II. Find the mean, variance and the standard deviation of the probability distribution of
the random variable X if for .

Ask by Carter Robbins. in the Philippines

Jan 02,2025

Question

II. Find the mean, variance and the standard deviation of the probability distribution of the random variable X if $$ \frac{1}{10} $$ for $$ \mathrm{X}=1,2,3, \ldots, 10 $$.

UpStudy AI · Accepted Answer

To find the **mean**, **variance**, and **standard deviation** of the probability distribution for the random variable $$ X $$, where $$ X $$ can take integer values from 1 to 10, each with a probability of $$ \frac{1}{10} $$, follow these steps:

### Given:
- **Random Variable $$ X $$**: Takes values $$ 1, 2, 3, \ldots, 10 $$
- **Probability $$ P(X = x) $$**: $$ \frac{1}{10} $$ for each $$ x $$

This is a **uniform discrete distribution** because each outcome has an equal probability.

---

### 1. Mean (Expected Value), $$ E[X] $$

The **mean** of a discrete random variable is calculated as:
$$
E[X] = \sum_{x=1}^{10} x \cdot P(X = x)
$$

Given that $$ P(X = x) = \frac{1}{10} $$ for all $$ x $$:
$$
E[X] = \frac{1}{10} \sum_{x=1}^{10} x
$$
$$
\sum_{x=1}^{10} x = \frac{10 \cdot (10 + 1)}{2} = 55
$$
$$
E[X] = \frac{55}{10} = 5.5
$$

**Mean (Expected Value):** $$ \mathbf{5.5} $$

---

### 2. Variance, $$ \text{Var}(X) $$

The **variance** measures the spread of the random variable around the mean and is calculated as:
$$
\text{Var}(X) = E[X^2] - (E[X])^2
$$

First, compute $$ E[X^2] $$:
$$
E[X^2] = \sum_{x=1}^{10} x^2 \cdot P(X = x)
$$
$$
E[X^2] = \frac{1}{10} \sum_{x=1}^{10} x^2
$$
$$
\sum_{x=1}^{10} x^2 = \frac{10 \cdot 11 \cdot 21}{6} = 385
$$
$$
E[X^2] = \frac{385}{10} = 38.5
$$

Now, calculate the variance:
$$
\text{Var}(X) = 38.5 - (5.5)^2 = 38.5 - 30.25 = 8.25
$$

**Variance:** $$ \mathbf{8.25} $$

---

### 3. Standard Deviation, $$ \sigma_X $$

The **standard deviation** is the square root of the variance:
$$
\sigma_X = \sqrt{\text{Var}(X)} = \sqrt{8.25} \approx 2.872
$$

**Standard Deviation:** $$ \mathbf{\approx 2.872} $$

---

### **Summary**
- **Mean (Expected Value):** $$ 5.5 $$
- **Variance:** $$ 8.25 $$
- **Standard Deviation:** $$ \approx 2.872 $$
**Mean:** 5.5 **Variance:** 8.25 **Standard Deviation:** ≈2.872

II. Find the mean, variance and the standard deviation of the probability distribution of
the random variable X if for .

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