Finding Mean (μ) and Standard Deviation (σ) from a Probability Distribution

Math Problem Statement

Find

muμ

mu equals Summation from nothing to nothing left bracket x times Upper P left parenthesis x right parenthesis right bracketμ=∑ [x•P(x)].

Then, find

sigmaσ

sigma squared equals Summation from nothing to nothing left bracket x squared times Upper P left parenthesis x right parenthesis right bracket minus mu squaredσ2=∑x2•P(x)−μ2.

P(x)

0.27070.2707

0.40430.4043

0.24150.2415

0.07210.0721

0.01080.0108

0.00060.0006

Solution

Step 1: Finding $\mu$

The given formula for $\mu$ is:

$\mu = \sum [x \cdot P(x)]$

This means we need to multiply each value of $x$ by the corresponding probability $P(x)$ , and then sum the results.

The values of $x$ and $P(x)$ are provided as follows:

\hline x & P(x) \\ \hline 0 & 0.2707 \\ 1 & 0.4043 \\ 2 & 0.2415 \\ 3 & 0.0721 \\ 4 & 0.0108 \\ 5 & 0.0006 \\ \hline \end{array}$$ Now, let's calculate $$\mu$$: $$\mu = (0 \cdot 0.2707) + (1 \cdot 0.4043) + (2 \cdot 0.2415) + (3 \cdot 0.0721) + (4 \cdot 0.0108) + (5 \cdot 0.0006)$$ I'll calculate this value. $$\mu = 0 + 0.4043 + 0.4830 + 0.2163 + 0.0432 + 0.0030$$ $$\mu = 1.1498$$ ### Step 2: Finding $$\sigma^2$$ and $$\sigma$$ The given formula for $$\sigma^2$$ is: $$\sigma^2 = \sum [x^2 \cdot P(x)] - \mu^2$$ First, we need to calculate $$\sum [x^2 \cdot P(x)]$$. We'll square each $$x$$ value, multiply by $$P(x)$$, and sum the results. $$\sum [x^2 \cdot P(x)] = (0^2 \cdot 0.2707) + (1^2 \cdot 0.4043) + (2^2 \cdot 0.2415) + (3^2 \cdot 0.0721) + (4^2 \cdot 0.0108) + (5^2 \cdot 0.0006)$$ I'll calculate this now. $$\sum [x^2 \cdot P(x)] = (0) + (1 \cdot 0.4043) + (4 \cdot 0.2415) + (9 \cdot 0.0721) + (16 \cdot 0.0108) + (25 \cdot 0.0006)$$ $$\sum [x^2 \cdot P(x)] = 0 + 0.4043 + 0.9660 + 0.6489 + 0.1728 + 0.015$$ $$\sum [x^2 \cdot P(x)] = 2.207$$ Now, using the formula for $$\sigma^2$$: $$\sigma^2 = 2.207 - (1.1498)^2$$ $$\sigma^2 = 2.207 - 1.321$$ $$\sigma^2 = 0.886$$ Finally, to find $$\sigma$$, we take the square root of $$\sigma^2$$: $$\sigma = \sqrt{0.886} \approx 0.941$$ ### Final Results: - $$\mu = 1.1498$$ - $$\sigma^2 = 0.886$$ - $$\sigma \approx 0.941$$ --- Would you like any details or further clarifications? Here are some related questions you can explore: 1. How does changing probabilities affect the mean $$\mu$$? 2. What happens to $$\sigma$$ if a higher $$P(x)$$ value is added for larger $$x$$? 3. What is the significance of variance $$\sigma^2$$ in probability distributions? 4. How would the mean and variance change if the values of $$P(x)$$ were more spread out? 5. Can you derive $$\mu$$ and $$\sigma^2$$ for other discrete distributions? **Tip:** Variance $$\sigma^2$$ measures how much the values of a random variable differ from the mean, while $$\sigma$$ gives you the average deviation in the original units.

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Math Problem Analysis

Mathematical Concepts

Probability
Statistics
Mean
Variance
Standard Deviation

Formulas

μ = ∑[x•P(x)]
σ^2 = ∑[x^2•P(x)] − μ^2
σ = √(σ^2)

Theorems

Law of Large Numbers (related to the expectation of random variables)
Variance and Standard Deviation Theorems

Suitable Grade Level

Grades 11-12, College

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