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Chapter 4: Problem 83

A random variable \(x\) can assume five values: 0,1,2,3,4 . A portion of the probability distribution is shown here: $$\begin{array}{l|lllll}x & 0 & 1 & 2 & 3 & 4 \\\\\hline p(x) & .1 & .3 & .3 & ? & .1\end{array}$$ a. Find \(p(3)\). b. Construct a probability histogram for \(p(x)\). c. Calculate the population mean, variance, and standard deviation. d. What is the probability that \(x\) is greater than \(2 ?\) e. What is the probability that \(x\) is 3 or less?

Short Answer

Expert verified
Based on the given probability distribution of the discrete random variable 饾懃, which can take five values: 0, 1, 2, 3, and 4: 1. The missing probability value is 饾憹(3) = 0.2. 2. The probability histogram is constructed as follows: Value of x | Histogram 0 | ########## 1 | #################### 2 | #################### 3 | ########## 4 | ########## 3. The population mean is 饾渿 = 1.9, the variance is 饾湈虏 = 1.21, and the standard deviation is 饾湈 鈮� 1.1. 4. The probability of 饾懃 > 2 is 0.3. 5. The probability of 饾懃 鈮� 3 is 0.9.

Step by step solution

01

Find \(p(3)\)

According to the properties of discrete probability distribution, the sum of all possible probabilities must equal 1, which can be expressed as: $$\sum_{i} p(x_i) = 1$$ Where \(p(x_i)\) is the probability of each possible value of the random variable. Using the given probabilities, we can calculate the probability value for \(p(3)\) as: $$p(3) = 1 - (0.1 + 0.3 + 0.3 + 0.1) = 1 - 0.8 = 0.2$$
02

Construct a probability histogram

A probability histogram for the given probability distribution will show the probabilities associated with each value of the variable on the y-axis and the values of the variable on the x-axis. The histogram can be represented as follows: Value of x | Histogram 0 | ########## 1 | #################### 2 | #################### 3 | ########## 4 | ########## (numbers of "#" representing the values of probabilities)
03

Calculate the population mean, variance, and standard deviation

The population mean \(\mu\) can be calculated as: $$\mu = E(x) = \sum_{i} x_i p(x_i)$$ Using the given probabilities and values, we have the following values for the population mean: $$\mu = 0 * 0.1 + 1 * 0.3 + 2 * 0.3 + 3 * 0.2 + 4 * 0.1 = 0 + 0.3 + 0.6 + 0.6 + 0.4 = 1.9$$ The variance, \(\sigma^2\), can be calculated as: $$\sigma^2 = E((x - \mu)^2) = \sum_{i} (x_i - \mu)^2 p(x_i)$$ Using the population mean calculated earlier, we get the following values for variance: $$\sigma^2 = (0 - 1.9)^2 * 0.1 + (1 - 1.9)^2 * 0.3 + (2 - 1.9)^2 * 0.3 + (3 - 1.9)^2 * 0.2 + (4 - 1.9)^2 * 0.1$$ $$\sigma^2 = 3.61 * 0.1 + 0.81 * 0.3 + 0.01* 0.3 + 1.21 * 0.2 + 4.41 * 0.1 = 1.21$$ The standard deviation, \(\sigma\), can be calculated as the square root of variance: $$\sigma = \sqrt{\sigma^2} = \sqrt{1.21} \approx 1.1$$
04

Find the probability of \(x > 2\)

We need to calculate the probability of \(x\) being greater than 2. This can be done by adding the probabilities associated with the values greater than 2, which are 3 and 4: $$P(x > 2) = p(3) + p(4) = 0.2 + 0.1 = 0.3$$
05

Find the probability of \(x \leq 3\)

We need to calculate the probability of \(x\) being equal to or less than 3. This can be done by adding the probabilities associated with the values less than or equal to 3: $$P(x \leq 3) = p(0) + p(1) + p(2) + p(3) = 0.1 + 0.3 + 0.3 + 0.2 = 0.9$$ Now, we have answers for all the required parts of the problem.

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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 statistics and probability theory. It is a variable that represents possible outcomes of a random phenomenon. In the context of our exercise, the random variable is represented by \(x\), which can assume the values 0, 1, 2, 3, and 4. Each of these values is associated with a probability that describes how likely it is for the random variable to take on that particular value.

Some key points about random variables are:
  • They can be discrete or continuous. Discrete random variables have specific, separate values like our example of 0 to 4.
  • The sum of all the probabilities of a discrete random variable must be 1.
By understanding random variables, you gain insight into how probability distributions are structured.
Population Mean
The population mean, often symbolized by \( \mu \), is a measure of the central tendency of a set of values. It represents the average or expected value of a random variable within the entire population. In probability, calculating the mean involves weighing each possible value of the random variable against its probability.

For the random variable in our exercise, the population mean is calculated as follows:
\[ \mu = 0 \times 0.1 + 1 \times 0.3 + 2 \times 0.3 + 3 \times 0.2 + 4 \times 0.1 = 1.9 \]

This means that, on average, the value of the random variable \(x\) is expected to be 1.9 given the probability distribution provided.
Variance
Variance is a statistical measure that tells us how much the values of a random variable differ from the population mean, in other words, it measures the dispersion of the data. It can be thought of as the average of the squared differences from the mean. The variance is denoted as \( \sigma^2 \).

To calculate variance for our random variable, we compute:
\[\sigma^2 = \sum (x_i - \mu)^2 \times p(x_i)\]

Substituting the values:
\[ \sigma^2 = (0 - 1.9)^2 \times 0.1 + (1 - 1.9)^2 \times 0.3 + (2 - 1.9)^2 \times 0.3 + (3 - 1.9)^2 \times 0.2 + (4 - 1.9)^2 \times 0.1 = 1.21 \]

A higher variance value indicates a greater dispersion of values around the mean.
Standard Deviation
The standard deviation is a key concept that quantifies the amount of variation or dispersion in a set of values. It is simply the square root of the variance \( \sigma^2 \), and is denoted by \( \sigma \). The standard deviation provides a measure of the average distance of each data point from the population mean.

For our example, the standard deviation is calculated as:
\[ \sigma = \sqrt{1.21} \approx 1.1 \]

This tells us that the values of the random variable \(x\) deviate, on average, by about 1.1 from the mean. Standard deviation is widely used because it is expressed in the same units as the data, which makes it more interpretable compared to variance.

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