Test 23 (Lessons 45-46): The Normal Curve Solutions

1. In an environmental science class, Anabeth recorded the height in millimeters of six plants before they were about to flower. After measuring, she wanted to know the average height as well as how much the plant heights varied. Calculate the mean and standard deviation. Show your work.
   $\left\{93, 95, 95, 98, 98, 100\right\}$

$$\begin{gathered} \mu =\frac{93+95+95+98+98+100}{6}=96.5 \\ \begin{array}{ccc}{\left(x–\mu \right)}^2=& & \sqrt{{\sigma }^2}=\sqrt{\frac{33.5}{6}}\\ {\left(93–96.5\right)}^2=& 12.25& \sigma =2.362\\ {\left(95–96.5\right)}^2=& 2.25& \sigma =2.4\\ {\left(95–96.5\right)}^2=& 2.25& \\ {\left(98–96.5\right)}^2=& 2.25& \\ {\left(98–96.5\right)}^2=& 2.25& \\ {\left(100–96.5\right)}^2=& 12.25& \\ \sum _{}{\left(x–\mu \right)}^2=& 33.5& \end{array} \end{gathered}$$

2. $$\begin{gathered} \mathrm{Mean}: 8.6 \\ \mathrm{M}\mathrm{edian}: 11 \\ \mathrm{M}\mathrm{ode}: 1 \end{gathered}$$

$$\begin{gathered} \mathrm{median} > \mathrm{mean} \\ \mathrm{left}‐\mathrm{skewed} \end{gathered}$$

3. $$\begin{gathered} \mathrm{M}\mathrm{ean}: 18.05 \\ \mathrm{M}\mathrm{edian}: 18 \\ \mathrm{M}\mathrm{ode}: 18 \end{gathered}$$

$$\begin{gathered} \mathrm{median} \approx \mathrm{mean} \\ \mathrm{symmetric} \end{gathered}$$

7. If 300 women were randomly selected, how many are expected to have a shoe size greater than 38.1?

$$\begin{gathered} 100–68=32 \\ 32÷2=16 \\ 68+16=84\% \\ 0.84\left(300\right)=252 \end{gathered}$$

There will be 252 women with a shoe size greater than 38.1.

8. Determine the likelihood that a randomly selected woman wears a size 38.5 or smaller to the nearest whole percent.

$$\begin{gathered} z=\frac{38.5–39.5}{1.4} \\ z=–0.714 \\ z=–0.71 \\ \mathrm{P}\left(X<\mathrm{size} 38.5\right)=0.2389 \end{gathered}$$

The likelihood of a randomly selected woman wearing a size 38.5 or smaller is 24%.

9. Determine the likelihood that a woman wears between a 38.5 and 40.5 shoe size depending on the style.

$$\begin{gathered} \mathrm{P}\left(X< \mathrm{size} 38.5\right)=0.2389 \\ z=\frac{40.5–39.5}{1.14} \\ z=0.877 \\ z=0.88 \\ \mathrm{P}\left(X<\mathrm{size} 40.5\right)=0.8106 \end{gathered}$$

$$\begin{gathered} \mathrm{P}\left(38.5<X<40.5\right) \\ 0.8106–0.2389=0.5717 \end{gathered}$$

The likelihood that a woman wears between a 38.5 and 40.5 shoe size is 57%.

10. A shoe manufacturer wants to produce a shoe that accommodates the majority of women purchasing shoes. What shoe size (whole or half) represents the 88th percentile of women’s shoes? 

$$\begin{gathered} z=1.18 \\ \begin{array}{c}1.18=\frac{X–39.5}{1.4}\\ \left(1.18\right)\left(1.4\right)=X–39.5\\ X=41.152\end{array} \end{gathered}$$ A size 41 shoe represents about 88% or less of women.