Questions: Consider the probability distribution shown for the random variable x found below. Complete part a through f. x 1 2 4 10 p(x) 0.4 0.2 0.2 0.2 a. Find μ=E(x). μ=3.6 (Round to the nearest tenth as needed.) b. Find σ²=E[(x-μ)²]. σ²=11.44 (Round to the nearest hundredth as needed.) c. Find σ. σ=3.3823 (Round to four decimal places as needed.) d. Interpret the value you obtained for μ. Choose the correct answer below. A. The average value of x over many trials is equal to μ. B. The average value of x over many trials will be more than μ. C. The average value of x over many trials will be less than μ.

Consider the probability distribution shown for the random variable x found below. Complete part a through f. x 1 2 4 10 p(x) 0.4 0.2 0.2 0.2 a. Find μ=E(x). μ=3.6 (Round to the nearest tenth as needed.) b. Find σ²=E[(x-μ)²]. σ²=11.44 (Round to the nearest hundredth as needed.) c. Find σ. σ=3.3823 (Round to four decimal places as needed.) d. Interpret the value you obtained for μ. Choose the correct answer below. A. The average value of x over many trials is equal to μ. B. The average value of x over many trials will be more than μ. C. The average value of x over many trials will be less than μ.
Transcript text: Consider the probability distribution shown for the random variable x found below. Complete part a through f. \begin{tabular}{lllll} $\mathbf{x}$ & 1 & 2 & 4 & 10 \\ $\mathbf{p}(\mathbf{x})$ & 0.4 & 0.2 & 0.2 & 0.2 \end{tabular} a. Find $\mu=E(x)$. $\mu=3.6$ (Round to the nearest tenth as needed.) b. Find $\sigma^{2}=E\left[(x-\mu)^{2}\right]$. $\sigma^{2}=11.44$ (Round to the nearest hundredth as needed.) c. Find $\sigma$. $\sigma=3.3823$ (Round to four decimal places as needed.) d. Interpret the value you obtained for $\mu$. Choose the correct answer below. A. The average value of $x$ over many trials is equal to $\mu$. B. The average value of $x$ over many trials will be more than $\mu$. C. The average value of $x$ over many trials will be less than $\mu$.
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Solution

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Solution Steps

Step 1: Calculate the Mean \( \mu = E(x) \)

The mean of the random variable \( x \) is calculated using the formula:

\[ \mu = E(x) = \sum_{i} x_i \cdot p(x_i) \]

Substituting the values:

\[ \mu = 1 \times 0.4 + 2 \times 0.2 + 4 \times 0.2 + 10 \times 0.2 = 3.6 \]

Step 2: Calculate the Variance \( \sigma^2 = E[(x - \mu)^2] \)

The variance is calculated using the formula:

\[ \sigma^2 = E[(x - \mu)^2] = \sum_{i} (x_i - \mu)^2 \cdot p(x_i) \]

Substituting the values:

\[ \sigma^2 = (1 - 3.6)^2 \times 0.4 + (2 - 3.6)^2 \times 0.2 + (4 - 3.6)^2 \times 0.2 + (10 - 3.6)^2 \times 0.2 = 11.44 \]

Step 3: Calculate the Standard Deviation \( \sigma \)

The standard deviation is the square root of the variance:

\[ \sigma = \sqrt{\sigma^2} = \sqrt{11.44} \approx 3.3823 \]

Step 4: Interpret the Mean \( \mu \)

The value obtained for \( \mu \) represents the average value of \( x \) over many trials. Therefore, the correct interpretation is:

A. The average value of \( x \) over many trials is equal to \( \mu \).

Final Answer

\[ \boxed{\mu = 3.6, \sigma^2 = 11.44, \sigma = 3.3823, \text{Interpretation: A}} \]

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