Chapter 10.4, Problem 1GP

To determine

To calculate: The centers and variations and write an inference about the two populations.

Answer to Problem 1GP

Second periods score centered around 16 with a variation of 0.8 and fifth period score centered around 17 with variation 1.4.

Fifth period center around a greater value but the scores are more spread out than second period.

Explanation of Solution

Given information: The double dot plots for showing the quiz score for the periods second and fifth.

Calculation:

Both the double dot plots are symmetric. Use the mean to compare the centers and the mean absolute deviation to compare the variations.

For the Second period:

  $mean=\frac{14+15+15+15+16+16+16+16+17+17+17+18}{12}$

  $mean=\frac{192}{12}$

  $mean=16$

Now finding the absolute variations

  $\begin{array}{l}|14-16|=2\\ |15-16|=1\\ |16-16|=0\\ |17-16|=1\\ |18-16|=2\end{array}$

Now,

  $\text{mean absolute deviation}=\frac{2+1+1+1+0+0+0+0+1+1+1+2}{12}$

  $\text{mean absolute deviation}=\frac{10}{12}$

  $\text{mean absolute deviation}=0.8$

For the fifth period:

  $mean=\frac{14+15+15+16+16+17+17+17+18+18+19+19+20}{13}$

  $mean=\frac{221}{13}$

  $mean=17$

Now finding the absolute variations

  $\begin{array}{l}|14-17|=3\\ |15-17|=2\\ |16-17|=1\\ |17-17|=0\\ |18-17|=1\\ |19-17|=2\\ |20-17|=3\end{array}$

Now,

  $\text{mean absolute deviation}=\frac{3+2+2+1+1+0+0+0+1+1+2+2+3}{13}$

  $\text{mean absolute deviation}=\frac{18}{13}$

  $\text{mean absolute deviation}=1.4$

  $$

	Second Period	Fifth period
Mean	16	17
Mean absolute deviation	0.8	1.4

Second period has a mean of 16 with mean absolute deviation 0.8 and the fifth period has a mean of 17 with mean absolute variation 1.4.

The mean of fifth period is greater than second period; however the variation of fifth period is also greater which means that the scores are more spread out.

Second periods score centered around 16 with a variation of 0.8 and fifth period score centered around 17 with variation 1.4.