Chapter 13.3, Problem 8PPS

a.

To determine

To calculate: The season having a greater range of points.

Answer to Problem 8PPS

In 2008 Cami had a greater range of points.

Explanation of Solution

Given data:

The following data is given in the form of a stem and leaf diagram.

A stem and leaf model diagram can be studied for providing a large number of data in concise manner.

The central stem shows the number in tens place and the number on the side (leaf) show the number in units place. The total number of data gives the number of matches played or number of times Cami scored.

For example in 2007 season, she scored between 0 and 9 five times $\left(06,07,08,09,09\right)$ as there are five numbers in leaf beside the 0 in stem.

Range of a sample of data refers to the difference between two extreme values (highest and lowest).

By looking at the sample it can be concluded that 2007 season had a range of $27-6=21.$

Whereas 2008 season had a range of $40-14=26.$

Conclusion:

Thus, Cami had a greater range of points in 2008 season which is 26.

b.

To determine

To find: An appropriate measure of central tendency for given two seasons.

Answer to Problem 8PPS

Median is the appropriate measure of central tendency.

Explanation of Solution

Given data:

The following data is given in the form of a stem and leaf diagram.

Central tendency of a data indicates the most likely value from it.

There are certain types of measures depending on the nature of data.

2007 season has a simple data pattern without any skewed region.

Hence, mean or average will be the best measure of central tendency for 2007 season.

Mean for season 2007 $=\frac{6+7+8+9+9+10+10+12+14+16+17+18+21+27}{14}$

  $=13.14$

2008 season has a bit different pattern because of an outlier (40).

Outlier is an unusually high or low value as compared to the majority of data.

For such a pattern mean will not be the best measure of central tendency, rather median will be a better measure.

If mean is used to find the central tendency of the given data it comes out to be 21,

However it should be noted that this is not the most likely score in 20008 season.

Mean comes out to be elevated by 2 because of an outlier as 40.

Although 2 is not a very big number but since Cami`s vary only by 1, 2 becomes valuable increase.

Median is obtained by simply finding out the middle value after arranging the data in ascending order.

The given data has an even number of value so the median can be found out by averaging the central 2 values.

Median $=\frac{18+22}{2}$

  $=20$

Conclusion:

Therefore, the median is appropriate measure of central tendency.

c.

To determine

To find: The effect on range of scores by including 40, which is an outlier.

Answer to Problem 8PPS

It simply increases the error in mean of the data.

Explanation of Solution

Given data:

The following data is given in the form of a stem and leaf diagram.

40 in the 2008 season is an outlier.

Most of the scoring in 2008 season was in range of 14 to 26 that is a range of just 12.

By inclusion of 40 the range increase to 26, however here it must be noted that there are no points scored between 26 and 40 so adding 40, an outlier, would simply increase the error in mean or average of the data.