1) A fair six-sided die is thrown 20 times and the number of sixes, $$ X $$, fair six-sided die is thrown 20 times and the number of odd-numbered scores, $$ Y $$, is recorded. Find the mean and standard deviation of $$ X+Y $$. [5] and $$ Y $$ respectively, where $$ X \sim N\left(5.1,0.2^{2}\right) $$ and $$ Y \sim N\left(2.5\right. $$, $$ \left.0.12^{2}\right) $$. Find the probability that 2) The masses, in grams, of large bags of sugar and small bags of sugar are denoted by $$ X $$ the mass of a randomly chosen large bag is less than twice the mass of a randomly chosen small bag. [5]

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UpStudy AI · Accepted Answer

### Problem 1

**Given:**
- A fair six-sided die is thrown 20 times.
- $$ X $$ = Number of sixes.
- $$ Y $$ = Number of odd-numbered scores.

**Objective:** Find the **mean** and **standard deviation** of $$ X + Y $$.

**Solution:**

1. **Determine Distributions:**
   - **For $$ X $$:** 
     - Probability of rolling a six, $$ p_X = \frac{1}{6} $$.
     - $$ X $$ follows a Binomial distribution: $$ X \sim \text{Binomial}(n=20, p=\frac{1}{6}) $$.
   - **For $$ Y $$:**
     - Probability of rolling an odd number (1, 3, 5), $$ p_Y = \frac{3}{6} = \frac{1}{2} $$.
     - $$ Y $$ follows a Binomial distribution: $$ Y \sim \text{Binomial}(n=20, p=\frac{1}{2}) $$.

2. **Calculate Expectations:**
   $$
   E(X) = n \cdot p_X = 20 \times \frac{1}{6} = \frac{20}{6} \approx 3.333
   $$
   $$
   E(Y) = n \cdot p_Y = 20 \times \frac{1}{2} = 10
   $$
   $$
   E(X + Y) = E(X) + E(Y) = 3.333 + 10 = 13.333
   $$

3. **Calculate Variances:**
   $$
   \text{Var}(X) = n \cdot p_X \cdot (1 - p_X) = 20 \times \frac{1}{6} \times \frac{5}{6} = \frac{100}{36} \approx 2.778
   $$
   $$
   \text{Var}(Y) = n \cdot p_Y \cdot (1 - p_Y) = 20 \times \frac{1}{2} \times \frac{1}{2} = 5
   $$

4. **Calculate Covariance:**
   - Since rolling a six and rolling an odd number are mutually exclusive:
     $$
     \text{Cov}(X, Y) = -\frac{n \cdot p_X \cdot p_Y}{1} = -\frac{20 \times \frac{1}{6} \times \frac{1}{2}}{1} = -\frac{20}{12} = -\frac{5}{3} \approx -1.667
     $$

5. **Calculate Variance of $$ X + Y $$:**
   $$
   \text{Var}(X + Y) = \text{Var}(X) + \text{Var}(Y) + 2 \cdot \text{Cov}(X, Y) = 2.778 + 5 + 2(-1.667) = 4.444
   $$

6. **Calculate Standard Deviation:**
   $$
   \text{SD}(X + Y) = \sqrt{\text{Var}(X + Y)} = \sqrt{4.444} \approx 2.108
   $$

**Answer:**
- **Mean of $$ X + Y $$:** $$ \frac{40}{3} $$ or approximately **13.33**.
- **Standard Deviation of $$ X + Y $$:** $$ \frac{\sqrt{40}}{3} $$ or approximately **2.11**.

---

### Problem 2

**Given:**
- Mass of large bags, $$ X \sim N(5.1, 0.2^2) $$.
- Mass of small bags, $$ Y \sim N(2.5, 0.12^2) $$.

**Objective:** Find the probability that a randomly chosen large bag is **less than twice** the mass of a randomly chosen small bag, i.e., $$ P(X < 2Y) $$.

**Solution:**

1. **Define the Inequality:**
   $$
   P(X < 2Y) \implies P(X - 2Y < 0)
   $$

2. **Determine Distribution of $$ Z = X - 2Y $$:**
   - **Mean of $$ Z $$:**
     $$
     E(Z) = E(X) - 2E(Y) = 5.1 - 2(2.5) = 5.1 - 5 = 0.1
     $$
   - **Variance of $$ Z $$:**
     $$
     \text{Var}(Z) = \text{Var}(X) + 4\text{Var}(Y) = (0.2)^2 + 4(0.12)^2 = 0.04 + 4(0.0144) = 0.04 + 0.0576 = 0.0976
     $$
   - **Standard Deviation of $$ Z $$:**
     $$
     \text{SD}(Z) = \sqrt{0.0976} \approx 0.312
     $$

3. **Standardize the Variable:**
   $$
   Z = X - 2Y \sim N(0.1, 0.0976)
   $$
   $$
   \text{Compute } z\text{-score for } Z = 0:
   $$
   $$
   z = \frac{0 - 0.1}{0.312} \approx -0.3205
   $$

4. **Find the Probability:**
   $$
   P(Z < 0) = P\left(Z_{\text{standard}} < -0.3205\right)
   $$
   - Using standard normal distribution tables or a calculator:
     $$
     P(Z < -0.3205) \approx 0.375 \text{ or } 37.5\%
     $$

**Answer:**
There is approximately a **37.5%** probability that a randomly chosen large bag is less than twice the mass of a randomly chosen small bag.
**Problem 1:** - **Mean of $$ X + Y $$:** $$ \frac{40}{3} $$ or approximately **13.33** - **Standard Deviation of $$ X + Y $$:** $$ \frac{\sqrt{40}}{3} $$ or approximately **2.11** **Problem 2:** - **Probability that a large bag is less than twice the mass of a small bag:** **37.5%**

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