STA642 ASSIGNMENT NO. 2 FALL 2022 || 100% RIGHT SOLUTION || PROBABILITY DISTRIBUTIONS || BY VuTech

STA642 ASSIGNMENT NO. 2 FALL 2022

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STA642 ASSIGNMENT NO. 2 FALL 2022 || 100% RIGHT SOLUTION || PROBABILITY DISTRIBUTIONS || BY VuTech

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Course Title	Probability Distributions
Course Code	STA642
Activity	Assignment # 2
Total Marks	10
Lessons included	108-127
Submission Deadline	Date:  25th January, 2023 Time: 23:59 Hours (Midnight)

Instructions to attempt assignment:

• Take help from video lectures/PPTs/Recommended book
• For writing symbols, use Math Type software. If you already did not download it, get it from here.
• Make your own attempt, any copy/past; cheating from other student/website may result in ZERO marks and/or assignment cancellation.

Question 1:

Drive the expression of MGF for the Binomial distribution also find it’s Mean and Variance?

Solution:

The binomial distribution models the number of successes in a fixed number of Bernoulli trials, where each trial has a probability p of success and a probability of (1-p) of failure. The random variable X represents the number of successes in n trials. The probability mass function (PMF) of X is given by:

`p(x) = (n choose x) * p^x * (1-p)^(n-x) for x = 0, 1, 2, ..., n`

where (n choose x) = `n! / (x! * (n-x)!)` is the binomial coefficient.

The moment generating function (MGF) of a random variable X is defined as `MGF(t) = E(e^(tx)),` where t is a real number and E is the expected value operator. To find the MGF of a binomial random variable X, we substitute the PMF into the definition of MGF and simplify:

`MGF(t) = E(e^(tx)) = SUM(x=0 to n) (e^(tx) * p(x)) = SUM(x=0 to n) (e^(tx) * (n choose x) * p^x * (1-p)^(n-x))`

= `SUM(x=0 to n) ( (e^t)^x * (1-p + pe^t)^(n-x) * (n choose x))`

= `(1-p + pe^t)^n`

This MGF tells us that the binomial distribution is completely determined by two parameters: n, the number of trials, and p, the probability of success.

The mean of a binomial random variable X is given by the expected value of X, which is the first moment of the distribution. We can find the mean by taking the derivative of the MGF with respect to t, and then setting t = 0:

`E(X) = d/dt MGF(t) | t=0 = d/dt ( (1-p + pe^t)^n ) | t=0 = n * pe^0 * (1-p + pe^0)^(n-1) = np`

So the mean of a binomial random variable X is equal to np, where n is the number of trials and p is the probability of success.

The variance of a binomial random variable X is given by `Var(X) = E(X^2) - [E(X)]^2.` The second moment of the distribution can be found by taking the second derivative of the MGF with respect to t and then setting t = 0:

E(X^2) = d^2/dt^2 MGF(t) | t=0 = d^2/dt^2 ( (1-p + pe^t)^n ) | t=0 = n * (n-1) * p^2 e^0 * (1-p + pe^0)^(n-2) = n * (n-1) * p^2 + np`

Therefore, the variance of a binomial random variable X is

`Var(X) = E(X^2) - [E(X)]^2 = n * (n-1) * p^2 + np - (np)^2 = np(1-p)`

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