Answer :
The question is asking for the lower bound of the 95% two tailed Confidence interval of the normally distributed population.
95% C.I. is given by 200 + or - 1.96(25) = 200 + or - 49 = (151, 249)
Therefore, the minimum weight of the middle 95% of players is 151 pounds.
95% C.I. is given by 200 + or - 1.96(25) = 200 + or - 49 = (151, 249)
Therefore, the minimum weight of the middle 95% of players is 151 pounds.
Option B is the answer. The minimum weight of the middle 95% of football players, with a mean of 200 pounds and a standard deviation of 25 pounds, is 151 pounds.
To find the minimum weight of the middle 95% of football players, given the weights are normally distributed with a mean of 200 pounds and a standard deviation of 25 pounds, we'll use the concept of z-scores and the 68-95-99.7 rule (also known as the empirical rule). The middle 95% corresponds to the weights between the 2.5th percentile and the 97.5th percentile. Since 95% is within two standard deviations of the mean in a normal distribution, we calculate:
Lower Z: Z = (X - μ) / σ = (X - 200) / 25
For the lower bound of middle 95%, Z is approximately -1.96.
X = μ + Z * σ = 200 + (-1.96 * 25) = 151
Therefore, the minimum weight of the middle 95% of the players is (Option B) 151 pounds.
