Chapter 12.2, Problem 25PPS

a.

To determine

Find the mean and median of the data.

Answer to Problem 25PPS

  $21.18and21.4$

Explanation of Solution

Given information:

ACT scores for a recent year:

  $\begin{array}{l}MeanACTScoresByState\\ \begin{array}{llllllllll}20.2\hfill & 21.3\hfill & 21.5\hfill & 20.4\hfill & 21.6\hfill & 20.3\hfill & 22.5\hfill & 21.5\hfill & 17.8\hfill & 20.5\hfill \\ 20.0\hfill & 21.7\hfill & 21.3\hfill & 20.2\hfill & 21.6\hfill & 22.0\hfill & 21.6\hfill & 20.3\hfill & 19.8\hfill & 22.6\hfill \\ 20.8\hfill & 22.4\hfill & 21.4\hfill & 22.2\hfill & 18.8\hfill & 21.5\hfill & 21.7\hfill & 21.7\hfill & 21.2\hfill & 22.5\hfill \\ 21.2\hfill & 20.1\hfill & 22.3\hfill & 20.3\hfill & 21.2\hfill & 21.4\hfill & 20.6\hfill & 22.5\hfill & 21.8\hfill & 21.9\hfill \\ 19.3\hfill & 21.5\hfill & 20.5\hfill & 20.3\hfill & 21.5\hfill & 22.7\hfill & 20.9\hfill & 22.5\hfill & 22.2\hfill & 21.4\hfill \end{array}\end{array}$

Calculation:

Here, we will calculate the mean of the given data:

  $\begin{array}{l}\mu =\frac{{\displaystyle \sum x}}{n}\\ \mu =\frac{1059}{50}\\ \mu =21.18\\ Mean=21.18\end{array}$

Now, we will calculate the median of the given data:

As, there are $50$ values (even) so the median will be the average of ${25}^{th}and{26}^{th}$ values:

  $\begin{array}{l}Median=\frac{21.4+21.4}{2}\\ Median=21.4\end{array}$

Hence, the mean and median is $21.18and21.4$ .

b.

To determine

Find if it is a sample or population.

Answer to Problem 25PPS

Population.

Explanation of Solution

Given information:

ACT scores for a recent year:

  $\begin{array}{l}MeanACTScoresByState\\ \begin{array}{llllllllll}20.2\hfill & 21.3\hfill & 21.5\hfill & 20.4\hfill & 21.6\hfill & 20.3\hfill & 22.5\hfill & 21.5\hfill & 17.8\hfill & 20.5\hfill \\ 20.0\hfill & 21.7\hfill & 21.3\hfill & 20.2\hfill & 21.6\hfill & 22.0\hfill & 21.6\hfill & 20.3\hfill & 19.8\hfill & 22.6\hfill \\ 20.8\hfill & 22.4\hfill & 21.4\hfill & 22.2\hfill & 18.8\hfill & 21.5\hfill & 21.7\hfill & 21.7\hfill & 21.2\hfill & 22.5\hfill \\ 21.2\hfill & 20.1\hfill & 22.3\hfill & 20.3\hfill & 21.2\hfill & 21.4\hfill & 20.6\hfill & 22.5\hfill & 21.8\hfill & 21.9\hfill \\ 19.3\hfill & 21.5\hfill & 20.5\hfill & 20.3\hfill & 21.5\hfill & 22.7\hfill & 20.9\hfill & 22.5\hfill & 22.2\hfill & 21.4\hfill \end{array}\end{array}$

Calculation:

Here, the data shows the average SAT scores for all states.

Hence, the data shows the population.

c.

To determine

Find the standard deviation of the data.

Answer to Problem 25PPS

  $1.02$

Explanation of Solution

Given information:

ACT scores for a recent year:

  $\begin{array}{l}MeanACTScoresByState\\ \begin{array}{llllllllll}20.2\hfill & 21.3\hfill & 21.5\hfill & 20.4\hfill & 21.6\hfill & 20.3\hfill & 22.5\hfill & 21.5\hfill & 17.8\hfill & 20.5\hfill \\ 20.0\hfill & 21.7\hfill & 21.3\hfill & 20.2\hfill & 21.6\hfill & 22.0\hfill & 21.6\hfill & 20.3\hfill & 19.8\hfill & 22.6\hfill \\ 20.8\hfill & 22.4\hfill & 21.4\hfill & 22.2\hfill & 18.8\hfill & 21.5\hfill & 21.7\hfill & 21.7\hfill & 21.2\hfill & 22.5\hfill \\ 21.2\hfill & 20.1\hfill & 22.3\hfill & 20.3\hfill & 21.2\hfill & 21.4\hfill & 20.6\hfill & 22.5\hfill & 21.8\hfill & 21.9\hfill \\ 19.3\hfill & 21.5\hfill & 20.5\hfill & 20.3\hfill & 21.5\hfill & 22.7\hfill & 20.9\hfill & 22.5\hfill & 22.2\hfill & 21.4\hfill \end{array}\end{array}$

Calculation:

Here, we will calculate the standard deviation of the data:

  $\begin{array}{l}\sigma =\sqrt{\frac{{\displaystyle \sum {\left(x-\mu \right)}^2}}{n}}\\ \sigma =1.019\approx 1.02\end{array}$

Hence, the standard deviation is $1.02$ .

d.

To determine

Find the mean and median if the mean score of $20.0$ was incorrect and $22.5$ was correct.

Answer to Problem 25PPS

The mean will increase and median will remain same.

Explanation of Solution

Given information:

ACT scores for a recent year:

  $\begin{array}{l}MeanACTScoresByState\\ \begin{array}{llllllllll}20.2\hfill & 21.3\hfill & 21.5\hfill & 20.4\hfill & 21.6\hfill & 20.3\hfill & 22.5\hfill & 21.5\hfill & 17.8\hfill & 20.5\hfill \\ 20.0\hfill & 21.7\hfill & 21.3\hfill & 20.2\hfill & 21.6\hfill & 22.0\hfill & 21.6\hfill & 20.3\hfill & 19.8\hfill & 22.6\hfill \\ 20.8\hfill & 22.4\hfill & 21.4\hfill & 22.2\hfill & 18.8\hfill & 21.5\hfill & 21.7\hfill & 21.7\hfill & 21.2\hfill & 22.5\hfill \\ 21.2\hfill & 20.1\hfill & 22.3\hfill & 20.3\hfill & 21.2\hfill & 21.4\hfill & 20.6\hfill & 22.5\hfill & 21.8\hfill & 21.9\hfill \\ 19.3\hfill & 21.5\hfill & 20.5\hfill & 20.3\hfill & 21.5\hfill & 22.7\hfill & 20.9\hfill & 22.5\hfill & 22.2\hfill & 21.4\hfill \end{array}\end{array}$

Calculation:

Here, as we can see the mean is increasing. But the median will remain same because the middle values of order data set does not changed.

Hence, the mean will increase and median will remain same.