5.3.1: Practice Using the z Table
- Page ID
- 22051
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\(\newcommand{\avec}{\mathbf a}\) \(\newcommand{\bvec}{\mathbf b}\) \(\newcommand{\cvec}{\mathbf c}\) \(\newcommand{\dvec}{\mathbf d}\) \(\newcommand{\dtil}{\widetilde{\mathbf d}}\) \(\newcommand{\evec}{\mathbf e}\) \(\newcommand{\fvec}{\mathbf f}\) \(\newcommand{\nvec}{\mathbf n}\) \(\newcommand{\pvec}{\mathbf p}\) \(\newcommand{\qvec}{\mathbf q}\) \(\newcommand{\svec}{\mathbf s}\) \(\newcommand{\tvec}{\mathbf t}\) \(\newcommand{\uvec}{\mathbf u}\) \(\newcommand{\vvec}{\mathbf v}\) \(\newcommand{\wvec}{\mathbf w}\) \(\newcommand{\xvec}{\mathbf x}\) \(\newcommand{\yvec}{\mathbf y}\) \(\newcommand{\zvec}{\mathbf z}\) \(\newcommand{\rvec}{\mathbf r}\) \(\newcommand{\mvec}{\mathbf m}\) \(\newcommand{\zerovec}{\mathbf 0}\) \(\newcommand{\onevec}{\mathbf 1}\) \(\newcommand{\real}{\mathbb R}\) \(\newcommand{\twovec}[2]{\left[\begin{array}{r}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\ctwovec}[2]{\left[\begin{array}{c}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\threevec}[3]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\cthreevec}[3]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\fourvec}[4]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\cfourvec}[4]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\fivevec}[5]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\cfivevec}[5]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\mattwo}[4]{\left[\begin{array}{rr}#1 \amp #2 \\ #3 \amp #4 \\ \end{array}\right]}\) \(\newcommand{\laspan}[1]{\text{Span}\{#1\}}\) \(\newcommand{\bcal}{\cal B}\) \(\newcommand{\ccal}{\cal C}\) \(\newcommand{\scal}{\cal S}\) \(\newcommand{\wcal}{\cal W}\) \(\newcommand{\ecal}{\cal E}\) \(\newcommand{\coords}[2]{\left\{#1\right\}_{#2}}\) \(\newcommand{\gray}[1]{\color{gray}{#1}}\) \(\newcommand{\lgray}[1]{\color{lightgray}{#1}}\) \(\newcommand{\rank}{\operatorname{rank}}\) \(\newcommand{\row}{\text{Row}}\) \(\newcommand{\col}{\text{Col}}\) \(\renewcommand{\row}{\text{Row}}\) \(\newcommand{\nul}{\text{Nul}}\) \(\newcommand{\var}{\text{Var}}\) \(\newcommand{\corr}{\text{corr}}\) \(\newcommand{\len}[1]{\left|#1\right|}\) \(\newcommand{\bbar}{\overline{\bvec}}\) \(\newcommand{\bhat}{\widehat{\bvec}}\) \(\newcommand{\bperp}{\bvec^\perp}\) \(\newcommand{\xhat}{\widehat{\xvec}}\) \(\newcommand{\vhat}{\widehat{\vvec}}\) \(\newcommand{\uhat}{\widehat{\uvec}}\) \(\newcommand{\what}{\widehat{\wvec}}\) \(\newcommand{\Sighat}{\widehat{\Sigma}}\) \(\newcommand{\lt}{<}\) \(\newcommand{\gt}{>}\) \(\newcommand{\amp}{&}\) \(\definecolor{fillinmathshade}{gray}{0.9}\)It's time to practice with the z-table!
Find the \(z\)-score that bounds the top 9% of the distribution.
Solution
Because we are looking for top 9%, we need to look for the p-value closest to p = .91000 (\(100\% - 9\% = 91\%\)) because the p-values (probabilities) in the z Table show the probability of score being lower, but this question is asking for top 9%, not the portion lower than \(9\%\). There should be \(91\%\) of scores lower than the top \(9\%\).
The closest p-value to p = .91000 (\(91\%\)) is 0.90988. The z-score for p = 0.90988 iz z=1.34.
The z-score for the top \(9\%\) of the distribution is z=1.34 (for p=0.90988, the closest probability to \(91\%\), which marks everyone lower than the top \(9\%\)).
Your turn!
Find the \(z\)-score that bounds \(25\%\) (p=0.25000) of the lower tail of the distribution.
Hint: You don't have to subtract anything for this one because the question is asking about the scores that are lower.
- Answer
-
The z-score for 25% of the lower tail of the distribution is\( z = -0.67\) (for p=0.25143, the closes probability to 0.25000 (\(25\%\))).
Now, let's try some scenarios...
The heights of women in the United States are normally distributed with a mean of 63.7 inches and a standard deviation of 2.7 inches. If you randomly select a woman in the United States, what is the probability that she will be between taller than 64 inches?
Solution
X (raw score) = 64 inches
\( \displaystyle \bar{X} \) = 63.7 inches
s=2.7 inches
\[z=\dfrac{x-\overline{X}}{s} = \dfrac{64-63.7}{2.7} = \dfrac{0.30}{2.7} = 0.11 \nonumber \]
Finding z=0.11 on the z Table, we see that p = 0.543860. This is the probability that a score will be lower than our raw score, but the question asked the proportion who would be taller.
\(1 - 0.54380 = 0.4562 \)
\(p \times 100 = 0.4562 \times 100 = 45.62\% \)
Final Answer (in words): The probability that a woman in the U.S. would be 64 inches or taller is 0.4562, or \(45.62\%\).
Your turn!
The heights of men in the United States are normally distributed with a mean of 69.1 inches and a standard deviation of 2.9 inches. What proportion of men are taller than 6 feet (72 inches)?
- Answer
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Final Answer: The probability that a man in the U.S. would be 72 inches or taller is 0.15866, or \(15.87\%\).
Last one, on something that you might find relevant!
Imagine that you scored 82 points on a final exam. After the final, you find out that the average score on the exam was 78 with a standard deviation of 7. What proportion (in a percentage) did worse than you (earned a lower score)?
- Answer
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The proportion of students in the class who did worse than you (earned a lower score) should be \(71.57\%\).
Contributors and Attributions
Foster et al. (University of Missouri-St. Louis, Rice University, & University of Houston, Downtown Campus)