STA301 ASSIGNMENT NO. 2 FALL 2022 || 100% RIGHT SOLUTION || STATISTICS AND PROBABILITY || BY VuTech

STA301 ASSIGNMENT NO. 2 FALL 2022

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STA301 ASSIGNMENT NO. 2 FALL 2022 || 100% RIGHT SOLUTION || STATISTICS AND PROBABILITY || BY VuTech

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Assignment No.2 (Course STA 301)

 Fall 2022 (Total Marks 20)

Deadline

Your Assignment must be uploaded/ submitted before or on January 23 , 2023, Time 23:59

(STUDENTS ARE STRICTLY DIRECTED TO SUBMIT THEIR ASSIGNMENT BEFORE OR BY DUE DATE. NO ASSIGNMNENT AFTER DUE DATE WILL BE ACCEPTED VIA E.MAIL).

Rules for Marking

It should be clear that your Assignment will not get any credit IF:

• The Assignment submitted, via email, after due date.
• The submitted Assignment is not found as MS Word document file.
• There will be unnecessary, extra or irrelevant material.
• The Statistical notations/symbols are not well-written i.e., without using MathType software.
• The Assignment will be copied from handouts, internet or from any other student’s file. Copied material (from handouts, any book or by any website) will be awarded ZERO MARKS. It is PLAGIARISM and an Academic Crime.
• The medium of the course is English. Assignment in Urdu or Roman languages will not be accepted.
• Assignment means Comprehensive yet precise accurate details about the given topic quoting different sources (books/articles/websites etc.). Do not rely only on handouts. You can take data/information from different authentic sources (like books, magazines, website etc.) BUT express/organize all the collected material in YOUR OWN WORDS. Only then you will get good marks.

Objective(s) of this Assignment:

The assignment is being uploaded to build up the concepts of Probability Distributions.

                

Assignment no. 2 (Lessons 25-30)

Question 1:                                                                                                                      Marks: 10

An experiment is repeated 100 times and a fitted binomial distribution is obtained as follows:

`\P(X = x) = C_x^5 (0.4)^x (0.6)^{n - x}, x = 0,1,2,3,4,5`

Calculate the Expected frequencies for the fitted binomial distribution.

Solution:

The expected frequency for a particular value of x in a binomial distribution is given by the product of the probability of that value and the total number of trials (in this case, 100). So for x = 0, the expected frequency is:

Expected frequency for x = 0 =` \P(X = 0) * 100 = C_0^5 (0.4)^0 (0.6)^{5 - 0} * 100 = 0.077 * 100 = 7.7`

Similarly, we can calculate the expected frequencies for the other values of x:

Expected frequency for x = 1 = `\P(X = 1) * 100 = C_1^5 (0.4)^1 (0.6)^{5 - 1} * 100 = 0.234 * 100 = 23.4`

Expected frequency for x = 2 = `\P(X = 2) * 100 = C_2^5 (0.4)^2 (0.6)^{5 - 2} * 100 = 0.312 * 100 = 31.2`

Expected frequency for x = 3 = `\P(X = 3) * 100 = C_3^5 (0.4)^3 (0.6)^{5 - 3} * 100 = 0.194 * 100 = 19.4`

Expected frequency for x = 4 =` \P(X = 4) * 100 = C_4^5 (0.4)^4 (0.6)^{5 - 4} * 100 = 0.053 * 100 = 5.3`

Expected frequency for x = 5 =` \P(X = 5) * 100 = C_5^5 (0.4)^5 (0.6)^{5 - 5} * 100 = 0.006 * 100 = 0.6`

So the expected frequencies for the fitted binomial distribution are 7.7, 23.4, 31.2, 19.4, 5.3, and 0.6 for x = 0, 1, 2, 3, 4, and 5, respectively.

 

Question 2:                                                                                                                        Marks: 5

Calculate the mean and the variance of the hypergeometric distribution `h(x;50,5,3)`.

Solution:

The mean of a hypergeometric distribution is given by:

mean = `(n * K) / N`

where n is the number of items to be drawn, K is the number of successful items in the population, and N is the total number of items in the population.

For `h(x;50,5,3),` the mean is:

mean = `(3 * 5) / 50 = 0.3``

The variance of a hypergeometric distribution is given by:

variance = `(n * K * (N - K) * (N - n)) / (N^2 * (N - 1))`

For ``h(x;50,5,3),` the variance is:

variance = `(3 * 5 * (50 - 5) * (50 - 3)) / (50^2 * (50 - 1)) = 0.45``

So the mean of the hypergeometric distribution h(x;50,5,3) is 0.3 and the variance is 0.45.

Question 3:                                                                                                                        Marks: 5

Suppose that the number of typing errors per page has a Poisson distribution with average 4 typing errors. What is the probability that in a given page the number of typing errors will at least 2?

 

Solution:

The probability that in a given page the number of typing errors will be at least 2 is the sum of the probabilities of having 2, 3, 4, 5, ... typing errors.

Using the Poisson probability formula:

`P(X = k) = (e^(-λ) * λ^k) / k!`

Where λ is the average number of typing errors (4 in this case) and k is the number of typing errors.

To find the probability of having 2 or more typing errors, we can use the cumulative distribution function (CDF) of the Poisson distribution, which is the sum of the probabilities of having 0, 1, 2, 3, ... typing errors up to a certain value.

So, to find the probability of having 2 or more typing errors:

`P(X ≥ 2) = 1 - P(X < 2) = 1 - (P(X = 0) + P(X = 1))`

Using the Poisson probability formula:

`P(X = 0) = (e^(-4) * 4^0) / 0! = e^(-4) = 0.0183 P(X = 1) = (e^(-4) * 4^1) / 1! = 4e^(-4) = 0.0732`

So:

`P(X ≥ 2) = 1 - (0.0183 + 0.0732) = 1 - 0.0915 = 0.9085`

So the probability of having at least 2 typing errors in a given page is 0.9085.

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