Question 3 Suppose the average (mean) weight of all male students is 60 kg and standard deviation is 25 kg . If a sample of 36 male students is selected at random, find the probability that the male students having average weight (i) more than 70 kg (ii) less than 55 kg (iii) between 50 kg and 65 kg

To solve this problem, we can use the Central Limit Theorem (CLT), which states that the sampling distribution of the sample mean will be normally distributed if the sample size is sufficiently large (which it is in this case, with a sample size of 36). Given: - Population mean (\( \mu \)) = 60 kg - Population standard deviation (\( \sigma \)) = 25 kg - Sample size (\( n \)) = 36 First, we need to calculate the standard error of the mean (SEM): \[ \text{Standard Error (SE)} = \frac{\sigma}{\sqrt{n}} = \frac{25}{\sqrt{36}} = \frac{25}{6} \approx 4.17 \text{ kg} \] Now, we can standardize the sample mean using the Z-score formula: \[ Z = \frac{\bar{X} - \mu}{SE} \] where \( \bar{X} \) is the sample mean. ### (i) Probability that the average weight is more than 70 kg 1. Calculate the Z-score for \( \bar{X} = 70 \): \[ Z = \frac{70 - 60}{4.17} \approx \frac{10}{4.17} \approx 2.40 \] 2. Look up the Z-score in the standard normal distribution table or use a calculator to find the probability: \[ P(Z > 2.40) = 1 - P(Z < 2.40) \approx 1 - 0.9918 = 0.0082 \] So, the probability that the average weight is more than 70 kg is approximately **0.0082**. ### (ii) Probability that the average weight is less than 55 kg 1. Calculate the Z-score for \( \bar{X} = 55 \): \[ Z = \frac{55 - 60}{4.17} \approx \frac{-5}{4.17} \approx -1.20 \] 2. Look up the Z-score in the standard normal distribution table: \[ P(Z < -1.20) \approx 0.1151 \] So, the probability that the average weight is less than 55 kg is approximately **0.1151**. ### (iii) Probability that the average weight is between 50 kg and 65 kg 1. Calculate the Z-scores for \( \bar{X} = 50 \) and \( \bar{X} = 65 \): For \( \bar{X} = 50 \): \[ Z = \frac{50 - 60}{4.17} \approx \frac{-10}{4.17} \approx -2.40 \] For \( \bar{X} = 65 \): \[ Z = \frac{65 - 60}{4.17} \approx \frac{5}{4.17} \approx 1.20 \] 2. Find the probabilities: \[ P(Z < -2.40) \approx 0.0082 \] \[ P(Z < 1.20) \approx 0.8849 \] 3. Calculate the probability that the average weight is between 50 kg and 65 kg: \[ P(50 < \bar{X} < 65) = P(Z < 1.20) - P(Z < -2.40) \approx 0.8849 - 0.0082 = 0.8767 \] So, the probability that the average weight is between 50 kg and 65 kg is approximately **0.8767**. ### Summary of Results: - (i) Probability of average weight > 70 kg: **0.0082** - (ii) Probability of average weight < 55 kg: **0.1151** - (iii) Probability of average weight between 50 kg and 65 kg: **0.8767**