Final answer:
For a normal population with mean 8 and standard deviation 6, approximately 97.72% of the population is less than 20, and there is a 69.15% probability that a randomly chosen value will be greater than 5.
Step-by-step explanation:
To find the proportion of the population less than 20 for a normal population with mean μ = 8 and standard deviation σ = 6, we would use the Z-score formula: Z = (X - μ) / σ. Plugging in the values, we get Z = (20 - 8) / 6 = 2. We then use the standard normal distribution table to find the proportion corresponding to a Z-score of 2, which is approximately 0.9772. Hence, about 97.72% of the population is less than 20.
For part (b), to find the probability that a randomly chosen value will be greater than 5, we calculate the Z-score for X = 5, which gives us Z = (5 - 8) / 6 = -0.5. Consulting the standard normal distribution table, we find the proportion corresponding to Z = -0.5 is approximately 0.3085. Since we want the probability that a value is greater, we subtract this from 1, yielding 1 - 0.3085 = 0.6915. Thus, there is a 69.15% chance that a randomly chosen value will be greater than 5.