5.2.1: Practice Calculating z-scores

Last updated
Save as PDF

Page ID: 18031

\( \newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} } \) \( \newcommand{\vecd}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash {#1}}} \)\(\newcommand{\id}{\mathrm{id}}\) \( \newcommand{\Span}{\mathrm{span}}\) \( \newcommand{\kernel}{\mathrm{null}\,}\) \( \newcommand{\range}{\mathrm{range}\,}\) \( \newcommand{\RealPart}{\mathrm{Re}}\) \( \newcommand{\ImaginaryPart}{\mathrm{Im}}\) \( \newcommand{\Argument}{\mathrm{Arg}}\) \( \newcommand{\norm}[1]{\| #1 \|}\) \( \newcommand{\inner}[2]{\langle #1, #2 \rangle}\) \( \newcommand{\Span}{\mathrm{span}}\) \(\newcommand{\id}{\mathrm{id}}\) \( \newcommand{\Span}{\mathrm{span}}\) \( \newcommand{\kernel}{\mathrm{null}\,}\) \( \newcommand{\range}{\mathrm{range}\,}\) \( \newcommand{\RealPart}{\mathrm{Re}}\) \( \newcommand{\ImaginaryPart}{\mathrm{Im}}\) \( \newcommand{\Argument}{\mathrm{Arg}}\) \( \newcommand{\norm}[1]{\| #1 \|}\) \( \newcommand{\inner}[2]{\langle #1, #2 \rangle}\) \( \newcommand{\Span}{\mathrm{span}}\)\(\newcommand{\AA}{\unicode[.8,0]{x212B}}\)

Example \(\PageIndex{1}\)

Assume the following scores represent a sample of statistics classes taken by five psychology professors: 2, 3, 5, 5, 6. If the standard deviation is 1.64, what is the z-score for each of these professor's?

Solution

The mean is 4.2 stats classes taken (\( \displaystyle \bar{X} = \dfrac{\sum X}{N} = \dfrac{21}{5} = 4.2 \)).

\[z_2=\dfrac{x-\overline{X}}{s} = \dfrac{2-4.2}{1.64} = \dfrac{-2.2}{1.64} = -1.34 \nonumber \]

\[z_3=\dfrac{x-\overline{X}}{s} = \dfrac{3-4.2}{1.64} = \dfrac{-1.2}{1.64} = -0.73 \nonumber \]

\[z_Both5=\dfrac{x-\overline{X}}{s} = \dfrac{5-4.2}{1.64} = \dfrac{0.80}{1.64} = 0.49 \nonumber \]

\[z_6=\dfrac{x-\overline{X}}{s} = \dfrac{6-4.2}{1.64} = \dfrac{1.8}{1.64} = 1.10 \nonumber \]

PS You might want to practice calculating the standard deviation yourself to make sure that you haven't forgotten how!

Your turn!

Exercise \(\PageIndex{1}\)

Calculate \(z\)-scores for the three IQ scores provided, which were taken from a population with a mean of 100 and standard deviation of 16: 112, 109, 88.

Answer

\(z_112\) = 0.75

\(z_109\) = 0.56

\(z_88\) = -0.75

This time, you'll get the z-score and will need to find the IQ scores. Remember, you can do this with the z-score formula that you used above and to algebra to find \(x|), or you an use the other z-score formula.

Exercise \(\PageIndex{2}\)

Use the \(z\)-scores provided to find two IQ scores taken from a population with a mean of 100 and standard deviation of 16:

\(z\) = 2.19

\(z\) = -0.06

Answer: The IQ scores are 135 (for \(z\) = 2.19) and 99 (\(z\) = -0.06) .

Contributors and Attributions

Foster et al. (University of Missouri-St. Louis, Rice University, & University of Houston, Downtown Campus)

Dr. MO (Taft College)