Solution Found!
Receivers 2013 NFL data from the 2013 football season
Chapter 5, Problem 41E(choose chapter or problem)
NFL data from the 2013 football season reported the number of yards gained by each of the league’s 181 wide receivers:
The mean is 426.98 yards, with a standard deviation of 408.34 yards.
a) According to the Normal model, what percent of receivers would you expect to gain more yards than 2 standard deviations above the mean number of yards?
b) For these data, what does that mean?
c) Explain the problem in using a Normal model here.
Questions & Answers
QUESTION:
NFL data from the 2013 football season reported the number of yards gained by each of the league’s 181 wide receivers:
The mean is 426.98 yards, with a standard deviation of 408.34 yards.
a) According to the Normal model, what percent of receivers would you expect to gain more yards than 2 standard deviations above the mean number of yards?
b) For these data, what does that mean?
c) Explain the problem in using a Normal model here.
ANSWER:Step 1 of 3
Given:
\(\mu=\text { Mean }=426.98\)
\(\sigma=\text { Standard deviation }=408.34\)
(a) The 68-95-99.7 rule tells us that about 95% of all data values are within 2 standard deviations of the mean. \(100 \%-95 \%=5 \%\) of all data values are then more than 2 standard deviations of the mean.
Since the normal distribution is symmetric about the mean, we then expect about \(\frac{5 \%}{2}=2.5 \%\) of all data values are more than 2 standard deviations above the mean.