Chapter 10.4, Problem 4IP

To determine

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

Answer to Problem 4IP

Girls have a mean height of 65 inches with a mean absolute deviation 0.8 inches and the boys have a mean height of 69 inches with mean absolute variation 1.4 inches.

The mean of boys is greater than girls; however the variation of boys is also greater which means that the heights of boys are more spread out.

Explanation of Solution

Given information: The double dot plots showing the heights of boys and girls.

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 Boys:

  $mean=\frac{66+67+67+68+68+69+69+69+70+70+71+71+72}{13}$

  $mean=\frac{897}{13}$

  $mean=69$

Now finding the absolute variations

  $\begin{array}{l}|66-69|=3\\ |67-69|=2\\ |68-69|=1\\ |69-69|=0\\ |70-69|=1\\ |71-69|=2\\ |72-69|=3\end{array}$

Now,

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

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

  $\text{mean absolute deviation}=1.4$

For the Girls:

  $mean=\frac{63+64+64+64+65+65+65+65+66+66+66+67}{12}$

  $mean=\frac{780}{12}$

  $mean=65$

Now finding the absolute variations

  $\begin{array}{l}|63-65|=2\\ |64-65|=1\\ |65-65|=0\\ |66-65|=1\\ |67-65|=2\end{array}$

Now,

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

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

  $\text{mean absolute deviation}=0.8$

  $$

	Girls	Boys
Mean	65	69
Mean absolute deviation	0.8	1.4

Girls have a mean height of 65 inches with a mean absolute deviation 0.8 inches and the boys have a mean height of 69 inches with mean absolute variation 1.4 inches.

The mean of boys is greater than girls; however the variation of boys is also greater which means that the heights of boys are more spread out.

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