5.2.1: Practice Calculating z-scores
- Page ID
- 22049
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\(\newcommand{\avec}{\mathbf a}\) \(\newcommand{\bvec}{\mathbf b}\) \(\newcommand{\cvec}{\mathbf c}\) \(\newcommand{\dvec}{\mathbf d}\) \(\newcommand{\dtil}{\widetilde{\mathbf d}}\) \(\newcommand{\evec}{\mathbf e}\) \(\newcommand{\fvec}{\mathbf f}\) \(\newcommand{\nvec}{\mathbf n}\) \(\newcommand{\pvec}{\mathbf p}\) \(\newcommand{\qvec}{\mathbf q}\) \(\newcommand{\svec}{\mathbf s}\) \(\newcommand{\tvec}{\mathbf t}\) \(\newcommand{\uvec}{\mathbf u}\) \(\newcommand{\vvec}{\mathbf v}\) \(\newcommand{\wvec}{\mathbf w}\) \(\newcommand{\xvec}{\mathbf x}\) \(\newcommand{\yvec}{\mathbf y}\) \(\newcommand{\zvec}{\mathbf z}\) \(\newcommand{\rvec}{\mathbf r}\) \(\newcommand{\mvec}{\mathbf m}\) \(\newcommand{\zerovec}{\mathbf 0}\) \(\newcommand{\onevec}{\mathbf 1}\) \(\newcommand{\real}{\mathbb R}\) \(\newcommand{\twovec}[2]{\left[\begin{array}{r}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\ctwovec}[2]{\left[\begin{array}{c}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\threevec}[3]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\cthreevec}[3]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\fourvec}[4]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\cfourvec}[4]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\fivevec}[5]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\cfivevec}[5]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\mattwo}[4]{\left[\begin{array}{rr}#1 \amp #2 \\ #3 \amp #4 \\ \end{array}\right]}\) \(\newcommand{\laspan}[1]{\text{Span}\{#1\}}\) \(\newcommand{\bcal}{\cal B}\) \(\newcommand{\ccal}{\cal C}\) \(\newcommand{\scal}{\cal S}\) \(\newcommand{\wcal}{\cal W}\) \(\newcommand{\ecal}{\cal E}\) \(\newcommand{\coords}[2]{\left\{#1\right\}_{#2}}\) \(\newcommand{\gray}[1]{\color{gray}{#1}}\) \(\newcommand{\lgray}[1]{\color{lightgray}{#1}}\) \(\newcommand{\rank}{\operatorname{rank}}\) \(\newcommand{\row}{\text{Row}}\) \(\newcommand{\col}{\text{Col}}\) \(\renewcommand{\row}{\text{Row}}\) \(\newcommand{\nul}{\text{Nul}}\) \(\newcommand{\var}{\text{Var}}\) \(\newcommand{\corr}{\text{corr}}\) \(\newcommand{\len}[1]{\left|#1\right|}\) \(\newcommand{\bbar}{\overline{\bvec}}\) \(\newcommand{\bhat}{\widehat{\bvec}}\) \(\newcommand{\bperp}{\bvec^\perp}\) \(\newcommand{\xhat}{\widehat{\xvec}}\) \(\newcommand{\vhat}{\widehat{\vvec}}\) \(\newcommand{\uhat}{\widehat{\uvec}}\) \(\newcommand{\what}{\widehat{\wvec}}\) \(\newcommand{\Sighat}{\widehat{\Sigma}}\) \(\newcommand{\lt}{<}\) \(\newcommand{\gt}{>}\) \(\newcommand{\amp}{&}\) \(\definecolor{fillinmathshade}{gray}{0.9}\)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 (\( \displaystyle \bar{X} = \frac{\sum X}{N} = \frac{21}{5} = 4.2 \)).
\[z_2=\dfrac{x-\overline{X}}{s} = \dfrac{2-4.2}{1.64} = \dfrac{-2.2}{1.64} = -1.34 \nonumber \]
\[z_3=\dfrac{x-\overline{X}}{s} = \dfrac{3-4.2}{1.64} = \dfrac{-1.2}{1.64} = -0.73 \nonumber \]
\[z_Both5=\dfrac{x-\overline{X}}{s} = \dfrac{5-4.2}{1.64} = \dfrac{0.80}{1.64} = 0.49 \nonumber \]
\[z_6=\dfrac{x-\overline{X}}{s} = \dfrac{6-4.2}{1.64} = \dfrac{1.8}{1.64} = 1.10 \nonumber \]
PS You might want to practice calculating the standard deviation yourself to make sure that you haven't forgotten how!
Your turn!
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.
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)