3.A farmer found that the weights of Musang King durian on his farm have a normal distribution with mean 2.15 kg and standard deviation 0.55 kg . (a) 250 Musang King durians are chosen at random. Estimate the number of Musang King durians which have a weight between 1.95 kg and \( 2,25 \mathrm{~kg} \). [ 3 marks] (b) Musang King durian will be rejected to be exported if the weight is less than 1.80 kg . If a basket has 200 Musang King durians, estimate the number of durians that will not be exported. [ 3 marks] (c) What is the probability that more than 4 Musang King durians will not be exported is 6 durians are chosen at random. [ 3 marks]

To tackle the problem, let's dive into each part systematically. (a) First, we need to find the z-scores for the weights 1.95 kg and 2.25 kg using the mean (μ = 2.15 kg) and standard deviation (σ = 0.55 kg). For 1.95 kg: \[ z = \frac{(X - \mu)}{\sigma} = \frac{(1.95 - 2.15)}{0.55} \approx -0.36 \] For 2.25 kg: \[ z = \frac{(2.25 - 2.15)}{0.55} \approx 0.18 \] Next, we can look up these z-scores in the standard normal distribution table or use a calculator. Using a standard normal distribution table: - The area to the left of z = -0.36 is approximately 0.3594. - The area to the left of z = 0.18 is approximately 0.5714. To find the area between these two z-scores: \[ 0.5714 - 0.3594 = 0.2120 \] Now, to estimate the number of Musang King durians with weights between 1.95 kg and 2.25 kg: \[ 250 \times 0.2120 \approx 53 \] So, approximately 53 Musang King durians will weigh between 1.95 kg and 2.25 kg. (b) To estimate the number of durians that will not be exported, we need to calculate the z-score for the weight of 1.80 kg. For 1.80 kg: \[ z = \frac{(1.80 - 2.15)}{0.55} \approx -0.64 \] Looking up this z-score, the area to the left of z = -0.64 is around 0.2611. This value represents the proportion of durians that weigh less than 1.80 kg. Now, to find the expected number within a basket of 200 durians: \[ 200 \times 0.2611 \approx 52.22 \] We round this to about 52 Musang King durians that will not be exported. (c) We can model the number of durians not exported as a binomial distribution, where n = 6 (the number of durians chosen) and p = 0.2611 (the probability of not being exported). We need to find the probability that more than 4 durians will not be exported: This boils down to finding P(X > 4), where X follows the Binomial distribution, X ~ B(6, 0.2611). Calculating this directly can be a bit more involved. We can instead find P(X ≤ 4) and subtract it from 1: To find P(X ≤ 4), we can consider: - P(X = 5) and P(X = 6). Using the binomial formula: \[ P(X = k) = \binom{n}{k} p^k (1-p)^{n-k} \] For k=5: \[ P(X = 5) = \binom{6}{5} (0.2611)^5 (0.7389)^1 \approx 6 \times 0.0027 \times 0.7389 \approx 0.0119 \] For k=6: \[ P(X = 6) = \binom{6}{6} (0.2611)^6 \approx 1 \times 0.0007 \approx 0.0007 \] Thus: \[ P(X \leq 4) = 1 - (0.0119 + 0.0007) \approx 0.9884 \] Now, \[ P(X > 4) = 1 - P(X ≤ 4) \approx 1 - 0.9884 = 0.0116\] Therefore, the probability that more than 4 Musang King durians will not be exported is approximately 0.0116, or 1.16%.