Chapter 12.3, Problem 21PPS

a.

To determine

To calculate : the mean and mean absolute deviation of the given data.

Answer to Problem 21PPS

The mean height of the players is $\underset{\_}{16.9\text{ minutes}}$ .

The mean absolute deviation of the given data is $\underset{\_}{35.1}$ .

Explanation of Solution

Given information :

Consider the information given in the question:

The time in minutes of top fifteen runners:

  $15:56,16:06,16:11,16:21,16:26,16:34,16:41,16:54,17:00,17:03,17:14,17:46,17:56,17:57,18:03$

Formula used :

Step1: Find the mean, $\overline{x}$

Step2: Find the sum of the absolute values of the difference between each value in the set of data and the mean.

Step3: divide the sum by the number of values in the set of data.

Calculation :

As per problem,

The data set:

  $15:56,16:06,16:11,16:21,16:26,16:34,16:41,16:54,17:00,17:03,17:14,17:46,17:56,17:57,18:03$ The number of values in the data set $\text{=15}$

Convert the time from minutes to seconds,

  $1\text{ min = 60 s}$

Therefore, the top fifteen running times in seconds are:

  $956,966,971,981,986,994,1001,1014,1020,1023,1034,1066,1076,1077,1083$

Step1: To find the mean, add the time (in seconds) and then divide by the number of values in the data set.

  $\begin{array}{l}\overline{x}=\frac{956+966+971+981+986+994+1001+1014+1020+1023+1034+1066+1076+1077+1083}{15}\\ \overline{x}=\frac{15248}{15}\approx 1016.5\text{ seconds or 16}\text{.9 minutes}\end{array}$

Therefore, the mean of the top fifteen running times is $16.9\text{ minutes}$

Step2: To find the sum of the absolute values, add the difference between each value and the mean.

The sum of absolute value

  $\begin{array}{l}=\left|956-1016.5\right|+\left|966-1016.5\right|+\left|971-1016.5\right|+\left|981-1016.5\right|+\left|986-1016.5\right|+\\ \left|994-016.5\right|+\left|1001-1016.5\right|+\left|1014-1016.5\right|+\left|1020-1016.5\right|+\left|1023-1016.5\right|+\\ \left|1034-1016.5\right|+\left|1066-1016.5\right|+\left|1076-1016.5\right|+\left|1077-1016.5\right|+\left|1083-1016.5\right|\\ =60.5+50.5+45.5+35.5+30.5+22.5+15.5+2.5+3.5+6.5+17.5+49.5+59.5+60.5+66.5\\ =526.5\end{array}$

Step 3: Divide the sum by the number of values

  $\text{mean absolute deviation =}\frac{526.5}{15}=35.1$

b.

To determine

To identify: the sample and the population.

Answer to Problem 21PPS

The sample is the top fifteen runners and the population is all the participants of the race.

Explanation of Solution

Given:

The result of a $5\text{K}$ race is published in the local paper. Over a thousand people participated, but only the time of top fifteen finishers are listed.

A sample is some portion of a larger group called population.

As per the given problem, over a thousand people participated in the race but only the time of top fifteen finishers are listed in the paper.

Therefore, the top fifteen runners is the sample and the population is all the people that participated in the race.

c.

To determine

To classify: the given data as qualitative or quantitative.

To explain if a statistical analysis of the sample can be applied to the population.

Answer to Problem 21PPS

The given data isquantitative.

Statistical analysis of the sample cannot be applied to the population as the sample is biased.

Explanation of Solution

Given:

The result of a $5\text{K}$ race is published in the local paper. Over a thousand people participated, but only the time of top fifteen finishers are listed.

The time in minutes of top fifteen runners:

  $15:56,16:06,16:11,16:21,16:26,16:34,16:41,16:54,17:00,17:03,17:14,17:46,17:56,17:57,18:03$

The given data are numerical values, hence classified as quantitative data.

Statistical analysis of the sample cannot be applied to the population as the sample in biased, that is, it favors the first fifteen finishers over the others. Since, the sample is not chosen at random, the measures of variation does not represent the entire population.