Question

In the 2009 NFL regular season, quarterback Philip Rivers of the San Diego Chargers had an average of 266 passing yards per game with a standard deviation of 79 yards. Between which two values do you expect the middle 68% of his performance to be?
Your answer
A. 108 and 424
B. 29 and 503
C. 187 and 345
D. 0 and 582

Answer 1

Answer:

C. 187 and 345

Step-by-step explanation:

The Empirical Rule states that, for a normally distributed random variable:

Approximately 68% of the measures are within 1 standard deviation of the mean.

Approximately 95% of the measures are within 2 standard deviations of the mean.

Approximately 99.7% of the measures are within 3 standard deviations of the mean.

In this problem, we have that:

Mean of 266 yards per game, standard deviation of 79 yards.

Between which two values do you expect the middle 68% of his performance to be?

By the Empirical Rule, within 1 standard deviation of the mean. So between 266 - 79 = 187 yards and 266 + 79 = 345 yards. The correct answer is given by option C.