Question

Use the probability distribution for the random variable \(x\) to answer the questions in Exercises 12-16. $$\begin{array}{l|rrrrrr}x & 0 & 1 & 2 & 3 & 4 & 5 \\\\\hline p(x) & .1 & .3 & .4 & .1 & ? & .05\end{array}$$ $$ \text { Find } \mu, \sigma^{2}, \text { and } \sigma \text { . } $$

Short answer

Answer: The mean (μ) is approximately 2.05, the variance (σ²) is approximately 0.6475, and the standard deviation (σ) is approximately 0.8053.

Step-by-step solution

Find the missing probability value for \(x=4\)

Since the sum of the probabilities in a probability distribution must equal 1, we can find the missing probability value for \(x=4\) by subtracting the sum of other probabilities from 1: \( p(4) = 1 - (p(0) + p(1) + p(2) + p(3) + p(5)) \) \( p(4) = 1 - (.1 + .3 + .4 + .1 + .05) \) \( p(4) = 1 - .95 \) \( p(4) = .05 \) Now we have the full probability distribution: $$\begin{array}{l|rrrrrr}x & 0 & 1 & 2 & 3 & 4 & 5 \\\\\hline p(x) & .1 & .3 & .4 & .1 & .05 & .05\end{array}$$

Compute the mean μ

To compute the mean (\(\mu\)) for a discrete random variable, use the formula: $$\mu = \sum_{i=1}^n x_i p(x_i)$$ Where \(x_i\) are the values of the random variable and \(p(x_i)\) are their respective probabilities. In our case, we have: $$ \mu = (0)(.1) + (1)(.3) + (2)(.4) + (3)(.1) + (4)(.05) + (5)(.05) $$ $$ \mu = 0 + .3 + .8 + .3 + .2 + .25 $$ $$ \mu = 2.05 $$

Compute the variance σ²

To compute the variance (\(\sigma^2\)), use the formula: $$\sigma^2 = E(x^2) - (\mu)^2$$ We first need to calculate \(E(x^2)\): $$ E(x^2) = \sum_{i=1}^n (x_i)^2 p(x_i) $$ For our distribution, $$ E(x^2) = (0^2)(.1) + (1^2)(.3) + (2^2)(.4) + (3^2)(.1) + (4^2)(.05) + (5^2)(.05) $$ $$ E(x^2) = 0 + .3 + 1.6 + .9 + 0.8 + 1.25 $$ $$ E(x^2) = 4.85 $$ Now we can calculate the variance: $$\sigma^2 = E(x^2) - (\mu)^2$$ $$\sigma^2 = 4.85 - (2.05)^2$$ $$\sigma^2 = 4.85 - 4.2025$$ $$\sigma^2 ≈ 0.6475$$

Compute the standard deviation σ

Finally, to compute the standard deviation (\(\sigma\)), take the square root of the variance: $$\sigma = \sqrt{\sigma^2}$$ $$\sigma = \sqrt{0.6475}$$ $$\sigma ≈ 0.8053$$ The mean (\(\mu\)), variance (\(\sigma^2\)), and standard deviation (\(\sigma\)) of the given probability distribution are approximately: $$\mu = 2.05$$ $$\sigma^2 ≈ 0.6475$$ $$\sigma ≈ 0.8053$$

Key concepts

Probability Distribution

Understanding a probability distribution is fundamental to statistical analysis. In essence, it's a mathematical function providing the probabilities of occurrence of different possible outcomes for an experiment. For example, a discrete probability distribution lists all possible discrete values of a random variable along with their corresponding probabilities.

Each probability is denoted by p(x), where x is a possible outcome, and it must be between 0 and 1. Moreover, the sum of all these probabilities must be 1, showcasing a complete representation of all possible outcomes. The table showing the distribution for the random variable, which in certain exercises might have missing values, is completed by ensuring the sum of probabilities equals 1. This adherence to principles ensures a coherent and comprehensive distribution that maps out the realm of possibility for the given random variable's outcomes.

Understanding the structure of discrete probability distributions helps in predicting and analyzing scenarios where the outcomes are distinct and finite, such as rolling a die, picking a card from a deck, or other scenarios with a set number of possible results.

Mean of Discrete Random Variable

The mean of a discrete random variable, often represented by μ (the Greek letter mu), is a measure of the central tendency of the probability distribution. You calculate the mean by multiplying each possible outcome of the variable by its probability and then summing all these products. This process is encapsulated by the formula:
\[\mu = \sum_{i=1}^n x_i p(x_i)\]
Here, x_i represents each discrete outcome and p(x_i) the probability of that outcome. The sum of these products gives the expected value or the average outcome of the random variable over many trials. The concept of mean as applied to a discrete probability distribution, gives insight into the 'center' of the distribution, acting as a balance point. It is a fundamental descriptor of any probability distribution and is often used as a baseline for comparison with actual results.

Variance and Standard Deviation

Moving beyond the mean, we explore how dispersed or spread out the probabilities are through variance (\(\sigma^2\)) and standard deviation (\(\sigma\)). Variance is the expected value of the squared deviation of a random variable from its mean, a measurement of the spread between numbers in a data set.

In simpler terms, it quantifies how much the outcomes vary from the average outcome. To compute the variance for a discrete random variable, you first need to calculate the expected value of the squares of the outcomes, then subtract the square of the mean from this value, using the formula: \[\sigma^2 = E(x^2) - (\mu)^2\].

The standard deviation is the positive square root of the variance and represents the average distance of each data point from the mean. A high standard deviation indicates that the data points are spread out over a wide range of values. The standard deviation is denoted by \(\sigma\) and provides an intuitive measure of the variability of the distribution. In any context involving probability distributions, while the mean locates the center of the distribution, the variance and standard deviation together describe its spread, which are critical in effective risk assessment and decision-making processes.