Answer:
(a) To find the percentage of students with body weight between 104 and 136 lb, we first need to convert these weights to z-scores. The z-score is calculated as z = (x - μ) / σ, where x is the value, μ is the mean, and σ is the standard deviation.
For 104 lb, the z-score is z1 = (104 - 120) / 16 = -1. For 136 lb, the z-score is z2 = (136 - 120) / 16 = 1.
Using standard normal distribution tables, we find that the percentage of data within -1 and 1 standard deviations from the mean is approximately 68.26%.
(b) The weight of 120 lb is the mean of the distribution. In a normal distribution, 50% of the data is below the mean and 50% is above. Therefore, the probability for a randomly selected student to have the weight below 120 lb is 0.5 or 50%.
(c) The 80th percentile of a standard normal distribution corresponds to a z-score of approximately 0.84. To convert this z-score back to a weight, we use the formula x = μ + zσ.
x = 120 + 0.84 * 16 = 120 + 13.44 = 133.44 lb.
Therefore, the 80th percentile of the body weight of this population is approximately 133.44 lb.
Explanation: