Unit E: Statistics and Probability

Chapter 1: Statistics

Practice

Instructions: Click the Download File button to download a printable PDF of the questions. Answer each of the following practice questions on a separate piece of paper. Step by step solutions are provided under the Solutions tab. You will learn the material more thoroughly if you complete the questions before checking the answers.

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Practice
Solutions

Determine the required measure of central tendency given the following set of data: 5, 10, 14, 5, 3, 6, 9, 12, 5, 9
1. the mean
2. the median
3. the mode
Determine the required measure of central tendency for the following set of data: 6, 4, 10, 9, 3
1. the mean
2. the median
3. the mode
Four nine-year-old boys have a mean height of 55.5 in. If the heights of three of the boys are 54, 57.25, and 53.75, what is the height of the fourth boy?

Question 1

Determine the required measure of central tendency given the following set of data: 5, 10, 14, 5, 3, 6, 9, 12, 5, 9

the mean
the median
the mode

Solutions

$\begin{array}{rcl} \bar{x} & = & \frac{sum of the values in the data set}{total number of values in the data set} \\ = & \frac{x_{1} + x_{2} + x_{3} + x_{4} + x_{5} + x_{6} + x_{7} + x_{8} + x_{9} + x_{10}}{n} \\ = & \frac{5 + 10 + 14 + 5 + 3 + 6 + 9 + 12 + 5 + 9}{10} \\ = & \frac{78}{10} \\ = & 7.8 \end{array}$

Note: The second line of the formula does not need to be included for your work on the assignment, so it will be excluded from the remaining practice questions.

The mean is 7.8.
Arrange the data in ascending order.

3, 5, 5, 5, 6, 9, 9, 10, 12, 14

There are a total of 10 data values in the set of data. Since there is an even number of data values, the median will be the average of the two middle values. In this set of data, calculate the average of the fifth and sixth values.

$3, 5, 5, 5, 6, 9, 9, 10, 12, 14$

$\begin{array}{rcl} median & = & \frac{sum of the two middle values}{2} \\ = & \frac{6 + 9}{2} \\ = & \frac{15}{2} \\ = & 7.5 \end{array}$

The median is 7.5.
Arrange the data in ascending order.

3, 5, 5, 5, 6, 9, 9, 10, 12, 14

$3, 5, 5, 5, 6, 9, 9, 10, 12, 14$

Since 5 occurs most frequently, the mode is 5.

Question 2

Determine the required measure of central tendency for the following set of data: 6, 4, 10, 9, 3

the mean
the median
the mode

Solutions

$\begin{array}{rcl} \bar{x} & = & \frac{sum of the values in the data set}{total number of values in the data set} \\ = & \frac{6 + 4 + 10 + 9 + 3}{5} \\ = & \frac{32}{5} \\ = & 6.4 \end{array}$

The mean is 6.4.
Arrange the data is ascending order.

3, 4, 6, 9, 10

There are a total of five values in the set of data. Since the number of data values is odd, the median is equal to the middle data value.

The median is 6.
Arrange the data in ascending order.

3, 4, 6, 9, 10

All the values only appear once, so there is no mode in the set of data.

Question 3

Four nine-year-old boys have a mean height of 55.5 in. If the heights of three of the boys are 54, 57.25, and 53.75, what is the height of the fourth boy?

Solution

Let x = the height of the fourth boy.

\begin{array}{rcl} \bar{x} & = & \frac{sum of the values in the data set}{total number of values in the data set} \\ 55.5 & = & \frac{54 + 57.25 + 53.75 + x}{4} \\ 55.5 \times 4 & = & (\frac{54 + 57.25 + 53.75 + x}{4}) \times 4 \\ 55.5 \times 4 & = & (\frac{54 + 57.25 + 53.75 + x}{4}) \times 4 \\ 222 & = & 54 + 57.25 + 53.75 + x \\ 222 & = & x + 165 \\ 222 - 165 & = & x + 165 - 165 \\ 57 & = & x \end{array}

The height of the fourth boy is 57 inches.