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Elementary Statistics Supplementary Notes

David W. Lyons

Section 1.4 Normal Distributions

Subsection 1.4.1 Using the Normal Table

Values for \(z\) in the left column of the standard normal table are given in increments of \(0.05\text{.}\) For \(z\) values that fall between rows of the table, use an area estimate that is between the corresponding values in the area column. For example, \(z=.77\) is between the row for \(z=.75\) with area \(54.67\) and the row for \(z=.80\) with area \(57.63\text{,}\) so a reasonable area estimate would be \(56\%\text{.}\) Use a similar estimation procedure if you want to find a \(z\) value that corresponds to an area that falls between rows of the table. For example, the area \(80\%\) falls between the row for area \(78.87\%\) for \(z=1.25\) and the row for the area \(80.64\%\) for \(z=1.30\text{,}\) so a reasonable \(z\) estimate would be \(z=1.27\) or \(z=1.28\text{.}\) The procedure of reading between the rows is not technical, and might be off by a little bit, but it is better than rounding off to the values in the nearest row.
When calculating a \(z\) value for normal data, you may round to the nearest 0.01, because that is the level of accuracy in the \(z\) column of the normal table. When calculating an area under the normal curve, you may usually round the final answer to the nearest whole percent. These practices are exceptions to the usual roundoff advice Subsection 1.1.1.

Subsection 1.4.2 Examples

Here are some basic problem types about the normal curve. Data called \(X\) is approximately normally distributed, with \(\AVE(X)=72.3\) and \(\SD(X)=6.7\text{.}\)
  1. (symmetric region) Estimate the percent of data in the range \(72.3\pm 3.9\text{.}\)
  2. (asymmetric region) Estimate the percent of data in the range \(69.0\) to \(81.0\text{.}\)
  3. (left tail region) Estimate the percentile rank for the data values \(69.0\) and \(81.0\text{.}\)
  4. (from area to \(z\)) Estimate the data values with 70th and 35th percentile ranks.
Here are the solutions. Here are the drawing method
 1 
images/normal_curve_sample_solutions.pdf
solution steps taught in the textbook.
  1. The \(z\) values for \(72.3\pm 3.9\) are \(\pm \frac{3.9}{6.7} \approx \pm .58\text{.}\) The normal table shows that the area between these \(z\) values is about \(44\%\text{.}\)
  2. The \(z\) value for \(69.0\) is \(z_L=\frac{69.0-72.3}{6.7}\approx -.49\text{,}\) and the \(z\) value for \(81.0\) is \(z_R = \frac{81.0-72.3}{6.7}\approx 1.30\text{.}\) The corresponding areas in the normal table are \(38\%\) and \(81\%\text{,}\) respectively. The area from \(z_L\) to \(0\) is about \(38/2=19\%\) and the area from \(0\) to \(z_R\) is about \(81/2=40.5\%\text{,}\) so the area from \(z_L\) to \(z_R\) is about \(19+40.5=59.5\%\approx 60\%\text{.}\)
  3. We have the same values \(z_L,z_R\) from the previous part. First, we consider \(z_L=-.49\text{.}\) Start with the observation that
    \begin{align*} 100\% \amp = (\text{area to the left of }-.49)\\ \amp + (\text{area between }-.49 \text{ and }+.49)\\ \amp + (\text{area to the right of }+4.9). \end{align*}
    By mirror symmetry, we have
    \begin{equation*} (\text{area to the left of }-.49) = (\text{area to the right of }+4.9). \end{equation*}
    Putting these together, we get
    \begin{equation*} (\text{area to the left of }-.49)= \frac{100\%-(\text{area between }-.49 \text{ and }+.49)}{2} \end{equation*}
    which is \((100 -38)/2=31\%\text{.}\) Next, we consider \(z_R=1.30\text{.}\) Start with
    \begin{align*} (\text{area to the left of }1.30)\amp = (\text{area left of }0)\\ \amp + (\text{area between }0 \text{ and }1.30)\\ \amp = 50\% + 81/2\\ \amp \approx 90\%. \end{align*}
  4. First, we find the \(z\) value that has 70th percentile rank. Let’s call that value \(A\text{.}\) Starting with the observation that
    \begin{align*} 100\% \amp = (\text{area to the left of }A)\\ \amp + (\text{area to the right of }A)\\ \amp = 70 + (\text{area to the right of }A) \end{align*}
    we conclude that the area to the right of \(A\) is \(30\%\text{.}\) By mirror symmetry, it must be that the area to the left of \(-A\) is also \(30\%\text{,}\) so the area between \(-A\) and \(+A\) must be \(100 - 2(30) = 40\%\text{.}\) Using the normal curve table, we find that \(A\approx .52\text{.}\) This converts to an \(X\) value of \(X=(.52)(6.7)+72.3\approx 75.8\text{.}\) Next, we find a \(z\) value that has 35th percentile rank. Let’s call that value \(-B\) (so that \(-B\) is to the left of average, and \(+B\) is to the right of average). Starting with
    \begin{align*} 100\% \amp = (\text{area to the left of }-B)\\ \amp + (\text{area between }-B \text{ and }B)\\ \amp + (\text{area to the right of }B) \end{align*}
    By mirror symmetry, we have
    \begin{equation*} (\text{area to the left of }-B) = (\text{area to the right of }B). \end{equation*}
    It follows that both sides of this equation are \(35\%\text{,}\) so we conclude that \((\text{area between }-B \text{ and }B)= 100 - 2(35)=30\%\text{.}\) From the normal table, we get \(-B\approx -.385\text{.}\) This converts to an \(X\) value of \(X=(-.385)(6.7)+72.3\approx 69.7\text{.}\)

Subsection 1.4.3 Normal curve practice problems (Ch 5)

Exercises Exercises

1. Convert between \(X\) and \(Z\) values.
Each row in the table below gives three of the four values \(\AVE(X)\text{,}\) \(\SD(X)\text{,}\) \(X\text{,}\) and \(Z=\frac{X-\AVE(X)}{\SD(X)}\) for some normal data called \(X\text{.}\) Find the missing entry in each row.
\begin{align*} \AVE(X) \amp\spacer\amp \SD(X) \amp\spacer\amp X \amp\spacer\amp Z\\ \rule{.7in}{.1ex} \amp \amp \rule{.6in}{.1ex} \amp \amp \rule{.5in}{.1ex} \amp \amp \rule{.5in}{.1ex}\\ 32.4 \amp\spacer\amp 2.43 \amp\spacer\amp 35.0 \amp\spacer\amp \\ 28.6 \amp\spacer\amp 8.62 \amp\spacer\amp \amp\spacer\amp -.45\\ 102.3 \amp\spacer\amp \amp\spacer\amp 110.0 \amp\spacer\amp .73\\ \amp\spacer\amp 3.71 \amp\spacer\amp 22.0 \amp\spacer\amp 1.42 \end{align*}
Answer.
\begin{align*} \AVE(X) \amp\spacer\amp \SD(X) \amp\spacer\amp X \amp\spacer\amp Z\\ \rule{.7in}{.1ex} \amp \amp \rule{.6in}{.1ex} \amp \amp \rule{.5in}{.1ex} \amp \amp \rule{.5in}{.1ex}\\ 32.4 \amp\spacer\amp 2.43 \amp\spacer\amp 35.0 \amp\spacer\amp 1.07\\ 28.6 \amp\spacer\amp 8.62 \amp\spacer\amp 24.7 \amp\spacer\amp -.45\\ 102.3 \amp\spacer\amp 10.55 \amp\spacer\amp 110.0 \amp\spacer\amp .73\\ 16.73 \amp\spacer\amp 3.71 \amp\spacer\amp 22.0 \amp\spacer\amp 1.42 \end{align*}
2. Convert between \(Z\) value and percentile rank.
Each row in the table below gives either the
Z
value or the percentile rank for some value of normally distributed data. Find the missing entry in each row.
\begin{align*} Z \amp\spacer\amp \text{percentile rank}\\ \rule{.3in}{.1ex} \amp \amp \rule{1.2in}{.1ex}\\ 1.2 \amp\spacer\amp\\ -.37 \amp\spacer\amp\\ \amp\spacer\amp 43\text{rd percentile}\\ \amp\spacer\amp 58\text{th percentile} \end{align*}
Answer.
\begin{align*} Z \amp\spacer\amp \text{percentile rank}\\ \rule{.3in}{.1ex} \amp \amp \rule{1.2in}{.1ex}\\ 1.2 \amp\spacer\amp 88\text{th percentile}\\ -.37 \amp\spacer\amp 36\text{th percentile}\\ -.18 \amp\spacer\amp 43\text{rd percentile}\\ .20 \amp\spacer\amp 58\text{th percentile} \end{align*}
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