Question 1174436
Absolutely! Let's break down the calculation of the variance of X step by step.

**1. Define the Random Variables and Probabilities**

* **Coin Toss:**
    * Let H be the event of getting heads, with P(H) = 1/3.
    * Let T be the event of getting tails, with P(T) = 2/3.
* **Choice of Numbers:**
    * If heads (H), we choose a number from {1, 2, ..., 10} uniformly. Let's call this random variable Y.
    * If tails (T), we choose a number from {-1, -2, ..., -6} uniformly. Let's call this random variable Z.
    * X is the random variable representing the final outcome.

**2. Calculate the Expected Values**

* **E[Y] (Expected value if heads):**
    * E[Y] = (1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 + 10) / 10 = 55 / 10 = 5.5
* **E[Z] (Expected value if tails):**
    * E[Z] = (-1 - 2 - 3 - 4 - 5 - 6) / 6 = -21 / 6 = -3.5
* **E[X] (Overall expected value):**
    * E[X] = E[X|H]P(H) + E[X|T]P(T) = E[Y]P(H) + E[Z]P(T)
    * E[X] = (5.5)(1/3) + (-3.5)(2/3) = 5.5/3 - 7/3 = -1.5/3 = -0.5

**3. Calculate the Second Moments**

* **E[Y²] (Second moment if heads):**
    * E[Y²] = (1² + 2² + ... + 10²) / 10 = (385) / 10 = 38.5
* **E[Z²] (Second moment if tails):**
    * E[Z²] = ((-1)² + (-2)² + ... + (-6)²) / 6 = (91) / 6 = 15.1667 (approximately)
* **E[X²] (Overall second moment):**
    * E[X²] = E[Y²]P(H) + E[Z²]P(T)
    * E[X²] = (38.5)(1/3) + (91/6)(2/3) = 38.5/3 + 91/9 = 12.8333 + 10.1111 = 22.9444 (approximately)

**4. Calculate the Variance**

* **Var(X) = E[X²] - (E[X])²**
* Var(X) = 22.9444 - (-0.5)² = 22.9444 - 0.25 = 22.6944

**5. Round to Two Decimal Places**

* Var(X) ≈ 22.69

Therefore, the variance of X is approximately 22.69.