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

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

Step 1: Write down the given information.
The given probability distribution is:

X	α	1	0	−3
P(X)	1/3	K	1/6	1/4

We are told that the mean = μ and standard deviation = σ, and that σ − μ = 2.
We must find σ + μ.

Step 2: Find K.
Since total probability = 1:
1/3 + K + 1/6 + 1/4 = 1.

Take LCM = 12:
(4 + 12K + 2 + 3)/12 = 1 ⇒ 9 + 12K = 12 ⇒ 12K = 3 ⇒ K = 1/4.

Step 3: Compute the mean μ.
μ = Σ[X·P(X)] = α(1/3) + 1(1/4) + 0(1/6) + (−3)(1/4).
= α/3 + 1/4 − 3/4 = α/3 − 1/2.

Step 4: Compute E(X²).
E(X²) = α²(1/3) + (1)²(1/4) + 0²(1/6) + (−3)²(1/4).
= α²/3 + 1/4 + 9/4 = α²/3 + 10/4 = α²/3 + 5/2.

Step 5: Compute variance σ².
σ² = E(X²) − μ² = (α²/3 + 5/2) − (α/3 − 1/2)².
Expand μ²: (α/3 − 1/2)² = α²/9 − α/3 + 1/4.
So, σ² = α²/3 + 5/2 − α²/9 + α/3 − 1/4.

Simplify:
α²(1/3 − 1/9) + α/3 + (5/2 − 1/4) = (2α²/9) + (α/3) + (9/4).

Hence σ² = (2α²/9) + (α/3) + (9/4).

Step 6: Given relation σ − μ = 2.
Let’s express μ and σ explicitly:
μ = α/3 − 1/2.
σ = √[(2α²/9) + (α/3) + (9/4)].

Substitute in σ − μ = 2:
√[(2α²/9) + (α/3) + (9/4)] − (α/3 − 1/2) = 2.
⇒ √[(2α²/9) + (α/3) + (9/4)] = 2 + α/3 − 1/2 = (α/3) + 3/2.

Step 7: Square both sides.
(2α²/9) + (α/3) + (9/4) = (α/3 + 3/2)² = α²/9 + α + 9/4.

Simplify:
(2α²/9 − α²/9) + (α/3 − α) + (9/4 − 9/4) = 0.
⇒ α²/9 − (2α/3) = 0.
⇒ α(α/9 − 2/3) = 0 ⇒ α(α − 6) = 0.
Since α ≠ 0, α = 6.

Step 8: Compute μ and σ.
μ = α/3 − 1/2 = 6/3 − 1/2 = 2 − 0.5 = 1.5.
σ − μ = 2 ⇒ σ = 3.5.

Step 9: Find σ + μ.
σ + μ = 3.5 + 1.5 = 5.

Final Answer: 5