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]

If you’re rolling a fair six-sided die, you have a probability of \( \frac{1}{6} \) for rolling a six, and since you're throwing it 20 times, the expected number of sixes \( E(X) \) would be \( \frac{1}{6} \times 20 = \frac{20}{6} \approx 3.33 \). For odd-numbered outcomes, you can get 1, 3, or 5, making it a total of 3 odd outcomes out of 6. Hence, the average number of odd scores in 20 throws is \( \frac{3}{6} \times 20 = 10 \). To find the mean and standard deviation of \( X + Y \), you add the means: \( E(X) + E(Y) = 3.33 + 10 = 13.33 \). The variances sum up too, so \( Var(X) + Var(Y) \) gives you the total variance. If \( X \sim N(5.1, 0.2^2) \) and \( Y \sim N(2.5, 0.12^2) \), calculate the variances \( 0.2^2 = 0.04 \) and \( 0.12^2 = 0.0144 \), yielding \( Var(X + Y) = 0.04 + 0.0144 = 0.0544 \). Therefore, you'd have \( X + Y \sim N(13.33, 0.0544) \). Regarding the comparison of mass between the large bags \( X \) and small bags \( Y \) of sugar, you're trying to find \( P(X < 2Y) \). This involves combining the distributions with the method of transformation. Calculate \( 2Y \) which has the mean \( 2 \times E(Y) \) and double the variance. Generally, you can standardize it using the z-scores to find this probability, and it might yield a bit of mathematical exploration along the way!