Chapter 1.1, Problem 38PPS

To determine

To find: B.L’s efficiency rating to the nearest tenth for the season statistics shown.

Answer to Problem 38PPS

The B.L’s efficiency rating to the nearest tenth for the season statistics is 93.166.

Explanation of Solution

Given information:

The following formula can be used to calculate a quarterback efficiency rating.

  $\left(\frac{\frac{C}{A}-0.3}{0.2}+\frac{\frac{Y}{A}-3}{4}+\frac{\frac{T}{A}}{0.05}+\frac{0.095-\frac{1}{A}}{0.04}\right)\cdot \frac{100}{6}$

And

• $C$ is the number of passes completed = 108.
• $A$is the number of passes attempted = 183.
• $Y$is passing yardage = 1159.
• $T$is the number of touchdown passes = 7.
• $I$is the number of interceptions = 5.

Calculation:

Consider the given expression,

  $\left(\frac{\frac{C}{A}-0.3}{0.2}+\frac{\frac{Y}{A}-3}{4}+\frac{\frac{T}{A}}{0.05}+\frac{0.095-\frac{1}{A}}{0.04}\right)\cdot \frac{100}{6}$

Now, substitute the value of $C,A,Y,T$ and $I$ we get,

  $\begin{array}{c}\left(\frac{\frac{C}{A}-0.3}{0.2}+\frac{\frac{Y}{A}-3}{4}+\frac{\frac{T}{A}}{0.05}+\frac{0.095-\frac{I}{A}}{0.04}\right)\cdot \frac{100}{6}=\left(\frac{C-0.3A}{0.2A}+\frac{Y-3A}{4A}+\frac{T}{0.05A}+\frac{0.095-I}{0.04A}\right)\cdot \frac{100}{6}\\ =\left(\frac{20\left(C-0.3A\right)}{4A}+\frac{Y-3A}{4A}+\frac{80T}{4A}+\frac{100\left(0.095-I\right)}{4A}\right)\cdot \frac{100}{6}\\ =\left(\frac{20C-6A}{4A}+\frac{Y-3A}{4A}+\frac{80T}{4A}+\frac{9.5A-100I}{4A}\right)\cdot \frac{100}{6}\\ \frac{=20C-6A+Y-3A+80T+9.5A-100I}{4A}\cdot \frac{100}{6}\\ =\frac{20C+0.5A+Y+80T-100I}{A}\cdot \frac{25}{6}\\ =\frac{7420+27.55+4002+2160-1200}{555}\cdot \frac{25}{6}\\ =\frac{12409.75}{555}\cdot \frac{25}{6}\\ =\frac{310243.75}{3.330}\\ \approx 93.166\end{array}$

Hence, the B.L’s efficiency rating to the nearest tenth for the season statistics is 93.166.