Discrete Random Variable Probability: A Step-by-Step Guide

by Henrik Larsen 59 views

Hey guys! Let's dive into the fascinating world of discrete random variables and probability functions. We're going to break down a specific problem step-by-step, making sure you grasp every concept along the way. So, buckle up and get ready to explore!

Problem Statement

We're given a discrete random variable, let's call it X, which has a probability function defined as follows:

P(X=x)={k(x2βˆ’9),x=4,5,60Β otherwiseΒ P(X=x)=\left\{\begin{array}{ll} k(x^2-9), & x=4,5,6 \\ 0 & \text { otherwise } \end{array}\right.

Where k is a positive constant. We have two main tasks ahead of us:

  • a. Show that k = 1/50
  • b. Find a certain value (which we'll get to later!).

Let's tackle these one by one.

Part a: Proving k = 1/50

The key concept here is that the sum of probabilities for all possible values of a discrete random variable must equal 1. Think of it like this: if you list out every single outcome and its probability, the total chance of something happening has to be 100%, or 1.

In our case, the possible values for X are 4, 5, and 6. So, we can write the following equation:

P(X=4)+P(X=5)+P(X=6)=1P(X=4) + P(X=5) + P(X=6) = 1

Now, let's substitute the probability function we were given:

k(42βˆ’9)+k(52βˆ’9)+k(62βˆ’9)=1k(4^2 - 9) + k(5^2 - 9) + k(6^2 - 9) = 1

Let's simplify this step-by-step. Remember the order of operations (PEMDAS/BODMAS)? Parentheses/Brackets first!

k(16βˆ’9)+k(25βˆ’9)+k(36βˆ’9)=1k(16 - 9) + k(25 - 9) + k(36 - 9) = 1

k(7)+k(16)+k(27)=1k(7) + k(16) + k(27) = 1

Now we can factor out the k:

k(7+16+27)=1k(7 + 16 + 27) = 1

And add the numbers inside the parentheses:

k(50)=1k(50) = 1

Finally, to isolate k, we divide both sides of the equation by 50:

k=150k = \frac{1}{50}

Boom! We've successfully shown that k = 1/50. That wasn't so bad, right? We used the fundamental property of probability distributions and some basic algebra to get there. Understanding this foundational principle is crucial for tackling more complex problems involving discrete random variables.

Digging Deeper: Why does the sum of probabilities equal 1?

It's worth pausing here to really internalize why this principle holds true. Imagine you're rolling a die. The possible outcomes are 1, 2, 3, 4, 5, and 6. Each outcome has a probability of 1/6. If you add up the probabilities of all these outcomes (1/6 + 1/6 + 1/6 + 1/6 + 1/6 + 1/6), you get 1. This represents the certainty that some outcome will occur when you roll the die. You're guaranteed to get one of the numbers! This concept extends to any discrete random variable: all possible outcomes collectively cover the entire probability space, which is why their probabilities must sum to 1. This is a cornerstone of probability theory and a critical concept to grasp.

Part b: Finding the Expected Value and Variance (Let's Assume the Question Asked for These)

Okay, so the original problem statement only had part (a). But let's make this even more valuable! Let's assume part (b) asked us to find the expected value and variance of X. This is a very common type of question when dealing with discrete random variables, and it's a great way to solidify our understanding.

Expected Value: The Average Outcome

The expected value, often denoted as E(X) or ΞΌ (mu), represents the average value we'd expect X to take over many trials. It's a weighted average, where each value of X is weighted by its probability. The formula for the expected value of a discrete random variable is:

E(X)=βˆ‘xP(X=x)E(X) = \sum xP(X=x)

This fancy sigma notation just means we're going to sum up the product of each value of x and its corresponding probability. In our case, we have:

E(X)=4βˆ—P(X=4)+5βˆ—P(X=5)+6βˆ—P(X=6)E(X) = 4 * P(X=4) + 5 * P(X=5) + 6 * P(X=6)

We already know that k = 1/50, so we can calculate the individual probabilities:

P(X=4)=150(42βˆ’9)=150(7)=750P(X=4) = \frac{1}{50}(4^2 - 9) = \frac{1}{50}(7) = \frac{7}{50}

P(X=5)=150(52βˆ’9)=150(16)=1650P(X=5) = \frac{1}{50}(5^2 - 9) = \frac{1}{50}(16) = \frac{16}{50}

P(X=6)=150(62βˆ’9)=150(27)=2750P(X=6) = \frac{1}{50}(6^2 - 9) = \frac{1}{50}(27) = \frac{27}{50}

Now we can plug these probabilities back into the expected value formula:

E(X)=4βˆ—750+5βˆ—1650+6βˆ—2750E(X) = 4 * \frac{7}{50} + 5 * \frac{16}{50} + 6 * \frac{27}{50}

E(X)=2850+8050+16250E(X) = \frac{28}{50} + \frac{80}{50} + \frac{162}{50}

E(X)=27050=5.4E(X) = \frac{270}{50} = 5.4

So, the expected value of X is 5.4. This means that, on average, we'd expect X to be around 5.4 if we observed it many times. Understanding expected value is fundamental in various fields, from finance to gambling! It provides a crucial measure of central tendency for random variables.

Variance: Measuring the Spread

The variance, often denoted as Var(X) or σ² (sigma squared), measures how spread out the values of X are from the expected value. A higher variance indicates that the values are more dispersed, while a lower variance indicates they are clustered closer to the mean. There are a couple of ways to calculate variance, but a common and convenient formula is:

Var(X)=E(X2)βˆ’[E(X)]2Var(X) = E(X^2) - [E(X)]^2

We already know E(X), which is 5.4. Now we need to calculate E(XΒ²). This is the expected value of X squared, and the formula is similar to the expected value formula:

E(X2)=βˆ‘x2P(X=x)E(X^2) = \sum x^2P(X=x)

So, in our case:

E(X2)=42βˆ—P(X=4)+52βˆ—P(X=5)+62βˆ—P(X=6)E(X^2) = 4^2 * P(X=4) + 5^2 * P(X=5) + 6^2 * P(X=6)

E(X2)=16βˆ—750+25βˆ—1650+36βˆ—2750E(X^2) = 16 * \frac{7}{50} + 25 * \frac{16}{50} + 36 * \frac{27}{50}

E(X2)=11250+40050+97250E(X^2) = \frac{112}{50} + \frac{400}{50} + \frac{972}{50}

E(X2)=148450=29.68E(X^2) = \frac{1484}{50} = 29.68

Now we have everything we need to calculate the variance:

Var(X)=E(X2)βˆ’[E(X)]2Var(X) = E(X^2) - [E(X)]^2

Var(X)=29.68βˆ’(5.4)2Var(X) = 29.68 - (5.4)^2

Var(X)=29.68βˆ’29.16Var(X) = 29.68 - 29.16

Var(X)=0.52Var(X) = 0.52

Therefore, the variance of X is 0.52. This tells us that the values of X are relatively close to the expected value of 5.4. A small variance implies a low degree of variability, suggesting that the outcomes are fairly consistent.

Standard Deviation: A More Intuitive Measure of Spread

Often, we use the standard deviation instead of the variance because it's in the same units as X, making it easier to interpret. The standard deviation is simply the square root of the variance:

SD(X)=Var(X)SD(X) = \sqrt{Var(X)}

SD(X)=0.52β‰ˆ0.721SD(X) = \sqrt{0.52} \approx 0.721

So, the standard deviation of X is approximately 0.721. This gives us a more concrete sense of the spread: the values of X typically deviate from the mean by about 0.721 units. Standard deviation is a critical metric in statistics, offering a readily interpretable measure of data dispersion.

Wrapping Up

We've successfully tackled a problem involving a discrete random variable, calculated probabilities, and even found the expected value and variance. Remember, the key takeaways are:

  • The sum of probabilities for all possible outcomes of a discrete random variable must equal 1.
  • The expected value represents the average outcome.
  • The variance and standard deviation measure the spread of the data.

These concepts are foundational in probability and statistics, and mastering them will open doors to understanding more advanced topics. Keep practicing, keep exploring, and you'll become a pro in no time! Remember, statistics can initially feel complex, but with consistent effort and focused practice, you'll unlock its power and see its applications in a multitude of real-world scenarios.