# Central Limit Theory and the Binomial Distribution Research Paper

**Pages:** 5 (1285 words) ·
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3 · **File:** .docx · **Level:** Master's · **Topic:** Mathematics · **Written:** March 16, 2018

SAMPLE EXCERPT . . .

6%.

c) When you invest in a large number of stocks chosen at random, your average return will approach a Normal distribution.

The correct statement is that if you invest in more and more stocks selected at random, then your average return on these stocks will get closer and closer to 10.6%. This is because the law of large numbers states that of the size of the sample is large, then there is the tendency that the mean will converge to a number, which is very close to the population mean.

Part 2

1. Ray Allen, who is one of the best 3-point shooters over the last 15 years in the NBA, has a chance to shoot four free throws. He was fouled shooting a 3-point shot, and he gets to take a fourth shot due to a technical foul. Suppose that the probability that he makes a free throw is .9, and his free throws are independent of each other. Let X be the random variable that gives the number of free throws made in 4 attempts.

a) Give the possible values for X. What type of distribution model is appropriate? Explain why.

The possible values for X are 0, 1, 2, 3, and 4

b) Obtain the probability distribution for x

c) Indicate what the probability is for Ray Allen to make three or more of the free throws

P ( X ? 3) =

P (0) = [4! / (0! (4 - 0)!] 0.90 (1 – 0.9)4-0

= [(4! / 0! 4!)] 1(0.1)4

=0.0001

P (1) = [4! / (1! (4 - 1)!] 0.91 (1 – 0.1)4-1

= [4! / (1! (3)!] 0.9 (0.1)3

= 0.0036

P (2) = [4! / (1! (4 - 1)!] 0.92 (1 – 0.1)4-2

= [4! / (1! (3)!] 0.92 (0.1)2

= 0.0324Buy full paper

for $19.77 Probability = 0.0001 + 0.0036 + 0.0324 = 0.0361

P ( X ? 3) = 1 – 0.0361

= 0.9639

2. Suppose that in a quality control study it was determined that for a perishable item the number X of defective items follows a binomial distribution with n= 20 and p = .10.

a) Find the probability that a sample of 20 contain no more than two defective items.## Research Paper on

b) If one were to take many samples of 20 and found the mean number of defectives in each sample, what should we expect for the mean and standard deviation of the number of defectives?

3. This problem is motivated by the study: http://site.iugaza.edu.ps/wdaya/files/2013/03/A-Random-Walk-Down-Wall-Street.pdf

The random walk theory of stock markets implies that an index of stock prices has probability 0.63 of increasing in any year. In addition the change in the index in any year is not influenced by whether it goes up or down in an earlier year. Suppose that X is the number of years among the next six years in which the index rises.

a) Assuming that X has a binomial distribution, give are the values of n and p.

N = 6 and P = 0.63

b) What is the set of values that X can attain?

X can be either 1, 2, 3, 4, 5 or 6

c) Calculate the probability of each of the values of X given in part b. You can use the Excel BINOM.DIST function. Include the Excel function used and its result in your Word document.

P (x = 1)

=BINOM.DIST(1,6,0.63, FALSE)

= 0.02621

P (x =2)

BINOM.DIST(2,6,0.63, FALSE)

= 0.1116

P (X = 3)

= BINOM.DIST(3,6,0.63, FALSE)

= 0.2533

P (X = 4)

= BINOM.DIST(4,6,0.63, FALSE)

= 0.3235

P (X= 5)

= BINOM.DIST(5,6,0.63, FALSE)

= 0.2203

P (X = 6)

= BINOM.DIST(6,6,0.63, FALSE)

= 0.0625

d) Use Excel to construct a bar chart for the probabilities. The values of X should be on the horizontal axis, and the probabilities on the vertical axis. For information on how to use… [END OF PREVIEW] . . . READ MORE

6%.

c) When you invest in a large number of stocks chosen at random, your average return will approach a Normal distribution.

The correct statement is that if you invest in more and more stocks selected at random, then your average return on these stocks will get closer and closer to 10.6%. This is because the law of large numbers states that of the size of the sample is large, then there is the tendency that the mean will converge to a number, which is very close to the population mean.

Part 2

1. Ray Allen, who is one of the best 3-point shooters over the last 15 years in the NBA, has a chance to shoot four free throws. He was fouled shooting a 3-point shot, and he gets to take a fourth shot due to a technical foul. Suppose that the probability that he makes a free throw is .9, and his free throws are independent of each other. Let X be the random variable that gives the number of free throws made in 4 attempts.

a) Give the possible values for X. What type of distribution model is appropriate? Explain why.

The possible values for X are 0, 1, 2, 3, and 4

b) Obtain the probability distribution for x

c) Indicate what the probability is for Ray Allen to make three or more of the free throws

P ( X ? 3) =

P (0) = [4! / (0! (4 - 0)!] 0.90 (1 – 0.9)4-0

= [(4! / 0! 4!)] 1(0.1)4

=0.0001

P (1) = [4! / (1! (4 - 1)!] 0.91 (1 – 0.1)4-1

= [4! / (1! (3)!] 0.9 (0.1)3

= 0.0036

P (2) = [4! / (1! (4 - 1)!] 0.92 (1 – 0.1)4-2

= [4! / (1! (3)!] 0.92 (0.1)2

= 0.0324Buy full paper

for $19.77 Probability = 0.0001 + 0.0036 + 0.0324 = 0.0361

P ( X ? 3) = 1 – 0.0361

= 0.9639

2. Suppose that in a quality control study it was determined that for a perishable item the number X of defective items follows a binomial distribution with n= 20 and p = .10.

a) Find the probability that a sample of 20 contain no more than two defective items.

## Research Paper on *Central Limit Theory and the Binomial Distribution* Assignment

b) If one were to take many samples of 20 and found the mean number of defectives in each sample, what should we expect for the mean and standard deviation of the number of defectives?3. This problem is motivated by the study: http://site.iugaza.edu.ps/wdaya/files/2013/03/A-Random-Walk-Down-Wall-Street.pdf

The random walk theory of stock markets implies that an index of stock prices has probability 0.63 of increasing in any year. In addition the change in the index in any year is not influenced by whether it goes up or down in an earlier year. Suppose that X is the number of years among the next six years in which the index rises.

a) Assuming that X has a binomial distribution, give are the values of n and p.

N = 6 and P = 0.63

b) What is the set of values that X can attain?

X can be either 1, 2, 3, 4, 5 or 6

c) Calculate the probability of each of the values of X given in part b. You can use the Excel BINOM.DIST function. Include the Excel function used and its result in your Word document.

P (x = 1)

=BINOM.DIST(1,6,0.63, FALSE)

= 0.02621

P (x =2)

BINOM.DIST(2,6,0.63, FALSE)

= 0.1116

P (X = 3)

= BINOM.DIST(3,6,0.63, FALSE)

= 0.2533

P (X = 4)

= BINOM.DIST(4,6,0.63, FALSE)

= 0.3235

P (X= 5)

= BINOM.DIST(5,6,0.63, FALSE)

= 0.2203

P (X = 6)

= BINOM.DIST(6,6,0.63, FALSE)

= 0.0625

d) Use Excel to construct a bar chart for the probabilities. The values of X should be on the horizontal axis, and the probabilities on the vertical axis. For information on how to use… [END OF PREVIEW] . . . READ MORE

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