Find

muμ

if

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

if

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.

x

0

1

2

3

4

5

​P(x)

0.27070.2707

0.40430.4043

0.24150.2415

0.07210.0721

0.01080.0108

0.00060.0006

### 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:

\[
\begin{array}{|c|c|}
\hline
x & P(x) \\
\hline
0 & 0.2707 \\
1 & 0.4043 \\
2 & 0.2415 \\
3 & 0.0721 \\
4 & 0.0108 \\
5 & 0.0006 \\
\hline
\end{array}
\]

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.

Find μ if μ = ∑[x•P(x)]. Then, find σ if σ^2 = ∑[x^2•P(x)]−μ^2. Given x: [0, 1, 2, 3, 4, 5] and P(x): [0.2707, 0.4043, 0.2415, 0.0721, 0.0108, 0.0006]

Learn how to calculate the mean (μ) and standard deviation (σ) of a probability distribution using summation formulas. Given x values [0, 1, 2, 3, 4, 5] and corresponding probabilities P(x) [0.2707, 0.4043, 0.2415, 0.0721, 0.0108, 0.0006], find μ and σ step-by-step.

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