Skip to main content
Library homepage
Loading table of contents menu...
Statistics LibreTexts

5.3.1: Practice Using the z Table

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

    It's time to practice with the z-table!

    Example \(\PageIndex{1}\)

    Find the \(z\)-score that bounds the top 9% of the distribution.


    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!

    Exercise \(\PageIndex{1}\)

    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.


    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...

    Example \(\PageIndex{2}\)

    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?



    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!

    Exercise \(\PageIndex{2}\)

    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)?


    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!

    Exercise \(\PageIndex{3}\)

    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)?


    The proportion of students in the class who did worse than you (earned a lower score) should be \(71.57\%\).

    Contributors and Attributions

    This page titled 5.3.1: Practice Using the z Table is shared under a CC BY-NC-SA license and was authored, remixed, and/or curated by Michelle Oja.

    • Was this article helpful?