Problem 9 - Calculus

Question

Find the mean value \(\mu=\Sigma n p_{n}\) or \(\mu=\int x p(x) d x\). $$ p_{1}=1 / 7, p_{2}=1 / 7, \ldots, p_{7}=1 / 7 $$

Step-by-Step Solution

Answer

The mean value is 4.

Step 1: Understand the Problem

We need to find the mean value using the given discrete probabilities. Since exact values for each outcome aren't provided, assume each outcome is evenly spaced.

Step 2: List Probabilities and Outcomes

We have probabilities for seven outcomes: \(p_1 = \frac{1}{7}, p_2 = \frac{1}{7}, \ldots, p_7 = \frac{1}{7}\). Let's assume outcomes are \(x_1, x_2, \ldots, x_7\).

Step 3: Assign Values to Outcomes

Since specific outcomes are not provided, assume \(x_n = n\) for simplicity, i.e.,\(x_1 = 1, x_2 = 2, \ldots, x_7 = 7\).

Step 4: Calculate Mean Using Discrete Formula

Use the formula \(\mu = \Sigma n p_{n}\). Calculate:\[\mu = 1\cdot\frac{1}{7} + 2\cdot\frac{1}{7} + 3\cdot\frac{1}{7} + 4\cdot\frac{1}{7} + 5\cdot\frac{1}{7} + 6\cdot\frac{1}{7} + 7\cdot\frac{1}{7}\].

Step 5: Simplify and Solve

\[\mu = \frac{1}{7}(1 + 2 + 3 + 4 + 5 + 6 + 7)\]. Calculate the sum inside the brackets: \(1 + 2 + 3 + 4 + 5 + 6 + 7 = 28\).\[\mu = \frac{28}{7} = 4\].

Key Concepts

Discrete Probability

Discrete probability deals with scenarios where outcomes of an experiment are distinct and countable. For example, rolling a die or flipping a coin. Each outcome has a probability, and all these probabilities together must sum up to one. In our original problem, we see this concept in action with discrete probabilities:

• The probabilities are given as \(p_1 = \frac{1}{7}\), \(p_2 = \frac{1}{7}\), ..., \(p_7 = \frac{1}{7}\).
• These probabilities are equally distributed across seven possible outcomes.
• The sum of these probabilities is \(1/7 + 1/7 + \ldots + 1/7 = 1\), satisfying the rule that the total probability in a discrete set must equal one.

Understanding discrete probability helps in calculating the mean value of a set by clearly defining all individual probabilities.

Probability Distribution

A probability distribution is a mathematical function that provides the probabilities of occurrence of different possible outcomes in an experiment. In discrete distributions, like in our example, each possible outcome has an associated probability that defines how likely it is for that outcome to occur. The example given can be considered as a uniform distribution because:

• All the outcomes \(x_1, x_2, \ldots, x_7\) have the same probability, \(\frac{1}{7}\).
• This means that the possibility of any one particular event occurring is the same as any other event, which is characteristic of a uniform distribution.

Such a uniform distribution simplifies the computation of the mean as it allows us to predict that each outcome contributes equally to the overall calculation.

Expected Value

The expected value, often referred to as the "mean" in statistics, is the average value of a random variable over a large number of experiments. To find the expected value \(\mu\) of a discrete distribution, you sum the products of each outcome and its probability, as shown in the formula \(\mu = \Sigma n p_n\). In the solution to our problem:

• We assigned outcomes \(x_1 = 1, ..., x_7 = 7\) to the probabilities provided.
• The calculation \(\mu = 1\cdot\frac{1}{7} + 2\cdot\frac{1}{7} + ... + 7\cdot\frac{1}{7}\) represents this principle.
• After simplifying, the expected value \(\mu = \frac{28}{7} = 4\).

This example underscores how the expected value gives us a central or "mean" understanding, summarizing the entire probability distribution with a single number.