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

Ask by Frank Page. 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 determine the **mean**, **variance**, and **standard deviation** of the random variable $$ X $$ with the given probability distribution, follow these steps:

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

This is a **uniform discrete distribution**, where each outcome is equally likely.

---

### 1. Mean ($$ \mu $$)
The **mean** of a discrete uniform distribution is calculated as:
$$
\mu = \frac{n + 1}{2}
$$
where $$ n $$ is the highest integer value of $$ X $$.

For $$ n = 10 $$:
$$
\mu = \frac{10 + 1}{2} = \frac{11}{2} = 5.5
$$

---

### 2. Variance ($$ \sigma^2 $$)
The **variance** of a discrete uniform distribution is given by:
$$
\sigma^2 = \frac{n^2 - 1}{12}
$$

For $$ n = 10 $$:
$$
\sigma^2 = \frac{10^2 - 1}{12} = \frac{100 - 1}{12} = \frac{99}{12} = 8.25
$$

---

### 3. Standard Deviation ($$ \sigma $$)
The **standard deviation** is the square root of the variance:
$$
\sigma = \sqrt{\sigma^2} = \sqrt{8.25} \approx 2.87298
$$

---

### **Summary:**
- **Mean ($$ \mu $$)**: 5.5
- **Variance ($$ \sigma^2 $$)**: 8.25
- **Standard Deviation ($$ \sigma $$)**: ≈ 2.873
- **Mean**: 5.5 - **Variance**: 8.25 - **Standard Deviation**: ≈ 2.873

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

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