91Ó°ÊÓ

The given statement is:

There are 435 representatives in the 113th session of the U.S. House of Representatives. On the website www.house.gov, you can find an alphabetized list of the 435 congresspersons. In 2013, the first representative listed is Robert Aderholt, a Republican from Alabama, and the last representative listed is Todd Young, a Republican from Indiana. Suppose that the alphabetized list is indexed 1 through 435.

Below is one of the possible solutions.

To obtain a sample, use a systematic random sampling procedure as given below:

1. The sample size is divided by the population size, and the result is rounded to the nearest whole number (m).
2. The sample size is 15 and the population size is 435.

$$\begin{gathered} k=23 \\ k+m=23+29 \\ =52 \\ k+2m=23+\left(2\right)\left(29\right) \\ =23+58 \\ =81 \\ k+3m=23+\left(2\right)\left(29\right) \\ =23+87 \\ =110 \\ k+4m=23+\left(4\right)\left(29\right) \\ =23+116 \\ =139 \\ k+5m=23+\left(5\right)\left(29\right) \\ =23+145 \\ =168 \\ k+6m \\ =23+\left(6\right)\left(29\right) \\ =23+174 \\ =197 \\ k+7m=23+\left(7\right)\left(29\right) \\ =23+203 \\ =226 \\ k+8m=23+\left(8\right)\left(29\right) \\ =23+232 \\ =255 \\ k+9m=23+\left(9\right)\left(29\right) \\ =23+261 \\ =284 \\ k+10m=23+\left(10\right)\left(29\right) \\ =23+290 \\ =313 \\ k+11m=23+\left(11\right)\left(29\right) \\ =23+319 \\ =342 \\ k+12m=23+\left(12\right)\left(29\right) \\ =23+348 \\ =371 \\ k+13m=23+\left(13\right)\left(29\right) \\ =23+377 \\ =400 \\ k+14m=23+\left(13\right)\left(29\right) \\ =23+406 \\ =429 \end{gathered}$$

$$\begin{gathered} k=12 \\ k+m=12+29 \\ =41 \\ k+2m=12+\left(2\right)\left(29\right) \\ =12+58 \\ =70 \\ k+3m=12+\left(2\right)\left(29\right) \\ =12+87 \\ =99 \\ k+4m=12+\left(4\right)\left(29\right) \\ =12+116 \\ =128 \\ k+5m=12+\left(5\right)\left(29\right) \\ =12+145 \\ =157 \\ k+6m=12+\left(6\right)\left(29\right) \\ =12+174 \\ =186 \\ k+7m=12+\left(7\right)\left(29\right) \\ =12+203 \\ =215 \\ k+8m=12+\left(8\right)\left(29\right) \\ =12+232 \\ =244 \\ k+9m=12+\left(9\right)\left(29\right) \\ =12+261 \\ =273 \\ k+10m=12+\left(10\right)\left(29\right) \\ =12+290 \\ =302 \\ k+11m=12+\left(11\right)\left(29\right) \\ =12+319=331 \\ k+12m=12+\left(12\right)\left(29\right) \\ =12+348 \\ =360 \\ k+13m=12+\left(13\right)\left(29\right) \\ =12+377 \\ =389 \\ k+14m=12+\left(13\right)\left(29\right) \\ =12+406 \\ =418 \end{gathered}$$

As a result, the numbers 12, 41, 70, 99, 128, 157, 186, 215, 244, 273, 302, 331, 360, 389, and 418 make up the size 15 sample.