Problem 35 - Algebra 2

Question

The formula for quarterback efficiency rating in the National Football League is \(\left(\frac{\frac{C}{A}-0.3}{0.2}+\frac{\frac{Y}{A}-3}{4}+\frac{\frac{T}{A}}{0.05}+\frac{0.095-\frac{I}{A}}{0.04}\right) \times \frac{100}{6},\) where \(C\) is the number of passes completed, A is the number of passes attempted, Y is passing yardage, T is the number of touchdown passes, and I is the number of interceptions. In 2005, Ben Roethlisberger of the Pittsburgh Steelers completed 168 of the 268 passes he attempted for 2385 yards. He threw 17 touchdowns and 9 interceptions. Find his efficiency rating for 2005.

Step-by-Step Solution

Answer

98.57

Step 1: Identify the Variables

First, let's identify the values for each variable in the formula: - The number of passes completed, \( C = 168 \).- The number of passes attempted, \( A = 268 \).- The passing yardage, \( Y = 2385 \).- The number of touchdowns, \( T = 17 \).- The number of interceptions, \( I = 9 \).

Step 2: Calculate Completion Percentage Term

Calculate the first term of the formula using \( \frac{C}{A} \):\[ \frac{C}{A} = \frac{168}{268} \approx 0.6269 \]Then apply it to the formula segment:\[ \frac{\frac{C}{A} - 0.3}{0.2} = \frac{0.6269 - 0.3}{0.2} = \frac{0.3269}{0.2} = 1.6345 \]

Step 3: Calculate Yardage Per Attempt Term

Calculate the second term using \( \frac{Y}{A} \):\[ \frac{Y}{A} = \frac{2385}{268} \approx 8.9015 \]Then solve this part of the formula:\[ \frac{\frac{Y}{A} - 3}{4} = \frac{8.9015 - 3}{4} = \frac{5.9015}{4} = 1.475375 \]

Step 4: Calculate Touchdown Percentage Term

Compute the third term with \( \frac{T}{A} \):\[ \frac{T}{A} = \frac{17}{268} \approx 0.063432 \]Substitute into the formula:\[ \frac{\frac{T}{A}}{0.05} = \frac{0.063432}{0.05} = 1.26864 \]

Step 5: Calculate Interception Percentage Term

Calculate the fourth term using \( \frac{I}{A} \):\[ \frac{I}{A} = \frac{9}{268} \approx 0.033582 \]Substitute into the formula:\[ \frac{0.095 - \frac{I}{A}}{0.04} = \frac{0.095 - 0.033582}{0.04} = \frac{0.061418}{0.04} = 1.53545 \]

Step 6: Sum Terms Together

Now sum all the intermediate terms:\[ 1.6345 + 1.475375 + 1.26864 + 1.53545 = 5.913965 \]

Step 7: Multiply by Final Factor

Finally, multiply the sum of the terms by \( \frac{100}{6} \):\[ \left(5.913965\right) \times \frac{100}{6} = 98.5661 \]

Step 8: Final Answer

Ben Roethlisberger's efficiency rating for 2005 was approximately 98.57.

Key Concepts

Pass Completion Percentage

Pass completion percentage is an important metric for measuring a quarterback's accuracy. It's calculated as the ratio of passes completed to passes attempted, multiplied by 100 to get a percentage.

Here's how you can think about it:

• A higher completion percentage generally means a quarterback is making better decisions and accurately throwing the ball to receivers.
• If a quarterback completes 168 out of 268 passes, the completion percentage is calculated as follows: \[ \frac{C}{A} = \frac{168}{268} \approx 0.6269 \]This means the completion percentage is about 62.7%.
• This figure plays a significant role in the overall efficiency rating.

Passing Yardage

Passing yardage measures the total number of yards gained through the air by a quarterback.

This is a key factor in evaluating both the quarterback's ability to push the ball downfield and the receiver's effectiveness in gaining yards after the catch.

Passing yardage is a direct contributor to the quarterback efficiency rating formula:

• The formula looks at the average yards per attempt: \[ \frac{Y}{A} = \frac{2385}{268} \approx 8.9015 \]
• This number is adjusted in the formula to help gauge effectiveness; in our example, an average of 8.9 yards per attempt suggests solid performance in gaining yardage.

Remember, higher yardage per attempt tends to improve the efficiency rating, highlighting the quarterback's ability to make significant plays.

Touchdown Passes

Touchdown passes are one of the most exciting and critical components of a quarterback's performance. They are a direct reflection of successful offensive drives that end with a score.

To calculate their impact on efficiency, we look at their proportion relative to attempts:

• The formula computes the ratio of touchdown passes to attempts: \[ \frac{T}{A} = \frac{17}{268} \approx 0.063432 \]
• Touchdowns significantly elevate a quarterback's rating, since they directly contribute to the game's score.
• The ratio is a crucial factor, since precise and consistent scoring opportunities boost a quarterback's efficiency.

Interceptions

Interceptions are the turnovers that occur when a quarterback throws a pass that is caught by the opposing team. They have a negative impact on a quarterback's efficiency.

Interceptions are calculated as a percentage of passing attempts:

• The interception ratio is given by: \[ \frac{I}{A} = \frac{9}{268} \approx 0.033582 \]
• Interceptions are detrimental because they not only stop offensive drives but also potentially give the opponent a chance to score.
• Thus, minimizing interceptions is crucial for a quarterback aiming for a high efficiency rating.

Countering interceptions with touchdowns is key to maintaining a favorable balance in the efficiency calculation.