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The random variable \(X\) has the following probability distribution: $$ \begin{array}{lllll} x & 2 & 3 & 5 & 8 \\ \text { Probability } & 0.2 & 0.4 & 0.3 & 0.1 \end{array} $$ Determine the following: (a) \(P(X \leq 3)\) (b) \(P(X > 2.5)\) (c) \(P(2.7 < X < 5.1)\) (d) \(E(\mathrm{X})\) (e) \(V(\mathrm{X})\)

Short Answer

Expert verified
(a) 0.6, (b) 0.8, (c) 0.7, (d) 3.9, (e) 3.09

Step by step solution

01

- Compute P(X ≤ 3)

To determine \( P(X \leq 3) \), we need to add the probabilities of all outcomes where \( X \) is less than or equal to 3. Here, these outcomes are \( x = 2 \) and \( x = 3 \). Thus,\[P(X \leq 3) = P(X = 2) + P(X = 3) = 0.2 + 0.4 = 0.6\]
02

- Compute P(X > 2.5)

To find \( P(X > 2.5) \), we consider all outcomes where \( X \) is greater than 2.5. This means we include \( x = 3, 5, 8 \). Thus,\[P(X > 2.5) = P(X = 3) + P(X = 5) + P(X = 8) = 0.4 + 0.3 + 0.1 = 0.8\]
03

- Compute P(2.7 < X < 5.1)

The probability \( P(2.7 < X < 5.1) \) includes outcomes where \( X = 3 \) and \( X = 5 \) since these are the values of \( X \) within the interval 2.7 to 5.1. Thus,\[P(2.7 < X < 5.1) = P(X = 3) + P(X = 5) = 0.4 + 0.3 = 0.7\]
04

- Calculate E(X)

The expected value \( E(X) \) is calculated using the formula:\[E(X) = \sum (x_i \times P(x_i))\]Calculating this gives:\[E(X) = (2 \times 0.2) + (3 \times 0.4) + (5 \times 0.3) + (8 \times 0.1) = 0.4 + 1.2 + 1.5 + 0.8 = 3.9\]
05

- Compute V(X)

The variance \( V(X) \) is calculated using the formula:\[V(X) = \sum (x_i^2 \times P(x_i)) - (E(X))^2\]Calculate \( \sum(x_i^2 \times P(x_i)) \):\[= (2^2 \times 0.2) + (3^2 \times 0.4) + (5^2 \times 0.3) + (8^2 \times 0.1) = 0.8 + 3.6 + 7.5 + 6.4 = 18.3\]Subtract the square of the expected value:\[V(X) = 18.3 - (3.9)^2 = 18.3 - 15.21 = 3.09\]

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Random Variable
A random variable is a fundamental concept in probability theory. It's essentially a variable whose value is determined by the outcome of a random phenomenon. In simple terms, think of a random variable as a rule for assigning numbers to every possible outcome of an experiment. For example, in our exercise, the random variable \( X \) can take on the values 2, 3, 5, and 8. Each value corresponds to a certain probability or likelihood of occurring.

Random variables can be discrete or continuous. A discrete random variable, like \( X \) in our exercise, has countable outcomes. Each outcome has a specific probability associated with it. This probability distribution lists all possible values the variable can take and the probability for each of those values.
Expected Value
The expected value is a key concept in probability that provides a measure of the "central tendency" of a distribution. It's essentially the average or mean value you would expect if an experiment were repeated many times. For our discrete random variable \( X \), the expected value \( E(X) \) is calculated by multiplying each possible value of the random variable by its probability, and then summing all those products.

The formula to calculate the expected value is:
  • \( E(X) = \sum (x_i \times P(x_i)) \)
In the exercise, this results in:
  • \( E(X) = 3.9 \)
This value helps us understand the long-term average or expected outcome of the random variable \( X \).
Variance
Variance is a measure of how much the values of a random variable deviate from the expected value. It tells us how spread out the values are. The larger the variance, the more spread out the values of \( X \) are around the mean.

To calculate the variance \( V(X) \) for a discrete random variable, you use the formula:
  • \( V(X) = \sum(x_i^2 \times P(x_i)) - (E(X))^2 \)
In the exercise, the variance came out to be \( 3.09 \), which indicates the level of dispersion from the expected value.

This means that the values of \( X \) don't stray too far from the mean of 3.9, but there is still some variation.
Discrete Probability
Discrete probability involves probabilities related to discrete random variables. A discrete probability distribution lists each possible value that a random variable can take, along with the probability associated with each value.

In our exercise, the discrete probability distribution for \( X \) was given as \( x = 2, 3, 5, 8 \) with corresponding probabilities \( 0.2, 0.4, 0.3, 0.1 \).

This means:
  • There's a 20% chance of \( X \) being 2.
  • A 40% chance of \( X \) being 3.
  • A 30% chance for \( X \) to be 5.
  • And a 10% chance for \( X \) to be 8.
The sum of these probabilities must equal 1, ensuring that the distribution accounts for all possible outcomes of \( X \). This set-up allows us to make calculations about probabilities and understand the behavior of the random variable.

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Most popular questions from this chapter

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