Problem 7 - Calculus Early Transcendentals: Pearson New International Edition

Question

A discrete probability distribution for a random variable \(X\) is given. Use the given distribution to find (a) \(P(X \geq 2)\) and (b) \(E(X)\). $$ p_{i}=(5-i) / 10, x_{i}=i, i=1,2,3,4 $$

Step-by-Step Solution

Answer

(a) 0.6, (b) 2.0

Step 1: Understand the given distribution

The random variable \(X\) takes values \(x_i = i\) where \(i = 1, 2, 3, 4\). The probability distribution function is given by \(p_i = \frac{5-i}{10}\). Evaluate the probabilities for each value of \(x_i\).

Step 2: Calculate individual probabilities

Substitute the values of \(i\) into the probability function to find \(p(1), p(2), p(3), p(4)\):- \(p(1) = \frac{4}{10} = 0.4\)- \(p(2) = \frac{3}{10} = 0.3\)- \(p(3) = \frac{2}{10} = 0.2\)- \(p(4) = \frac{1}{10} = 0.1\)

Step 3: Calculate \(P(X \geq 2)\)

Add the probabilities of the events that are greater than or equal to 2:\[P(X \geq 2) = P(2) + P(3) + P(4) = 0.3 + 0.2 + 0.1 = 0.6\]

Step 4: Calculate the Expected Value \(E(X)\)

Apply the formula for the expected value: \[E(X) = \sum_{i=1}^{4} x_i \cdot p(x_i) = 1 \cdot 0.4 + 2 \cdot 0.3 + 3 \cdot 0.2 + 4 \cdot 0.1\]Calculate each term and sum them:\[E(X) = 0.4 + 0.6 + 0.6 + 0.4 = 2.0\]

Key Concepts

Discrete Random Variable

Discrete random variables are integral to understanding probability distributions. They can only take on a specific, countable number of values.
This means there are gaps between possible values. Unlike continuous random variables, you can't have infinite options between two points.
In our exercise, the random variable \(X\) can take values of \(1, 2, 3,\) or \(4\). Each of these values corresponds to a probability, described by the function \(p(x_i) = \frac{5-i}{10}\). This association of discrete values and their probabilities forms a discrete probability distribution.

Expected Value

The expected value is a key concept in probability, offering the average outcome you'd anticipate.
It is like a weighted average where each outcome is weighted by its probability of occurring. You calculate it using the formula

• \(E(X) = \sum x_i \cdot p(x_i)\).
  

In our example, you perform the calculation by plugging in values:

• \(E(X) = 1 \times 0.4 + 2 \times 0.3 + 3 \times 0.2 + 4 \times 0.1\).

This gives \(E(X) = 2.0\), meaning that over many trials, the average value of \(X\) should approach 2.0. This concept of expected value is fundamental in predicting outcomes in probabilistic scenarios.

Probability Calculation

Probability calculation helps determine the likelihood of certain outcomes. The probability of any specific event is the sum of the probabilities of the individual events that comprise it.
In the given exercise, we calculated \(P(X \geq 2)\), meaning the probability that \(X\) is at least 2.
To find this, you add together the probabilities of all values that meet this condition, \(P(2) + P(3) + P(4)\).

• \(P(X \geq 2) = 0.3 + 0.2 + 0.1 = 0.6\)
  

This calculation provides us with a cumulative probability, which is often used to assess the range of outcomes over a specified threshold.