Question

Let the mean and the standard deviation of the probability distribution be be $\mu$ and $\sigma$, respectively. If $\sigma - \mu = 2$, then $\sigma + \mu$ is equal to ________.

Answer 1

Correct Answer: 5

Assuming the total probability equals 1:

\[ \frac{1}{3} + K + \frac{1}{6} + \frac{1}{4} = 1. \]

Solving for K yields:

\[ K = \frac{1}{4}. \]

Step 1: Mean Calculation \(\mu\) The mean \(\mu\) is calculated as:

\[ \mu = \alpha \cdot \frac{1}{3} + 1 \cdot K + 0 \cdot \frac{1}{6} + (-3) \cdot \frac{1}{4}. \]

With substituted values:

\[ \mu = \frac{\alpha}{3} + \frac{1}{4} \cdot 1 + 0 - \frac{3}{4} = \frac{\alpha}{3} - \frac{1}{2}. \]

Step 2: Variance Calculation \(\sigma^2\) The variance \(\sigma^2\) is calculated as:

\[ \sigma^2 = \left(\alpha^2 \cdot \frac{1}{3} + 1^2 \cdot K + 0^2 \cdot \frac{1}{6} + (-3)^2 \cdot \frac{1}{4} \right) - \mu^2. \]

After substituting values:

\[ \sigma^2 = \frac{\alpha^2}{3} + \frac{1}{4} + \frac{9}{4} - \left(\frac{\alpha}{3} - \frac{1}{2}\right)^2. \]

Simplification results in:

\[ \sigma^2 = \frac{\alpha^2}{3} + \frac{1}{4} + \frac{9}{4} - \left(\frac{\alpha^2}{9} - \alpha + \frac{1}{4}\right). \]

Further simplification yields:

\[ \sigma^2 = \frac{2\alpha^2}{9} + \alpha + \frac{9}{4}. \]

Step 3: Condition \(\sigma - \mu = 2\) Given the condition:

\[ \sigma = \mu + 2. \]

Substituting this condition and solving for \(\alpha\):

\[ \sigma^2 = (\mu + 2)^2. \]

Equating and simplifying leads to:

\[ \alpha = 6 \quad \text{(rejecting \(\alpha = 0\))}. \]

Step 4: Calculation of \(\sigma + \mu\) Substituting \(\alpha = 6\) into the expressions for \(\mu\) and \(\sigma\):

\[ \sigma + \mu = 2\mu + 2 = 5. \]

The result is $5$.