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

Given that the total probability sums to 1:

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

Simplifying:

$K = \frac{1}{4}.$

Step 1: Calculate the Mean $\mu$ The mean $\mu$ is given by:

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

Substituting the values:

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

Step 2: Calculate the Variance $\sigma^2$ The variance $\sigma^2$ is given by:

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

Substituting the values:

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

Simplifying:

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

Further simplification gives:

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

Step 3: Given Condition $\sigma - \mu = 2$ Given:

$\sigma = \mu + 2.$

Substituting this condition and solving for $\alpha$:

$\sigma^2 = (\mu + 2)^2.$

Equating and simplifying:

$\alpha = 6 \quad \text{(since $\alpha = 0$ is rejected)}.$

Step 4: Calculate $\sigma + \mu$ Substitute $\alpha = 6$ into the expressions for $\mu$ and $\sigma$:

$\sigma + \mu = 2\mu + 2 = 5.$

Therefore, the correct answer is $5$.

Answer 2

Given that the total probability sums to 1:

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

Simplifying:

$K = \frac{1}{4}.$

Step 1: Calculate the Mean $\mu$ The mean $\mu$ is given by:

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

Substituting the values:

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

Step 2: Calculate the Variance $\sigma^2$ The variance $\sigma^2$ is given by:

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

Substituting the values:

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

Simplifying:

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

Further simplification gives:

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

Step 3: Given Condition $\sigma - \mu = 2$ Given:

$\sigma = \mu + 2.$

Substituting this condition and solving for $\alpha$:

$\sigma^2 = (\mu + 2)^2.$

Equating and simplifying:

$\alpha = 6 \quad \text{(since $\alpha = 0$ is rejected)}.$

Step 4: Calculate $\sigma + \mu$ Substitute $\alpha = 6$ into the expressions for $\mu$ and $\sigma$:

$\sigma + \mu = 2\mu + 2 = 5.$

Therefore, the correct answer is $5$.