Answer:
Explained below.
Step-by-step explanation:
The question is:
Compare the distributions using either the means and standard deviations or the five-number summaries. Justify your choice.
Set A: {36, 51, 37, 42, 54, 39, 53, 42, 46, 38, 50, 47}
Set B: {22, 57, 46, 24, 31, 41, 64, 50, 28, 59, 65, 38}
The five-number summary is:
- Minimum
- First Quartile
- Median
- Third Quartile
- Maximum
The five-number summary for set <em>A</em> is:
Variable Minimum Q₁ Median Q₃ Maximum
Set A 36.00 38.25 44.00 50.75 54.00
The five-number summary for set <em>B</em> is:
Variable Minimum Q₁ Median Q₃ Maximum
Set B 22.00 28.75 48.00 58.50 65.00
Compute the mean for both the data as follows:
![Mean_{A}=\frac{1}{12}\times [36+51+37+...+47]=44.58\approx 44.6\\\\Mean_{B}=\frac{1}{12}\times [22+57+46+...+38]=44.58\approx 44.6](https://tex.z-dn.net/?f=Mean_%7BA%7D%3D%5Cfrac%7B1%7D%7B12%7D%5Ctimes%20%5B36%2B51%2B37%2B...%2B47%5D%3D44.58%5Capprox%2044.6%5C%5C%5C%5CMean_%7BB%7D%3D%5Cfrac%7B1%7D%7B12%7D%5Ctimes%20%5B22%2B57%2B46%2B...%2B38%5D%3D44.58%5Capprox%2044.6)
- Both the distribution has the same mean.
Compare mean and median for the two data:
![Mean_{A}>Median_{A}\\\\Mean_{B}>Median_{B}](https://tex.z-dn.net/?f=Mean_%7BA%7D%3EMedian_%7BA%7D%5C%5C%5C%5CMean_%7BB%7D%3EMedian_%7BB%7D)
- This implies that set A is positively skewed whereas set B is negatively skewed.
Compute the standard deviation for both the set as follows:
![SD_{A}=\sqrt{\frac{1}{12-1}\times [(36-44.6)^{2}+...+(47-44.6)^{2}]}=6.44\approx 6.4\\\\SD_{B}=\sqrt{\frac{1}{12-1}\times [(22-44.6)^{2}+...+(38-44.6)^{2}]}=15.56\approx 15.6](https://tex.z-dn.net/?f=SD_%7BA%7D%3D%5Csqrt%7B%5Cfrac%7B1%7D%7B12-1%7D%5Ctimes%20%5B%2836-44.6%29%5E%7B2%7D%2B...%2B%2847-44.6%29%5E%7B2%7D%5D%7D%3D6.44%5Capprox%206.4%5C%5C%5C%5CSD_%7BB%7D%3D%5Csqrt%7B%5Cfrac%7B1%7D%7B12-1%7D%5Ctimes%20%5B%2822-44.6%29%5E%7B2%7D%2B...%2B%2838-44.6%29%5E%7B2%7D%5D%7D%3D15.56%5Capprox%2015.6)
- The set B has a greater standard deviation that set A. Implying set B has a greater variability that set B.