248.Follow the requirement to compose a 350words essay using the given background

Follow the requirement to compose a 350words essay using the given backgrounUse the
background information below to create the essay.Please read requirements very carefully.All the work has to be 100percent original.
gbs_essay6.docx

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Essay
1. Use the background information below to create the essay.
Case Study
Scenario
Land and Agua Insurance Company has a call center in Tempe, Arizona. The business
was originally established in Phoenix, Arizona in 1972 as a small business, and it has
grown with the population of the area. The insurance company specializes in bundling
insurance for cars, off-road vehicles, and watercraft (e.g., jet skis and boats). The
company has 150,000 clients in Arizona.
Marjorie Jones, Vice President of Operations, is concerned about customer complaints
and the amount of time representatives are taking to resolve the calls. You are part of
the team investigating the data to determine the probabilities of errors and call times.
Ms. Jones also wants to understand the approximate range around the average for call
times.
2. Answer the questions below in essay format. Your essay must include an introduction,
a body, and a conclusion. It must address all relevant parts of each question. Your
response should be a minimum of 500 words in length, and it should include your
analysis of the probability calculations. Make sure to cite any source you use. Proper
citation format for a source includes the name of the author(s), the title of the work,
the date of the publication, and the page number if you directly quote the source.
Essay: Probability
Using the Quality Summary and Call Center Data, provide a summary report for the vice
president including the following information in an essay with a minimum of 500 words:
1. Based on the probability of an error provided in the quality summary under call quality
using a sample size of 15, predict the probability of both < 2 errors or errors using the correct discrete probability distribution. Assume calls are either correct or incorrect. 2. Using the call time mean and standard deviation from the quality sample, find the probability of a call time < 7min, between 7 and 9 min, and > 9 min.
3. Calculate and evaluate the 95% confidence interval for the mean from the call time
data.
Answer
The central limit theorem and the typicality of the disseminations are basic components
for insights and utilized for the rest of the lessons. “The Central Limit Theorem
expresses that the example methods for huge estimated tests will be ordinarily
disseminated paying little heed to the state of their populace appropriations” (Donnelly,
2015, p. 301). For this situation, we have to know the room for mistakes and the
standard blunder are two unique ideas. The room for give and take, or the width of the
interim, is the basic z-score duplicated by the standard mistake for the mean. Be that as
it may, a bigger example lessens the standard mistake and, along these lines, the room
for give and take.
Typical conveyances utilize the z-score estimation, which you found out about in Lesson
2, to recognize the likelihood. To additionally develop this idea, there are times when
you should discover the probabilities that are > < or some place in the middle of two distinctive z-scores (Donnelly, 2015). In light of the likelihood of a blunder gave in the quality synopsis under call quality utilizing an example size of 15, we have n = 15 calls. The consequent phase is to determine the likelihood of <2 blunders. P (mistake) = 0.15. At that point we make utilization of binomial likelihood recipe: (x < = 2) = P(x=0) + P(x=1) + P(x=2); P(x=0) = 15C0 * 0.15^0 * 0.85^15 = 0.08735; P(x=1) = 15C1 * 0.15^1 * 0.85^14 = 0.23123; P(x=2) = 15C2 * 0.15^2 * 0.85^13 = 0.28564. In this manner, we have P(X<=2) = 0.28564 + 0.23123 + 0.08735. The last answer will be P(x < = 2) = 0.604225. The calls are perfect. Binomial Probabilities Data Sample size Probability of an event of interest 15 0.15 Statistics Mean Variance Standard deviation 2.25 1.9125 1.3829 Binomial Probabilities Table X P(X) 0.0874 0.2312 0.2856 0.2184 0.1156 0.0449 0 1 2 3 4 5 <2 0.3186 >=5
0.0617
6
7
8
9
10
11
12
13
14
15
0.0132
0.0030
0.0005
0.0001
0.0000
0.0000
0.0000
0.0000
0.0000
0.0000
We have the quantities of test size, mean and standard deviation figured through
exceed expectations. The example estimate is 15, Mean is 12.05, Standard Deviation is
4.502. The second inquiry is to discover P(X<7). Since (x-mean)/standard deviation=2, P(2<(7-12.05)/4.502)=P(z<-1.12)=P(z>1.12)=1-P(z>1.12). With a z appropriation
table, the last answer will be P(x<7) =1-0.869=0.131. The subsequent stage is to ascertain the likelihood in the vicinity of 7 and 9. P(70.68)- 0.131=1-P(z<0.68)- 0.131=0.1169. P(x>9)
=P (z> (9-12.05)/4.502) =P (z>-0.68) =P (z<0.68) =0.752. When computing the certainty interim populace standard deviation is known, we utilize the z conveyance, with the recipe of mean +-Za /2* standard blunder. Standard mistake is a critical piece of utilizing the central limit theorem. Standard mistake is unique in relation to examining blunder. This measurement centers on the standard deviation versus the mean. We require standard mistake to discover the certainty interim. Standard error= standard deviation/sqrt test size=4.502/sqrt15= 1.16. The estimation of 1.96 depends on the way that 95% of the territory of a typical circulation is inside 1.96 standard deviations of the mean; 1.16 is the standard blunder of the mean. LCL=12.05 (1.96) (1.16) = 9.776, UCL=12.05 + (1.96) (1.16) = 14.32. All in all, the likelihood of either < 2 mistakes or = 5 blunders is 0.604. The likelihood of a call time < 7 min is 0.131, the likelihood of a call time in the vicinity of 7 and 9 min is 0.1169, and the likelihood of a call time > 9 min is 0.752. Moreover, we should know
about the meaning of the certainty interim for the mean – an interim gauge around the
example imply that gives a range of where the genuine populace means
falsehoods.
•
Calculate and evaluate the 95% confidence interval for the mean from the call
time data.
o
Problem uses t critical value since you do not know population standard
deviation and you are using the sample standard deviation.
11.07 +/- (1.98)(3.38/sqrt(150)= 10.52, 11.61

Reference:
Donnelly, R. A. (2015). Business statistics (2nd ed.). Upper Saddle River, NJ:
Pearson.

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