Section 3.2 Normal Curve Problems
Subsection 3.2.1 Examples
Here are some basic problem types about the normal curve. Data called \(X\) is approximately normally distributed, with \(\AVE(X)=72.3\) and \(\SD(X)=6.7\text{.}\)
(symmetric region) Estimate the percent of data in the range \(72.3\pm 3.9\text{.}\)
(asymmetric region) Estimate the percent of data in the range \(69.0\) to \(81.0\text{.}\)
(left tail region) Estimate the percentile rank for the data values \(69.0\) and \(81.0\text{.}\)
(from area to \(z\)) Estimate the data values with 70th and 35th percentile ranks.
Here are the solutions. Here are the
drawing method 1 solution steps taught in the textbook.
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 \(44\%\text{.}\)
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 \(19+40.5=59.5\%\approx
60\%\text{.}\)
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*}
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 3.2.2 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*}
images/normal_curve_sample_solutions.pdf
