistician whenever you try to evaluate the
worth of any scientific research. But if you know at least a little about standard deviation going in,
that will make your talk with him or her much more productive.
Okay, because so many of you asked nicely...
Here is one formula for computing the standard deviation. A warning, this is for math geeks only!
Writers and others seeking only a basic understanding of stats don 't need to read any more in this
chapter. Remember, a decent calculator and stats program will calculate this for you...
Terms you 'll need to know x = one value in your set of data avg (x) = the mean (average) of all
values x in your set of data n = the number of values x in your set of data
For each value x, subtract the overall avg (x) from x, then multiply that result by itself (otherwise
known as determining the square of that value). Sum up all those squared values. Then divide that
result by

Standard Deviation and High School Seniors
Part 1 of 1 – |
Question 1 of 10 | 1.0 Points |
Consider the following scenario in answering questions 1 through 4.
Results from previous studies showed 79% of all high school seniors from a certain city plan to
attend college after graduation. A random sample of 200 high school seniors from this city reveals
that 162 plan to attend college. Does this indicate that the percentage has increased from that of
previous studies? Test at the 5% level of significance.
State the null and alternative hypotheses. | | | A. H0:  = .79, H1:  > .79 | | | | B. H0: p = .79,
H1: p ≠ .79 | | | | C. H0: p ≤ .79, H1: p > .79 | | | | D. H0: = .79, H1: > .79 | |
Reset Selection | Question 2 of 10 | ... Show more content on Helpwriting.net ...
Compute the z or t value of the sample test statistic. | | | A. z = 1.916 | | | | B. t = –1.916 | | | | C. t =
1.916 | | | | D. z = 1.645 | |
Reset Selection | Question 7 of 10 | 1.0 Points |
Consider the following scenario in answering questions 5 through 7. In an article appearing in
Today's Health a writer states that the average number of calories in a serving of popcorn is 75. To
determine if the average number of calories in a serving of popcorn is different from 75, a
nutritionist selected a random sample of 20 servings of popcorn and computed the sample mean
number of calories per serving to be 78 with a sample standard deviation of 7.
At the  = .05 level of significance, does the nutritionist have enough evidence to reject the writer's
claim? | | | A. No | | | | B. Yes | | | | C. Cannot Determine | |
Reset Selection | Question 8 of 10 | 1.0 Points |
Consider the following in answering questions 8 through 10. A lab technician is tested for her
consistency by taking multiple measurements of cholesterol levels from the same blood sample. The
target accuracy is a variance in measurements of 1.2 or less. If the lab technician takes 16
measurements and the variance of the measurements in the sample is 2.2, does this provide enough
evidence to reject the claim that the lab technician's accuracy is within the target accuracy?
State the null

Portfolio Theory
PORTFOLIO THEORY
Let us begin our discussion on Portfolio Theory with an example of two investments (assets or
securities) – Ace and Bravo. Their return expectations are given in the table below. You will notice
that both Ace and Bravo are risky investments because they do not offer a certain return. You can
begin by comparing the expected return and risk of Ace and Bravo:
State of Probability Return
Economy of occurrence Ace Bravo
Boom 0.2 +20 –15
Growth 0.6 +5 +5
Recession 0.2 –10 +25
1. What kind of a correlation do you observe between the two securities? 2. Calculate the expected
return and standard deviation of both Ace and Bravo. Interpret the results.
Expected Return = E(R) = ∑ ri.pi ... Show more content on Helpwriting.net ...
Portfolio expected returns are a weighted average of the expected returns of the constituent
investments. If the two investments are A and B, with a being invested in A and (1–a) in B, then
[pic]
Portfolio standard deviation is less than the weighted average of the standard deviation of the
constituent investments (expect for perfectly +vely correlated investments where it is the weighted
average). The risk of a portfolio is measured by its standard deviation and calculated as:
[pic]
Covariance means the extent to which the returns on two investments move together. It is related to
correlation coefficient, but can take on any –ve or +ve value.
Let us learn the above concepts with the help of a solved example:
Let there be two shares A and B, with the following returns for alternative economic states:

|State of Economy |Probability |Returns on A |Returns on B |
|Boom |0.3 |20% |3% |
|Growth |0.4 |10% |35% |
|Recessions |0.3 |0% |–5%

Hypothesis: Standard Deviation and Critical Value
Study Guide – Testing the Difference Between Two Means, Two Variances, and Two Proportions
1.|If the test value in the figure below is 2.57 when the critical value is 1.96, what decision about the
hypothesis should be made?|
A)|reject the null hypothesis |
B)|accept the null hypothesis |
C)|reject the alternative hypothesis |
D)|not enough information |
2.|The standard error of difference is .|
A)|True|
B)|False|
3.|In the figure below, if the –test value is 1.43, the null hypothesis should not be rejected. |
A)|True|
B)|False|
4.|When hypothesizing a difference of 0, if the confidence interval does not contain 0, the null
hypothesis is rejected.|
A)|True|
B)|False|
5.|For normally distributed ... Show more content on Helpwriting.net ...
The variance of nine business dinners was $6.12 and the variance of 12 business lunches was $0.87.
What is the test value? |
A)|3.1 |
B)|9.61 |
C)|49.5 |
D)|7.03 |
25.|Samples are independent when they are not related.|
A)|True|
B)|False|
26.|A pooled estimate of the variance is a weighted average of the variance using the two sample

variances and the __________ of each variance as the weights.|
Use the following to answer questions 27–29:
Mauricio Cruz, a wine merchant for Cruz 's Spirits Emporium, wants to determine if the average
price of imported wine is less than the average price of domestic wine. The data obtained is shown
in the table below.
|Imported Wine|Domestic Wine| |7.03|9.78| |2.31|3.62| |15|16 |
27.|What is the null hypothesis? Use .|
A)| |
B)| |
C)| |
D)| |
28.|What is the critical value? Use .|
A)|–1.761 |
B)|–2.045 |
C)|–1.697 |
D)|–1.703 |
29.|What is the test value? Use .|
A)|–6.97 |
B)|–2.50 |
C)|–4.53 |
D)|–2.54 |
30.|Determine the value of as shown in the figure below, if the degrees of freedom were seven and
nine.|
A)|0.01 |
B)|0.025 |
C)|0.05 |
D)|0.1 |
31.|If the variances are not known and one or both sample sizes are less than 30, the –test must be
used.|

Standard Deviation and Double Degree Essay
1.0 Introduction This report will cover the distribution of final exam results for BSB123 and what
factors influence the results. Factors that will be considered are the gender of the student, whether
the student is studying a double or single degree, the results from the weekly quiz's and the grade
achieved on the mid semester report. The presence of outliers will be determined to help analyse the
accuracy of the data. There are an infinite number of internal and external factors that contribute to
the outcome of a single exam result. Beaty, & Barling (1982) explains how factors such as stress
and anxiety can contribute to low test results and they give several self help solutions of how to
boost ones success. This report will ... Show more content on Helpwriting.net ...
4.3 Quiz Results Analysing the effect that the weekly quiz results have on the final exam results can
help to understand if high quiz results mean high exam results. Figure 5.0 The data in the above
scatter plot shows that there is a correlation between the quiz results and the exam results. R² =
0.536, which indicates that about 54% of the variation in the average of the quiz is accounted for the
linear relationship with the exam results. In other words, about 46% of the variation is not explained
by the least–squares regression line. 4.4 Report Results The last exam factor that will be addressed
in the bivariate section is the impact of the results from the mid semester report on the exam results.
Figure 6.0 The correlation between the report results and the final exam results in not over
whelming storage. The trend line in the graph can be seen as slopping upward from left to right
showing that. It can be seen that the majority of students that achieved well on the report also
achieved well on the exam 5.0 Trivariate analysis Trivariate analysis compares tree factors against
each other in order to help better understand the exam out comes. 5.1 Degree vs. Exam & Quiz
Figure 7.0 Figure 7.0 states that students who are doing a double degree achieve higher marks
compared to the students who are doing a single degree. On average it can be seen that over

Descriptive Analysis Of Descriptive And Descriptive Analysis
5.6 DESCRIPTIVE ANALYSIS TEST RESULTS
Descriptive analysis can be used to summarize the data, the data was processed in SPSS and
descriptive analysis test was done in order to produce the summary of the data as shown in table 6.
The descriptive analysis in the table 6 shows the total number of valid and missing variavles along
with its mean, standard deviations.
It shall be noted that no question has any missing entry and all elements were analyzed on equal
parameter, the survey forms with missing entry were rejected at data screening stage. The maximum
and minimum value of all questions lie between 1 to 5, 1 being the highest and 5 being the lowest
value. The mean of questions categorized the high resource people and low resource people ... Show
more content on Helpwriting.net ...
Q. 2(D) Mean = 2.31 and Standard Deviation = 1.009
Mean is less than 2.5 which tells us that respondent are more inclined towards innovators which are
high resources in VALS framework and standard deviation is almost equal to 1 which indicates that
the respondent are equally spread out over a wider range of values(mean).
Q. 2 (E) Mean = 2.46 and Standard Deviation = 1.080
Mean is less than 2.5 which tells us that respondent are more inclined towards experiencers which
are high resources in VALS framework and standard deviation is almost equal to 1 which indicates
that the respondent are equally spread out over a wider range of values(mean).
Q. 3 Mean = 2.5 and Standard Deviation = 1.186
Mean is equal to 2.5 which tells us that respondent are neither inclined towards innovators and
survivors in VALS framework and standard deviation is greater than 1 which indicates that the
respondent are spread out over a wider range of values(mean).
Mean is less than 2.5 which tells us that respondent are more inclined towards survivors which are
low resources in VALS framework and standard deviation is slightly greater than 1 which indicates
that the respondent are spread out over a wider range of values(mean).
Mean is

Standard Deviations Use In The Business World
Standard Deviation use in the Business World
Abstract
This paper evaluates the role of standard deviation in business. As part of the evaluation, a brief
summary of five different peer reviewed papers has been presented. Topics such as, the purpose of
the study, the research questions, the hypothesis of the study, and the main findings of the study for
the five papers, have been summarized by each of the learning team members.
Standard Deviation use in the Business World
Standard Deviation is a statistical measurement that shows how data are spread above and below the
mean. The square root of the variance is the standard deviation (Cleaves, Hobbs, & Noble, 2012). It
plays a key role in business management, with one of its ... Show more content on Helpwriting.net
...
3. Is the delayed reaction to bad news a manifestation of their lower degree of earnings persistence?
Hypothesis
The hypothesis is that good news announcements are associated with positive returns and bad news
is associated with negative returns. Announcements of bad news have generally been established to
have lower earnings response coefficients. The conditions of changing volatility, the ISD of an at–
the–money option can be interpreted as an estimate of the expected standard deviation of the return
over the life of that option, and can therefore be used to analyze the pattern of volatility, which the
market expects to occur around an announcement. Announcements of earnings per share (eps)
figures with a high transitory component, whose implications for the future are more difficult to
assess, should be associated with a delayed volatility reaction.
Findings of the study
1. If the day of the of the anticipated volatility increase is known, then by measuring the ISD at two
points before that day, the `basic' volatility and the amount increase can be deduced.
2. The ISDs tend to rise before the announcement date and fall after it. The day 10 ISDs suggest that
volatility rises again roughly two weeks after the announcement.
3. Announcing bad news and announcing news that is difficult to interpret both have an incremental
effect on delaying the volatility reaction, but the effect of bad news appeared to be dominant.
4.

Normal Distribution and Significance Level
Math 221
Week 6 Lab
Submitted by: Merima Ceric
Part 1. Normal Distributions and Birth Weights in America
1) What percent of the babies born with each gestation period have a low birth weight (under 5.5
pounds)? a) Under 28 = 99.88% The NORMDIST formula was used to calculate:
=NORMDIST(5.5,1.88,1.99,True) X= 5.5 Mean= 1.88 Standard Deviation=1.19
b) 32 to 35 weeks = 43.83% The NORMDIST formula was used to calculate: =NORMDIST
(5.5,5.73,1.48,True) X= 5.5 Mean= 5.73 Standard Deviation=1.48
c) 37 to 39 weeks = 4.66% The NORMDIST formula was used to calculate:
=NORMDIST(5.5,7.33,1.09, True) X= 5.5 Mean= 7.33 ... Show more content on Helpwriting.net ...
3) No, the ages are not normally distributed because with the view of the histogram, it is not bell
shaped.
4)
| | | | | | | |
Histogram is approximately bell shaped and symmetric and agrees with the results predicted by the
Central Limit Theorem which says that whatever the population shape, the sampling distribution is
approximately normal
5) Standard deviation = 3.5518 ( = STDEV formula was used) Standard deviation of population =
22.57 Standard deviation of sample mean=Standard deviation / sqauare root of sample size = 40
Therefore Standard deviation of sample mean= 3.567841 These two values are close; thus the result
predicted by central limit theorem is true.
Part 3. Finding z– and t–scores for Confidence Intervals
1. Using Excel, find the z–score that corresponds to the following Confidence Levels:

a. 80%
Confidence level = 80% or Significance

Population Standard Deviation
REVIEW EXERCISES CHAPTER 8 AND 9 PROFESSOR JONAS WIU–RES BY DEBRA
JAMES CHAPTER 8 1. High temperature in the United States a meteorologist claims that the
average of the highest temperatures in the united states in 98. A random sample of 50 cities is
selected, and the highest temperatures are recorded. The data are shown. At a=0.05 can the claim be
rejected? a=7.7 97, 101, 99, 99, 100, 94, 87, 99, 108, 93, 96, 88, 98, 97,88, 105, 97, 96, 98, 102, 99,
94, 96, 114, 99, 96, 98, 97, 91, 98, 80, 95, 98, 96, 80, 95, 88, 99, 102, 95, 101, 94, 92, 99, 101, 97,
94, 97, 102, 61. The claim can be rejected; correct answer may be either above 98 or below it. 2.
Salaries for Actuaries nationwide graduates entering the actuarial ... Show more content on
Helpwriting.net ...
0.025), Z = 2.24 Since Δ > Z at the 2.5% significance level, the null hypothesis is rejected
meaning that the data does not support the idea that a Rhode Island Teacher's mean salary is $3,000
more than a New York Teacher's mean salary. 4. High and Low Temperatures – March is a month of
variable weather in the northeast. The chart below records the actual high and low temperatures for
a selection of days in March from the weather reports for Pittsburgh Pennsylvania. At the 0.01 level
of significance is there sufficient evidence to conclude that there is more than a 10% difference
between average highs and lows? Professor Jonas, I have not had a good time with this assignment,
it is very hard working with one

EXERCISE 18 Mean Standard Deviation And 95 And 99 Of The...
Name: Ashley Lee
Class: HLT–362 Applied Statistics for Healthcare Professionals
Date: 04/01/2015
EXERCISE 18 Mean, Standard Deviation, and 95% and 99% of the Normal Curve
1. Assuming that the distribution is normal for weight relative to the ideal and 99% of the male
participants scored between (–53.68, 64.64), where did 95% of the values for weight relative to the
ideal lie? Round your answer to two decimal places.
In order to find where 95% of the values for the weight of relative to the ideal lies you would use
the formula that is presented in the text on page 132 of Exercise 18. This formula is:. The = MEAN
(5.48) and the (SD) =Standard Deviation (22.93). These numbers were derived from table 1 on
pg.133 under the column ... Show more content on Helpwriting.net ...
To find the 95% of the men's scores you would again use the formula:. The Mean=52.53 and the
SD=30.90. These scores were found on pg. 134 in table 2 column labeled Male in the pain category.
Formula: 52.53±1.96(30.90)
52.53–1.96(30.90) = 52.53–60.56
52.53–60.56= –8.03
52.53+1.96(30.90) = 52.53+60.56
52.53+60.56= 113.09
ANSWER= (–8.03,113.09)
5. Were the body image scores significantly different for women versus men? Provide a rationale for
your Answer.
On pg. 133 of the workbook Exercise 18 the body image scores were found to be significantly
higher in women versus men. The rationale for this information was stated in the information
provided on pg.133 stating that women had a score of (73.1±17.0) and men had a score of
(60.2±17.0).
6. Assuming that the distribution of Mental Health scores for men is normal, where are 99% of the
men's mental health scores around the mean in this distribution? Round your answer to two decimal
places.
To find the 99% of the men's mental health scores you would use the formula:. This formula was

found on page 132 of Exercise 18 in the second paragraph. The Mean=57.09 and the SD=23.72.
These scores were found on pg. 134 in table 2 column labeled Male in the mental health category.
Formula: 57.09±2.58(23.72)
57.09–2.58(23.72) = 57.09–61.20
57.09–61.20= –4.11
57.09+2.58(23.72) = 57.09+61.20
57.09+61.20= 118.29
ANSWER=

PSY315 Week Two Practice Problems
University of Phoenix Material
Week Two Practice Problems
Prepare a written response to the following questions.
Chapter 2
12. For the following scores, find the mean, median, sum of squared deviations, variance, and
standard deviation:
1,112; 1,245; 1,361; 1,372; 1,472
Mean is 1312
Median is 1361
Sum of squared deviations is 76089.2
Variance is 15218
Standard deviation is 123.361
16. A psychologist interested in political behavior measured the square footage of the desks in the
official office for four U.S. governors and of four chief executive officers (CEOs) of major U.S.
corporations. The figures for the governors were 44, 36, 52, and 40 square feet. The figures for the
CEOs were 32, 60, 48, 36 square feet.
a. Figure the ... Show more content on Helpwriting.net ...
The amount of time it takes to recover physiologically from a certain kind of sudden noise is found
to be normally distributed with a mean of 80 seconds and a standard deviation of 10 seconds. Using
the 50%–34%–14% figures, approximately what percentage of scores (on time to recover) will be:
Above 100? 2%
Below 100? 98%
Above 90? 84%
Below 90? 16%
Above 80? 50%
Below 80? 50%
Above 70? 84%

Below 70? 16%
Above 60? 98%
Below 60? 52%
18. Suppose that the scores of architects on a particular creativity test are normally distributed.
Using a normal curve table, what percentage of architects have Z scores:
Above .10? 100%+.10=.10%
Below .10? 100%–.10%=99.99%
Above .20? 100%+.20%=.20%
Below .20? 100%–.20%=99.80%
Above 1.10? 100%+1.10%=101.10%
Below 1.10? 100%–1.10=98.90%
Above –.10? 100%+.10=.10%
Below –.10? 100%–.10%=99.99%
Using a normal curve table 2% of the architects have Z scores.
21. Suppose that you are designing an instrument panel for a large industrial machine. The machine
requires the person using it to reach 2 feet from a particular position. The reach from this position
for adult women is known to have a mean of 2.8 feet with a standard deviation of .5. The reach for
adult men is known to have a mean of 3.1 feet with a standard deviation of .6. Both women's and
men's reach from this position is normally distributed. If this design is implemented:
What percentage of women will not be able to work on this instrument

Standard Deviation Essay
In investigating the impact of the regime changes on league competitiveness, the overall change in
league equilibrium is charted based on the standard deviation of the team winning percentages for a
particular season. One can compare the winning percentages' actual standard deviation (ASD) with
the ideal standard deviation (ISD) in a given year, assuming a perfectly competitive league. This
approach was first used by Roger Noll and Scully in 1988 in their studies where they argued that the
dispersion of wins could be measured by comparing the actual performance of a league to the
performance that would have occurred if the league had the maximum degree of competitive
balance in the sense that all teams were equal in playing strengths. The ... Show more content on
Helpwriting.net ...
In a perfectly competitive league, the probability of any team winning (or losing) any given game is
given by the equation P(W) = P(L) = 1 – P(W) = 0.5. Thus, for an 82–game season, ISD=0.5/(√82).
The ASD is calculated by utilizing the following formula ASD=√(∑_(i=1)^N▒〖(p_i– p ̅)〗^2/N)
where p ̅ is the league mean point ratio (in this case is equal to 0.5) while p_i is each teams actual
winning record during the given season. N instead is the number of teams during the season. The
ratio of standard deviations (RSD) is calculated by finding the ratio of ASD to ISD for a given year
RSD = ASD / ISD. RSD takes the value of unity if the league is perfectly balanced, with higher
values representing greater levels of imbalance. Zero, corresponds to a situation in which all teams
end up with the same points ratio. John Gobok (2012) in his research paper found out that between
1976–2010 RSD diverged away from 1, signifying that the league became more unequal. The
imposition of the soft cap in 1984 initially improved league competitiveness, as RSD reverted back
towards 1. Nevertheless, between 1985–2000, league competitiveness went back to its original state
and even worsened in the mid–90s. The exceptionally high competitive imbalance in the league in
the 1990s can be attributed to the Jordan–era Bulls, where the team achieved the two winningest

The Effect Of A Catalyst On Reaction Rate
Data Collection and Processing/Conclusion and Evaluation IA
THE EFFECT OF A CATALYST ON REACTION RATE
2.0 Data Collection and Processing
2.1 Presenting Raw Data
Table 1 – Raw data of the varying hydrogen peroxide concentration, in percentage, compared to the
total volume in the test tube and the volume of foam, across multiple groups
Table 2 – Observations of the foam created
Amount of hydroxide peroxide solution Observations
0.0 No reaction occurred as distilled water was used, hence having no hydrogen peroxide solution.
0.5 A small amount of bubbles were produced as oxygen was converted from hydrogen peroxide
and water by the catalase.
1.0 A moderate of bubbles were produced as oxygen was converted from hydrogen peroxide and
water by the catalase.
2.0 A moderate of bubbles were produced as oxygen was converted from hydrogen peroxide and
4.0 The most amount of bubbles was produced in this concentration as oxygen was converted from
hydrogen peroxide and water by the catalase.
6.0 A large amount of bubbles were produced as oxygen was converted from hydrogen peroxide and
2.2 Processing Raw Data
Experimental data, that is where measurements are taken and manipulated, always involves
variation. To assert this statistical tools such as the mean and standard and standard deviation as a
percentage of the mean for the volume of foam created and the final volume to find a central
tendency and dispersion.

Quiz: Standard Deviation and Confidence Interval Estimate
Chapter 9 Quiz
1 |2 |3 |4 |5 |6 |7 |8 |9 |10 |11 |12 |13 |14 |15 | | | | | | | | | | | | | | | | | |
1 Compute the critical value za/2 that corresponds to a 94% level of confidence.
A. 1.88
B. 1.66
C. 1.96
D. 2.33
2. In a sample of 10 randomly selected women, it was found that their mean height was 63.4 inches.
Form previous studies, it is assumed that the standard deviation, σ, is 2.4. Construct the 95%
confidence interval for the population mean.
A. (61.9, 64.9)
B. (58.1, 67.3)
C. (59.7, 66.5)
D. (60.8, 65.4)
3. Suppose a 95% confidence interval for µ turns out to be (120, 310). To make more useful
inferences from the data, it is desired to reduce the width of the confidence ... Show more content on
Helpwriting.net ...
C. The percentage of the entire population of students who plan to go to college is between 77% and
87%.
D. It is unlikely that the given sample proportion result would be obtained unless the true percentage
was between 77% and 87%.
E. Between 77% and 87% of he population were surveyed.

13. One month of actual unemployment rate in France was 13.4%. If during that month you took a
simple random sample of 100 Frenchmen and constructed a confidence interval estimate of the
unemployment rate, which of the following would have been true?
I. The center of the interval was 13.4%
II. The interval contained 13.4%
III. A 99% confidence interval estimate contained 13.4%
A. I and II
B. I and III
C. II and III
D. I, II, and III
E. None of the above gives the complete set of true responses.
14. The margin of error in a confidence interval estimate using z–scores covers which of following?
I. Random sampling errors
II. Errors due to under–coverage and non–response in obtaining sample surveys
III. Errors due to using sample standard deviations as estimates for population standard deviations
A. I only
B. II only
C. III only
D. I and II
E. I and III
15. Compute the critical value ta/2 that corresponds to a 95% level of confidence and degree of
freedom = 10.
A. 2.228
B.

Linear Regression Analysis Paper
Welcome Back LV Team,
Today, we are going to look at variability. Before we do that, let us review our last week session.
The central tendency can be measured in three ways: mean, median, mode. The mean is the more
precise, then median and lastly mode. Each of them provides different type of information about a
distribution, depending on the data that needs to be analyzed and what we want to explain to the
audience. The wrong measurement of central tendency could misrepresent the data and distort it
systematically.
Variability displays how scores differ, distribute or spread from one another. For example, if we
have the same number, then there is zero variation. If the number keeps on repeating then there
would be less variable (ex. 3,4,4,5,4).
Just like the central tendency, there are three measures of variability: range, standard deviation and
the variance.
Range is ... Show more content on Helpwriting.net ...
As the sample size increases, there would be less difference between the biased (using n) and the
unbiased (n–1) estimates of the standard deviation.
Here is an example, if n=10 and my numerator of the SD= 900, then
Biased estimate = 9.48 unbiased estimate = 10
Now, n= 10000, then
Biased estimate= 0.30 unbiased estimate= 0.300015
The difference is less as the sample size increased.
Variability is best seen when using a visual aid. Earlier, I mentioned about normal distribution and
plotting the values would make it easier to see, whether it is normal or not.
It is not a perfect symmetry, but you could easily see the shape of the curve. The shaded area
represents a total of 68% (34% on each side of the mean) of the price sold between $7.54 and
$18.06. This is considered as a normal or bell shape distribution and there is no kurtosis or skewed.
Skewed is present when the curve has a one side tail and in the absence of symmetry. Kurtosis is
seen in a flat or a peak distribution (Salkind, 2004,

Individual Assignments from the Text
11. For the following scores, find the (a) mean, (b) median, (c) sum of squared deviations, (d)
variance, and (e) standard deviation:
2, 2, 0, 5, 1, 4, 1, 3, 0, 0, 1, 4, 4, 0, 1, 4, 3, 4, 2, 1, 0
Answer:
A) Mean = 2
B) Median = 2
C) Sum of squared deviations =–52
D). Variance = –2.47
E). Standard deviation = 1.57
12. For the following scores, find the (a) mean, (b) median, (c) sum of squared deviations, (d)
variance, and (e) standard deviation:
1,112; 1,245; 1,361; 1,372; 1,472
Answer:
A) Mean = 1.361
B) Median = 1.3124
C) Sum of squared deviations = 0.07608920.0190223
D). Variance = 0.0190223
E). Standard deviation =0.1379213544
13. For the following scores, find the (a) mean, (b) median, (c) sum of squared deviations, ... Show
more content on Helpwriting.net ...
Once we see the mean for 'Gun' after a 'Black' face was shown is less than that for 'Tool' after the
'Black' was shown. This shows that participants were more prompt to press button for Gun, then as a
result of the mindset created by the black person's picture. In fact, the reaction time for 'Tool' after
'Black' was largest in both experiments showing they were least expecting for the tool's picture after
a black person's image. Therefore in experiment 2, the mean decreased as the participants had to
press button as fast as possible, decreasing response times. Standard deviation shows the variability
of reaction times within a given group. This also decreased in second experiment.
But overall, the difference was only of a few milliseconds between means for different groups for
both experiments, meaning they were able to recognize gun or tool in almost same time whether the
previous image shown was black or white person. So, the prime didn't work or had very small
effect.
Chapter 3.
14. On a standard measure of hearing ability, the mean is 300 and the standard deviation is 20. Give
the Z scores for persons who score (a) 340, (b) 310, and (c) 260. Give the raw scores for persons

whose Z scores on this test are (d) 2.4, (e) 1.5, (f) 0, and (g)–4.5.
Answer:
A).

Is The Dependent Variable?
Means
The dependent variable was defined as how people rated themselves when presented with, "I am
good at managing unpredictable situations." The scale was numbered 1–5, 5 meaning strong
agreement and 1 meaning strong disagreement. The overall mean of the dependent variable was a
3.46. This places the mean above the neutral point and indicates that the group was in a slight
agreement with the statement. The mean of the group that had none to few restrictions was a 3.57. In
relation to the scale of measurement, this group was somewhat in agreement with the statement.
This was also the highest mean of the predictor variables. This group agreed most that they are with
the statement. The Several restrictions group had a mean of 3.48. This ... Show more content on
Helpwriting.net ...
They had the largest standard deviation of the group at 1.075. This means that on average, the data
points in that group fell about 1.075 above or below the mean, making a larger span of the data. All
of the groups were below the gold standard 50% or less than the mean for having outliers. However,
the group that was closest to having minor outliers was the many retractions group. The standard
deviation was 33% of the mean, the highest out of the other categories. The some restrictions group
that was farthest away from having minor outliers with a standard deviation of 26.6% of the mean.
The lowest percent of the groups.
Estimating with SDs and SEs
When evaluating standard error scores of the dependent variable the range was between 3.393 and
3.525. When evaluating the range from the standard deviation the range was between 2.503 and
4.414. The standard error provided a stronger degree of precision when evaluating the mean than the
standard deviation because the standard error had a smaller range of scores than the standard
deviation. If these were population statistics, then the minimum score needed to be equal to or
greater than 1.548. This score is the second standard deviation below the mean, scores lower than
this would mean that it is significant. On the higher end, scores must be equal to or below 5.370.
This score is the second standard deviation above the mean, scores above this number would also
result in significant data. If someone

Determining The Standard Error, Or Standard Deviation Of...
Assignment
Ans 1.1
a) The standard error, or standard deviation of the sample mean, = population standard
deviation/square root of sample size = $10.75/14
b) z = (60–68.30)/10.75 = –0.77
from the z tables we see that the probability that z is less than –0.77 is .2206, so the probability that
x is GREATER than 60 is 1–.2206 = .7794.
c) Now we are dealing with variation of the sample mean, so the standard error is 10.75/sqrt(100) =
1.075
Z = (66–68.30)/1.075 = –2.14, and the associated probability is .0162.
5. 150–138=12 = 1 standard deviation. By the empirical rule, we know that 68% of the data fall
within 1 standard deviation of the mean, above and below. Since the normal distribution is
symmetrical, 34% fall within 1 standard deviation below the mean (i.e. between 138 and 150). So a
score of 138 is the equivalent of a z score of –1. Similarly, a score of 162 is the equivalent of a z
score of 1.
By the Empirical Rule, approximately 95% of the data fall within 2 standard deviations above and
below the mean, so that would be from 126 to 174.
Same principle as above, but this time it 's plus or minus 3 standard deviations, from 114 to 186.
Ans 1.2:
a) The standard error of the mean, also called the standard deviation of the mean, is a method used
to estimate the standard deviation of a sampling distribution. To understand this, first we need to
understand why a sampling distribution is required.
In statistics, point estimation involves the use of sample data to

Measure Of Dispersion And Nonprofit Administration
Measure of Dispersion is a descriptive statistic that complements the measures of central tendency
as it indicates how data relates to a variable and its mean. As we examine the different types of
measure of dispersion within public and nonprofit administration, we will also cross–examine this
idea among other academic disciplines to discover how this statistical concept is used among other
researchers. In addition, we will learn the importance of the proper utilization of identifying
measures of dispersion and how to fully explain this idea. This is important because if we fail to
properly use the measures of dispersion, we can jeopardize the reliability of our research and the
validity of our data collected. Measures of dispersion reveals to a researcher how parts of data may
gather around the mean of the variable of interest. Using measures of dispersion will also allow the
researcher understand the population or sample of interest to help researchers draw better
conclusions. To understand measures of dispersion, we must first identify the statistical concepts
that help us measure dispersion. As we progressed through primary and secondary schooling, we
probably learned different mathematical concepts such as mean, median, mode, and range. These
mathematical concepts may be considered easy mathematical concepts, but they are part of complex
statistical ideas. Meier et al. (2015) discusses in his textbook the usage of range and other variety of
dispersion measures. He

Standard Deviation and Percentage Essay
Remington's Steakhouse Project Brian Jones Research Methods & Applications Dr. Jones August
25, 2011 Table of Contents Table of Contents 2 List of Tables 3 Introduction 4 The Research
Objectives 4 The Research Questions 5 Literature Review 6 Answers to Research Questions 8
Recommendations to Remington's 15 References 18 Annotated Bibliography 19 Appendix(ces) 22
List of Tables Table 1 Demographic Description of the Average Remington's Patron9 Table 2
Reported Income by Remington's Questionnaire Respondents9 Table 3 Importance Ratings when
Selecting a Restaurant11 Table 4 Perception Measures of Remington's Steakhouse Patrons12 Table
5 Relationship Measures of Remington's Steakhouse Patrons12 Table 6 Patron ... Show more
content on Helpwriting.net ...
4. What are Remington's Steakhouse scores on each of the six perception measures of: has large
portions, has competent employees, has excellent food quality, has quick service, has a good
atmosphere and has reasonable prices? 5. Is Remington's delivering on the characteristics that
patrons want when selecting a restaurant? 6. Is there a relationship between "satisfaction with
Remington's" and each of the following perceptions; has large portions, has competent employees,
has excellent food quality and has quick service? Literature Review According to the National
Restaurant Association (2011), 2011 Florida restaurant sales are projected to be $30.1 billion in
sales with Florida having 34,035 restaurants in 2009. The NRA (2011) states that in 1955 25% of
each dollar spent on food was in a restaurant compared to 49% presently. Given this growth trend,
restaurant owners are faced with ever increasing demand being offset by concurrent competition
growth. To drive customer growth and repeat business a detailed examination of the literature is
necessary to determine what others have found to be dominate factors in accomplishing this goal.
Employees must first align their perceptions of service quality to what the customer's perceptions
are (Becker & Wellins 1990). Understanding what drives the customer establishment
relationship is important to turning one time customers into loyal

Standard Deviation and Minimum Order
Sport Obermeyer
Sport Obermeyer started to make firm commitments for producing its 1993–1994 line of fashion
skiwear with scant information about how the market would react to the line. Inaccurate forecasts of
retailer demand had become a growing problem at Obermeyer: in recent years greater product
variety and more intense competition had made accurate predictions increasingly difficult. Also,
another issue Sport Obermeyer faced was how to allocate production between factories in Hong
Kong and China. They need consider all aspects in a short–term period and also a long–term period.
Questions:
1. Using the sample data given in Table 2–20, make a recommendation for how many units of each
style Wally should make during the ... Show more content on Helpwriting.net ...
Figure 2 shows we calculate the minimum order by using formula Q*=average–2*SD. Because we
assume that all of the 10 styles in the sample are made in Hong Kong and there is a minimum
production quantity 600 units so we changed all less than 600 units to 600. Then we can get the total
Figure 2,
|Style |Price |Average forecast|Standard deviation |2*standard |Q*=average–2*SD | |
|

Types Of Distributions Module 2
03–Types of Distributions Module 2 Slide 1: Welcome back! Congratulations for all the work that
you have done! While I'm sure you are anxious to learn about the technique to answer research
questions and hypotheses, social workers and researchers need to understand the basic. Think of
your first bike ride. Did you need training wheels? Or did you need that gentle push from a parent or
relative? These were the foundations that prepared you for many years of enjoyable bicycling! As
stated in the previous powerpoints, it is important to analyze each variable, especially our
independent and dependent variables to determine the type of statistical techniques to use in answer
our research questions and/or hypotheses. One analysis that we perform on each variable is the
distribution of their scores. As with other powerpoints, this powerpoint supplements your readings
for the purpose of providing a better understanding of the module's concepts. Yet, it does not replace
the need for reading your textbook. Additionally, embedded in the powerpoint are questions to test
your assimilation of these concepts. So, let's begin. Slide 2: As humans, we acquire knowledge with
visual applications or pictures. After looking at picture or graph, we can instantly interpret its
meaning. Frequency polygons convey pictures of different shapes of distributions for interval and
ratio level variables only and enhances our understanding of central tendency and dispersion,
concepts that we discussed

Does height matter in the NBA?
Introduction:
For the people who don't know, the acronym NBA stands for National Basketball Association.
NBA's mission is to be the most respected and successful sports league and sports marketing
organization in the world. Basketball popularity has been increasing through out the years and many
consider the sport as second most popular in the world. Like every sport basketball, has its share of
myths and misconceptions. One of the most popular myths that surround this sport is the height
factor. The general population tends to think that height is a key factor when playing this sport. This
report's purpose is to clear up the misconception of the height factor that the general population has.
Hypothesis: Basketball myths such as "You ... Show more content on Helpwriting.net ...
"The formula Player Efficiency Rating (PER) sums up a players' accomplishments (the statistical
ones, anyway) for a season. The ratings are not intended to be the final word on how a player
performs, but are designed to inform the debate." (http://www.alleyoop.com). The PER ratings will
be taken for the top ten players from the last 3 seasons. This comparison is to show whether height
is a factor to become a top–level player or not. There is also a player ranking to show which player
fits what category. In this research method, the mean and the standard deviation of the heights and
the PER ratings will be taken. The standard deviation of height shows if there is a wide range of
variation in the data. If its small then it might indicate that height affects the game. If the range of
data is large, then height does not affect it that much.
NBA Draft: Picks VS Height:
This method is done to find the answer to research question 3. Teams often see good players in
college anywhere from 6'7" to 6'9" who play in the paint. "However, teams are often leery of
picking such a player because they are afraid that he will not be able to make the transition from
college to the pros. They often label these types of players, "tweeners." In other words, they feel that
the player does not really have a set position in the pros. This leaves GMs in a bit of a quandary as
to what to do."(nbadraft.net)
In this research method, the mean and the standard deviation of the heights will be

Australian Multinational Bank Of Australia ( Cba ) And...
1. Stocks selection
In this report, three companies, BHP Billiton Limited (BHP), Commonwealth Bank of Australia
(CBA) and Telstra Corporation Limited (TLS) are chosen to form part of the investment portfolio.
They are all from ASX 100.
BHP is a leading global resources company and one of the world's largest mining companies (BHP
2014). CBA is an Australian multinational bank providing various financial services in New
Zealand, Fiji, Asia and the United Kingdom (CBA 2014). TLS is Australia's largest
telecommunications and media company, and it offers a full range of communications services (TLS
2014). Based on above information, these three companies are from different industries.
2. Determination of individual stock expected returns and standard deviations
2.1 The choice of data
The 2008 global financial crisis (GFC) (Shah 2009, p. 25) has caused the share prices of BHP, CBA
and TLS to drop rapidly from 2008 to 2009, apparent from the 10–year historical data of the three
companies' share price (Appendix 1). This shows that the all three companies' share prices are
influenced by the GFC, which means the GFC has the similar result in the calculation of three
stocks' expected return. Hence, the data from that period are included in the data choice.
The 10–year, monthly historical raw data of the stocks' closing prices and dividends are obtained
from Yahoo Finance (2015). The closing price is chosen for calculation purposes, as the closing
price of the trading day could

Investigating The Statistical Properties Of An Individual...
Reaction Time
Aim:
To determine the statistical properties (mean, standard deviation and tendencies in data) of an
individual's reaction time when tested with audial and visual stimuli, as well as the importance of
these properties.
Notes and Comments
In this experiment students will use a reaction time test to determine the delay between audial and
visual stimuli being displayed and the student reacting. Once a large (N>>3) sample size is taken,
statistical calculations can be made to determine the mean, and standard deviation; these
calculations help us build a mathematical model that can then help predict the possibilities of data
points lying on a number. This is if the recorded data is of a normal distribution and follows a
Gaussian ... Show more content on Helpwriting.net ...
The equation for calculating the standard deviation is as follows:
When substituting values for deviation and sample size the standard deviation comes out to be:
Now that we have a visual representation of the data in graph form, as well as calculating the mean
and standard deviation it is clear and undisputable that this data set is of a normal distribution and
will follow all trends set by the Gaussian curve, this is because it has slightly over 68% of data
points within one standard deviation and slightly over 95% of all data points within two standard
deviations.
The same calculations can be done with the data from the sound test to calculate the mean and
standard deviation for the sound reaction test, the mean and standard deviations come out to:
Although each result carries uncertainty, these uncertainties some somewhat reliant upon the
computer hardware that is being used when undertaking the experiment. The hardware that is used
on a computer has a response time of anywhere from 1–100 milliseconds. For the light test monitor
response time is a big factor, a delay between the monitor sending information to the screen and that
information being displayed is called display lag and many professional computer gamers use IPS
monitors to reach a response time of 1

Standard Deviation Essay
For ACT
Standard deviation=17.77 For NSW
Standard deviation=18.66
For NT
For QLD
For VIC
For WA
For TAS
For SA
Range = largest value –smallest value
For ACT range= 155.30– 96.00 =59.3 For NSW
Range= 151.8– 96.60
= 55.20
For NT
Range= 157.60– 107.30 =50.3
For QLD
Range=153.80–86.60
=67.2
For VIC
Range=148.90–94.20
=54.7

For WA
Range=151.30–93.20
=58.1
For TAS
Range=156.70–104.50
=52.2
For SA
Standard deviation=148.90–97.50 =51.4
Calculation of coefficient of variation can be calculated using 13.764 14.139 13.335 16.573 13.743
14.712 13.292 12.968
For ACT
CV=13.764
NSW
CV=14.139 ... Show more content on Helpwriting.net ...
Hence it is apparent that mental and behavioural problems on the whole are gender sensitive since
the % of females affected by it is more than their natural composition in the population. Further
there is variation across the various mental and behavioural problems as per gender.
Solution of Question 4.a
Determination of (meu)
Number of days when rainfall takes place = 131 days
Total number of weeks in a given year (2014) = 52 weeks
Hence the average number of raining days = 131/52 = 2.519
According to Poisson distribution Meu = 2.519
And now the probability on any given week in a year there would be no rainfall is

Standard Deviation
Deviation
Definition: Behavior commonly seen in children that is the result of some obstacle to normal
development such behavior may be commonly understand as negative (a timid child, a destructive
child) or positive (a quite child), both positive and negative deviation will disappear once the child
begins to concentrate on a piece of work freely chosen by him. The physical deforms are easier to
identify. This can be by birth due to an accident etc... and most such physical deforms can be either
cured. However, deforms that take place in development of psychological aspects of a child are not
only threat to building the character and the personality of the child also you find certain physical
deforms in curable in ... Show more content on Helpwriting.net ...
The normalized children possess a unique character and personality not recognized in young
children. Normalization appears through the repetition of a three step cycle. The building of
character and the formulation of personality that we call normalization comes about children follow
the cycle of work. Preparation for an activity which involve gathering together the material
necessary to do the activity. The movement and the thought involved in the preparation serves to
call the attention of the mind to begin to focus on the activity. An activity which so engross the child
that the reaches a deep level of concentration characterized by a general feeling of satisfaction and
well–being.
Characteristics: 1. Love for work: include the ability to choose work freely and to find joy in his
work. 2. Love of order 3. Profound spontaneous concentrate. 4. Attachment to reality. 5. Love of
silence and working alone. 6. Subblimission of possession. 7. Real choice. 8. Obedience. 9.
Independence and initiative. 10. Spontaneous self discipline. 11. Joy.
Three folds of phenomenon: 1. Polarization of attention 2. Spontaneous repetition 3. Leading to a
state of concentration.
The Prepared Environment:
The prepared environment must be in every way prepared. The environment plays an important role
the aesthetics of the rooms is not all that is needed in the

Essay about Introduction for Learn Formula for Standard...
Introduction for learn formula for standard deviation: In Statistics, the standard deviation is named
as the determination of indecision of a chance variable, expressed as the average deviation of a
group of data from its arithmetic mean and planned as the positive square root of the inconsistency.
In this article we are discussing about learn formula for standard deviation problems.
The following formula for solving standard deviation. S = sqrt ((sum(X –M)^2)/N–1)) M – Mean of
given values. N – Total number of values. X – Specified value.
Example problems for learn formula for standard deviation:
Learn formula for standard deviation – Example 1
Find the standard deviation for ... Show more content on Helpwriting.net ...
Solution:
(i) Find the mean and deviation for the given values. X = 32, 40, 25, 30, 26 and 39. M = (32 + 40 +
25 + 30 + 26 + 39)/ 6 = 192/6 = 32
(ii) Then we can find the sum of (X – M) 2
X X–M (X–M)2
32 32–32 = 0 0
40 40–32 = 8 64
25 25–32 =–7 49
30 30–32 =–2 4
26 26–32 =–6 36
39 39–32= 7 49 Total 202 N = 6 is the total number of given values. Then N–1 = 6 – 1 = 5
(iii) Apply values in standard deviation formula: S = sqrt ((sum(X –M)^2)/(N–1)) = sqrt (202)/sqrt
(5) = 14.21/2.23 = 6.37 The answer is: 6.37
Learn formula for standard deviation – Example 3
Solve the standard deviation for the values 41, 39, 30, 32 and 28.
Solution:
(i) Find the mean and deviation for the given values. X = 41, 39, 30, 32 and 28. M = (41 + 39 + 30 +
32 + 28)/5 = 170/5 = 34
(ii) Then we can find the sum of (X – M) 2
X X–M (X–M)2
41 41–34 = 7 49
39 39–34 = 5 25
30 30–34 = –4 16
32 32–34 = –2 4
28 28–34 = –6 36 Total

The Events Factors Caused By Driving And Low Standard...
As the preceding Tables 5–8 depicted, five scenario factors were investigated. First, drivers at Path 1
have higher speed and low standard deviation of speed, standard deviation of lane deviation, and
acceleration noise than other paths. Vehicles from the on–ramp before toll plaza (i.e., Paths 4 and 5),
as shown in Figure 3 (a), have lower speed and higher dangerous behavior (i.e., standard deviation
of speed, acceleration noise, and standard deviation of lane deviation) compared to Path 1 at all
zones. It can be explained that vehicles from the on–ramp perform sudden lane changing maneuvers
before and after the toll plaza. Additionally, vehicles from Path 3 which is driving through the
express lanes to the off–ramp after the toll plaza have also lower speed and riskier behavior. This is
because drivers are reducing vehicle's speed to change lanes after the toll plaza. Furthermore,
vehicles from Path 2 which is driving from mainline to cash toll payment and back to mainline have
lower speed at and after the toll plaza, and riskier indicators at all zones due to speed variation and
sudden lane changing before and after the toll plaza. Second, for signage, drivers who navigate in
scenarios with Case 2 signage, which is adding the DMS sign at the on–ramp, removing the third
overhead sign, and relocate the second overhead sign as shown in Figure 3 (b), exhibited lower
speed variation after the on–ramp than drivers who perform scenarios of the base condition with the
second and

Spss Code Book Sample
Your temporary usage period for IBM SPSS Statistics will expire in 12 days. GET FILE=
'C:UsersadminDesktopMGSM816 – Market Research_Indivitual assign ament_42403693.sav '.
DATASET NAME DataSet1 WINDOW=FRONT. CODEBOOK Qnaire_no [n] A1 [s] A2 [s] A3 [s]
A4 [s] A5 [s] A6 [s] A7 [s] A 8 [s] A9 [s] A10 [s] A11 [s] A12 [s] A13 [s] A14 [s] A15 [s] A16 [s]
A17 [ s] A18 [s] A19 [s] A20 [s] A21 [s] A22 [s] B1 [s] B2 [s] B3 [s] B4 [s] B5 [s] B6 [s] B7 [s] B8
[s] B9 [s] B10 [s] B11 [s] B12 [s] C1 [s] C2 [s] C3 [s] C4 [s] D1 [s] D2 [s] D3 [s] D 4 [s] E1 [s] E2
[s] E3 [s] E4 [s] E5 [s] E6 [s] E7 [s] E8 [s] E9 [s] E10 [ s] F1 [s] F2 [s] F3 [s] F4 [s] G1a [s] G1b [s]
G1c [s] G1d [s] G2a [s] G2b [s] G2c [s] G2d [s] G2e [s] G3a [s] G3b [s] G3c [s] G3d ... Show more
to buy recycle paper product Numeric F11 Scale Input 8 0 2.25 .707 2.00 2.00 3.00 Never true
Slightly true Neutral True Always true 1 4 3 0 0 12.5% 50.0% 37.5% 0.0% 0.0% Count Percent
Type Format Measurement Role N Valid Missing Central Tendency and Dispersion Mean Standard
Deviation Percentile 25 Percentile 50 Percentile 75 Labeled Values 1 2 3 4 5
Page 11
A11 Value Standard

Ops/571 Week 5 Paper
Statistical Process Control
OPS/571
Over the past five weeks, data has been collected from the process of getting my daughter, Sophie,
ready for daycare in the morning. I have tracked six key areas, or steps, in the process: The time it
takes to wake her up, The time it takes to get her to go to the bathroom, The time it takes to get her
stuff ready, The time it takes to get her dressed, The time it takes to brush her teeth and hair, and The
time it takes to get her into the car. In this paper, I will discuss what I have discovered based on this
data, I will identify roadblocks to the process and recommend strategies to overcome them, and I
will discuss the variables which affect the steps in the process. Finally I will discuss ... Show more
The following step is getting her dressed. This is also the step that has been identified as the main
bottleneck in the process. This step has a mean time of 16.5 minutes and a standard deviation of
5.16. The extreme deviation in this step is based on a single primary factor: fussiness. On some
mornings, Sophie is happy with the clothes I have chosen and eager to get dressed. On other
mornings she does not like what I have chosen and/or does not want to get dressed. On good days, it
generally takes from ten to fifteen minutes to get her dressed, on days when she doesn't want to get
ready, it can take up to twenty–seven minutes. Since this step is the primary bottleneck in the
process, this is the step I have focused on in regards to improving efficiency. The strategy I have
recommended in order to improve efficiency is simple. On the night before, prior to Sophie's
bedtime, I will pick out her clothes for the following day, with her help. This will allow her to have
some level of control over what she wears and she will also know ahead of time what outfit she will
be putting on. Doing this should eliminate the problem of her not liking the clothes I have picked
out for her and it should improve the mean time and decrease the standard deviation of the time it
takes to get her dressed. Brushing her teeth and combing her hair is generally an easy process as she
enjoys

Standard Deviation and Cash Flow
Chapter 12 Problems
1. Cash flow (LO2) Assume a corporation has earnings before depreciation and taxes of $100,000,
depreciation of $50,000, and that it has a 30 percent tax bracket. Compute its cash flow using the
format below.
Earnings before depreciation and taxes _____ Depreciation _____ Earnings before taxes _____
Taxes @ 30% _____ Earnings after taxes _____ Depreciation _____
12–1. Solution:
Earnings before depreciation and taxes $100,000 Depreciation – 50,000 Earnings before taxes
50,000 Taxes @ 30% 15,000 Earnings after taxes 35,000 Depreciation + ... Show more content on
Helpwriting.net ...
12–6. Solution:
Payback for Product X Payback for Product Y
$200,000 – 60,000 1 year $200,000 – 40,000 1 Year 140,000 – 90,000 2 years 160,000 – 70,000 2
years 50,000 – 50,000 3 years 90,000 – 80,000 3 years 10,000/20,000 .5 years
Payback Product X = 3.00 years
Payback Product Y =3.50 years
Product X would be selected because of the faster payback.
7. Payback method (LO3) Assume a $50,000 investment and the following cash flows for two
alternatives.
|Year |Investment A |Investment B |
|1 |$10,000 |$20,000 |
|2 |11,000 |25,000 |
|3 |13,000 |15,000 |
|4 |16,000 |– |

|5 |30,000 |– |
Which alternative would you select under the payback method?
12–7. Solution:
Payback for Investment A Payback for Investment B
$50,000 –$10,000 1 year $50,000 – $20,000 1 year 40,000 – 11,000 2 years 30,000 – 25,000 2 years
29,000 –

What Is Standard Deviation
Standard Deviation
The Standard deviation is calculated from the raw data and the means of each golf club, illustrating
the variance of each data point for each handicap. The greater the standard deviation, the further
away the data is from the mean, showing the data to be unreliable. On the other hand, the smaller
the standard deviation the more reliable the data from each trial is. However, as the difference in
standard deviation is due to the different golf club face angles, the standard deviation will allow for
comparison within each golf club face angle, individually determining if a relationship between the
two variables exists. The Standard deviation will be calculated using the following formula:
Each of the symbols represents

Week 7 Quiz Notes

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