Mastery Check Solutions

1. Explain the type of distribution for each group.

The Pennsylvania data is normal because the $\mathrm{median}=\mathrm{mean}$.

The Mexico data is left-skewed because $\mathrm{mean}<\mathrm{median}$.

3. Scientists measured the wingspan of eight Monarch butterflies found on a plant in Florida, USA. Calculate the mean and standard deviation. 
   $\left\{8.8, 9.2, 9.4, 9.5, 9.5, 9.7, 9.8, 10\right\}$

$$\begin{gathered} \mu =\frac{8.8+9.2+9.4+9.5+9.5+9.7+9.8+10 }{8}=\frac{75.9}{8}=9.4875=9.49 \\ {\left(x_i–\mu \right)}^2 \\ {(8.8–9.49)}^2=0.4761 \\ {\left(9.2–9.49\right)}^2=0.0841 \\ {\left(9.4–9.49\right)}^2=0.0081 \\ {\left(9.5–9.49\right)}^2=0.0001 \\ {\left(9.5–9.49\right)}^2=0.0001 \\ {\left(9.7–9.49\right)}^2=0.0441 \\ {\left(9.8–9.49\right)}^2=0.0961 \\ {\left(10–9.49\right)}^2=0.2601 \\ \sum _{}{\left(x_i–\mu \right)}^2=0.9688 \\ \sigma =\sqrt{\frac{0.9688}{8}}=0.3480=0.35 \end{gathered}$$

4. What is the interval for 99.7% of Florida Monarchs? Explain if they vary more than the Pennsylvania butterflies.

$$\begin{gathered} 9.49–3\left(0.35\right)=8.44 \\ 9.49+3\left(0.35\right)=10.54 \\ \left[8.44, 10.54\right] \end{gathered}$$

The Florida (FL) Monarchs vary more than the Pennsylvania butterflies because the FL standard deviation is larger and the interval is larger.