SOLUTION: Let X be the random variable denoting the number of forms required to be filled out by a contractor for participation in contract bids, where the values of X are 1, 2, 3, 4, and

Question 1173951: Let X be the random variable denoting the number of forms required to be filled out by a
contractor for participation in contract bids, where the values of X are 1, 2, 3, 4, and 5, and
suppose that the respective probabilities are proportional to X, that is ,
. Then calculate a) The value of the constant c
b) d) Mean and variance of the distribution.
e) Cumulative distribution function (cdf)
Answer by CPhill(2189) (Show Source): You can put this solution on YOUR website!
Let's break down this problem step-by-step.
**1. Understanding the Problem**
* X is a random variable representing the number of forms (1, 2, 3, 4, 5).
* The probabilities are proportional to X: P(X=x) = cx, where c is a constant.
**a) Finding the Value of the Constant c**
* The sum of all probabilities must equal 1.
* Therefore: P(1) + P(2) + P(3) + P(4) + P(5) = 1
* Substitute P(x) = cx: c(1) + c(2) + c(3) + c(4) + c(5) = 1
* Simplify: c(1 + 2 + 3 + 4 + 5) = 1
* c(15) = 1
* c = 1/15
**b) Probability Distribution**
* Now that we know c = 1/15, we can find the probabilities:
* P(X=1) = (1/15) * 1 = 1/15
* P(X=2) = (1/15) * 2 = 2/15
* P(X=3) = (1/15) * 3 = 3/15 = 1/5
* P(X=4) = (1/15) * 4 = 4/15
* P(X=5) = (1/15) * 5 = 5/15 = 1/3
**d) Mean (μ) and Variance (σ^2)**
* **Mean (μ):**
* μ = E(X) = Σ[x * P(X=x)]
* μ = (1 * 1/15) + (2 * 2/15) + (3 * 3/15) + (4 * 4/15) + (5 * 5/15)
* μ = (1 + 4 + 9 + 16 + 25) / 15
* μ = 55 / 15 = 11/3 ≈ 3.6667
* **Variance (σ^2):**
* σ^2 = E(X^2) - μ^2
* First, find E(X^2):
* E(X^2) = Σ[x^2 * P(X=x)]
* E(X^2) = (1^2 * 1/15) + (2^2 * 2/15) + (3^2 * 3/15) + (4^2 * 4/15) + (5^2 * 5/15)
* E(X^2) = (1 + 8 + 27 + 64 + 125) / 15
* E(X^2) = 225 / 15 = 15
* Now, calculate σ^2:
* σ^2 = 15 - (11/3)^2
* σ^2 = 15 - 121/9
* σ^2 = (135 - 121) / 9
* σ^2 = 14 / 9 ≈ 1.5556
**e) Cumulative Distribution Function (CDF)**
* The CDF, F(x) = P(X ≤ x), is the probability that X is less than or equal to x.
* F(1) = P(X ≤ 1) = P(X=1) = 1/15
* F(2) = P(X ≤ 2) = P(X=1) + P(X=2) = 1/15 + 2/15 = 3/15 = 1/5
* F(3) = P(X ≤ 3) = P(X=1) + P(X=2) + P(X=3) = 3/15 + 3/15 = 6/15 = 2/5
* F(4) = P(X ≤ 4) = P(X=1) + P(X=2) + P(X=3) + P(X=4) = 6/15 + 4/15 = 10/15 = 2/3
* F(5) = P(X ≤ 5) = P(X=1) + P(X=2) + P(X=3) + P(X=4) + P(X=5) = 10/15 + 5/15 = 15/15 = 1
* Therefore, the CDF is:
* F(x) = { 0, if x < 1
{ 1/15, if 1 ≤ x < 2
{ 1/5, if 2 ≤ x < 3
{ 2/5, if 3 ≤ x < 4
{ 2/3, if 4 ≤ x < 5
{ 1, if x ≥ 5

RELATED QUESTIONS
Consider the event of getting a head in 4tosses of a fair coin. Let X be the random... (answered by greenestamps,math_tutor2020)

The 36 investigators in an agency are each required to submit 25 investigation reports a... (answered by ikleyn)

Suppose three iphones are tested random. We want to find out the number of defective... (answered by math_tutor2020)

Let x be a random variable that denotes the number of tails obtained when three coins are (answered by ewatrrr)

Let a be the last digit of your social security number if it is not zero. In the latter... (answered by stanbon)

Do what is indicated. 2. Two balls are drawn in succession without replacement from an (answered by Boreal)

A box consists of 5 green and 3 red balls. Three balls are drawn at random from it. If X... (answered by Fombitz)

Let x be a discrete random variable that possesses a binomial distribution with n equals... (answered by rothauserc)

In a quarter-pound bag of red pistachio nuts, some shells are too difficult to pry open... (answered by stanbon)