Question

the most appropriate measure of center for this data is the
done
mean
median
mode
weight of male dogs (lb.)
4 5
5 0 2 8
6 2 8
7 2 8
8 1
9 5

Explanation:

Step1: Analyze the data distribution

First, we need to list out the data points from the stem - and - leaf plot. The stem - and - leaf plot for the weight of male dogs (lb.) is:

• Stem 4, leaf 5: 45
• Stem 5, leaves 0, 2, 8: 50, 52, 58
• Stem 6, leaves 2, 8: 62, 68
• Stem 7, leaves 2, 8: 72, 78
• Stem 8, leaf 1: 81
• Stem 9, leaf 5: 95

So the data set is \(45,50,52,58,62,68,72,78,81,95\)

Step2: Check for skewness

To determine the appropriate measure of center, we check the skewness of the data. Let's calculate the mean, median, and see the distribution.

First, calculate the median. Since there are \(n = 10\) data points (even number), the median is the average of the \(\frac{n}{2}=5^{th}\) and \((\frac{n}{2}+ 1)=6^{th}\) ordered values.

The ordered data set is \(45,50,52,58,62,68,72,78,81,95\)

The \(5^{th}\) value is \(62\) and the \(6^{th}\) value is \(68\). The median \(=\frac{62 + 68}{2}=\frac{130}{2}=65\)

Now, calculate the mean. The mean \(\bar{x}=\frac{\sum_{i = 1}^{n}x_{i}}{n}\)

\(\sum_{i=1}^{10}x_{i}=45 + 50+52 + 58+62+68+72+78+81+95\)

\(45+50 = 95\); \(95+52=147\); \(147 + 58 = 205\); \(205+62 = 267\); \(267+68 = 335\); \(335+72 = 407\); \(407+78 = 485\); \(485+81 = 566\); \(566+95=661\)

Mean \(\bar{x}=\frac{661}{10}=66.1\)

Now, let's check the skewness. The data has a tail on the right (since 95 is a relatively large value compared to the other values on the left). When data is skewed (especially right - skewed), the median is a more appropriate measure of center than the mean because the mean is affected by the extreme values (like 95 in this case). The mode is the most frequently occurring value. Let's check the frequency of each value: all values occur once, so there is no mode (or all values are modes with frequency 1). So, for a skewed data set, the median is the most appropriate measure of center.

Answer:

median