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Week 3 DQ 2 – Data Analysis and Business Intelligence | Westcliff University

Wk3 DQ2 | Discussion Question 2 | Checked by Turnitin

Define Central Limit theorem and its important in statistics. Suppose your statistics instructor gave six examinations during the semester. You received the following exam scores (percent correct): 79, 64, 84, 82, 92, and 77. To compute your final course grade, the instructor decided to randomly select two exam scores, compute their mean, and use this score to determine your final course grade.

a. Compute the population mean.

b. How many different samples of two test grades are possible?

c. List all possible samples of size 2 and compute the mean of each.

d. Compute the mean of the sample means and compare it to the population mean.

e. If you were a student, would you like this arrangement? Would the result be different from dropping the lowest score? Write a brief report

Ans:

This week 3 DQ 2 is mainly based on the concept and importance of Central Limit Theorem along with numerical analysis.

Central Limit Theorem

The central limit theorem states that the distribution of sample means approximates a normal distribution as the sample size gets larger (assuming that all samples are identical in size), regardless of population distribution shape (Ganti, 2019). The Central Limit Theorem is exactly what the shape of the distribution of means will be when we draw repeated samples from a given population. Specifically, as the sample sizes get larger, the distribution of means calculated from repeated sampling will approach normality (Mishra, 2018).

Importance of Central Limit Theorem in Statistics

Central Limit theorem is important in Statistics because it allows us to use the normal distribution to make inferences concerning the population mean.

It is important in Statistics because it guarantees that, when it applies, the samples that are drawn are always randomly selected.

It is important in Statistics because it enables reasonably accurate probabilities to be determined for events involving the sample average when the sample size is large enough regardless of the distribution of the original value.

Numerical Solution:

Given Information;

Exam Scores: 79, 64, 84, 82, 92 and 77

1. Numerical Answer (a) :

Population mean (µ) =∑X/N = (79 + 64+ 84 +82+ 92 + 77) = 478/6 =79.66

Therefore, the Population mean is 79.66.

1. Numerical Answer (b) :

Number of samples for selecting two test grades = nCr = 6C2 = 15

Therefore, the number of samples for selecting two test grades is 15.

 Possible samples of size 2 and their mean value Samples Sample Mean (x̅) 79,64 71.5 79,84 81.5 79,82 80.5 79,92 85.5 79,77 78 64,84 74 64,82 73 64,92 78 64,77 70.5 84,82 83 84,92 88 84,77 80.5 82,92 87 82,77 79.5 92,77 84.5 Total 1195

1. Numerical Answer (d) :

Calculated Population mean (µ) = ∑X/N =478/6= 79.66

Sample mean (µx) = ∑means/ samples = 1195/15 =79.66

Therefore both the Population mean and sample mean are equal.

1. Numerical Answer (e):

If I were a student, I would not like this arrangement because if we take into account the concept of Central Limit Theorem which says that as the number of samples considered go on increasing, the tendency of the sample being more representative of the population would go higher i.e. the sample distribution has higher tendency to follow the normal distribution.

After dropping of the lowest score, population mean is given by:

Population mean (µ) = (79 + 84 +82+ 92 + 77) / 5 = 82.8.

References

Ganti, A. (2019, April 19). Central Limit Theorem (CLT). Retrieved from Investopedia: https://www.investopedia.com/terms/c/central_limit_theorem.asp

Mishra, M. (2018, June 19). Understanding The Central Limit Theorem. Retrieved from https://towardsdatascience.com/understanding-the-central-limit-theorem-642473c63ad8

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