Final answer:
The probability a wide receiver gained 455 yards or less is determined using the mean and standard deviation of yards gained, finding the z-score, and then the corresponding cumulative probability. To be in the top 5% for a bonus, find the z-score for the 95th percentile and use it to solve for the required yards.
Step-by-step explanation:
To answer Question 1 regarding the probability that a wide receiver gained 455 yards or less during the season, we would need to know the mean and the standard deviation of the yards gained by wide receivers. We would use the normal distribution model for this. Once we have the mean (let's call it μ) and the standard deviation (let's call it σ), we can find the z-score for 455 yards using the formula:
Z = (X - μ) / σ
where X is the number of yards (455) in this case. After finding the z-score, we would use the standard normal distribution table or a calculator to find the cumulative probability corresponding to that z-score, which gives us the probability that a wide receiver gained 455 yards or less.
For Question 2 about a receiver needing to be in the top 5% of yards gained to get a bonus, we can use the inverse of the normal distribution function. We look for the z-score that corresponds to the 95th percentile and then use the z-score formula in reverse to solve for X:
X = (Z * σ) + μ
Once we have X, we round to the nearest yard to find the number of yards the receiver needs to gain to receive the bonus.