Question

Compare the distributions using either the means and standard deviations or the five-number summaries. Justify your choice. Set A Set B The distributions are symmetric, so use the means and standard deviations. The mean for Set A is about 44.6 with standard deviation of about 6.2. The mean for Set B is about 42.8 with standard deviation of about 1.86. While the average low temperatures for the cities are approximately equal, the greater standard deviation for Set B means that Set A’s low temperatures have a greater variability than Set B temperatures. The distributions are symmetric, so use the means and standard deviations. The mean for Set B is about 41.56 with standard deviation of about 6.07. The mean for Set A is about 43.8 with standard deviation of about 14.8. While the average low temperatures for the cities are approximately equal, the greater standard deviation for Set A means that Set A’s low temperatures have a greater variability than Set B temperatures. The distributions are symmetric, so use the means and standard deviations. The mean for Set A is about 44.6 with standard deviation of about 6.4. The mean for Set B is about 41.5 with standard deviation of about 6.7. While the average low temperatures for the cities are approximately equal, the greater standard deviation for Set B means that Set B’s low temperatures have a greater variability than Set A temperatures. The distributions are symmetric, so use the means and standard deviations. The mean for Set A is about 44.6 with standard deviation of about 6.2. The mean for Set B is about 43.8 with standard deviation of about 14.8. While the average low temperatures for the cities are approximately equal, the greater standard deviation for Set B means that Set B’s low temperatures have a greater variability than Set A temperatures.

Answer 1

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:

1. Minimum
2. First Quartile
3. Median
4. Third Quartile
5. Maximum

The five-number summary for set A 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 B 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:

$Mean_{A}=\frac{1}{12}\times [36+51+37+...+47]=44.58\approx 44.6\\\\Mean_{B}=\frac{1}{12}\times [22+57+46+...+38]=44.58\approx 44.6$

• Both the distribution has the same mean.

Compare mean and median for the two data:

$Mean_{A}>Median_{A}\\\\Mean_{B}>Median_{B}$

• This implies that set A is positively skewed whereas set B is negatively skewed.

Compute the standard deviation for both the set as follows:

$SD_{A}=\sqrt{\frac{1}{12-1}\times [(36-44.6)^{2}+...+(47-44.6)^{2}]}=6.44\approx 6.4\\\\SD_{B}=\sqrt{\frac{1}{12-1}\times [(22-44.6)^{2}+...+(38-44.6)^{2}]}=15.56\approx 15.6$

• The set B has a greater standard deviation that set A. Implying set B has a greater variability that set B.