2) The masses, in grams, of large bags of sugar and small bags of sugar are denoted by \( X \) and \( Y \) respectively, where \( X \sim N\left(5.1,0.2^{2}\right) \) and \( Y \sim N\left(2.5,0.12^{2}\right) \). Find the probability that the mass of a randomly chosen large bag is less than twice the mass of a randomly chosen small bag. [5] 3) The masses, in kilograms, of cartons of sugar and cartons of flour have the distributions N(78.8, \( \left.12.6^{2}\right) \) and \( N\left(62.0,10.0^{2}\right) \) respectively. a) The standard load for a certain crane is 8 cartons of sugar and 3 cartons of flour. The maximum load that can be carried safely by the crane is 900 kg. Stating a necessary assumption, find the percentage of standard loads that will exceed the maximum safe load. [5]

### Problem 2 **Given:** - Mass of a large bag of sugar, \( X \sim N(5.1\, \text{g},\, 0.2^2\, \text{g}^2) \) - Mass of a small bag of sugar, \( Y \sim N(2.5\, \text{g},\, 0.12^2\, \text{g}^2) \) **Objective:** Find the probability that the mass of 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 Event:** \[ P(X < 2Y) = P(X - 2Y < 0) \] 2. **Determine the Distribution of \( Z = X - 2Y \):** Since \( X \) and \( Y \) are independent normal random variables, \( Z \) will also follow a normal distribution. - **Mean of \( Z \):** \[ \mu_Z = E[X] - 2E[Y] = 5.1 - 2 \times 2.5 = 5.1 - 5.0 = 0.1\, \text{g} \] - **Variance of \( Z \):** \[ \sigma_Z^2 = \text{Var}(X) + 4\,\text{Var}(Y) = 0.2^2 + 4 \times 0.12^2 = 0.04 + 4 \times 0.0144 = 0.04 + 0.0576 = 0.0976\, \text{g}^2 \] - **Standard Deviation of \( Z \):** \[ \sigma_Z = \sqrt{0.0976} \approx 0.312\, \text{g} \] Thus, \( Z = X - 2Y \sim N(0.1\, \text{g},\, 0.312^2\, \text{g}^2) \). 3. **Standardize the Variable to Find the Probability:** We need to find \( P(Z < 0) \). - **Calculate the Z-score:** \[ Z_{\text{score}} = \frac{0 - \mu_Z}{\sigma_Z} = \frac{0 - 0.1}{0.312} \approx -0.3205 \] - **Find the Probability Using Standard Normal Distribution:** \[ P(Z < 0) = P\left(Z < -0.3205\right) = \Phi(-0.3205) \] Where \( \Phi \) is the cumulative distribution function (CDF) of the standard normal distribution. Using standard normal distribution tables or a calculator: \[ \Phi(-0.3205) \approx 0.3745 \text{ or } 37.45\% \] **Conclusion:** The probability that the mass of a randomly chosen large bag of sugar is less than twice the mass of a randomly chosen small bag is approximately **37.5%**.