In the first part of this problem, we must compute some statistic variables of two distributions:
0. the mean value,
,1. the median,
,2. the standard deviation.
,3. the interquartile range.
1. The mean of a data set is the sum of all the data divided by the count n:
[tex]\mu=\frac{x_1+x_2+\cdots+x_n}{n}\text{.}[/tex]2. The median is the data value separating the upper half of a data set from the lower half, it is computed following these steps:
• arrange data values from lowest to the highest value,
,• the median is the data value in the middle of the set
,• if there are 2 data values in the middle the median is the mean of those 2 values.
3. The standard deviation for a sample data set is given by the following formula:
[tex]\sigma=\sqrt[]{\frac{(x_1-\mu)^2+(_{}x_2-\mu)^2+\cdots+(x_n-\mu)^2}{n-1}_{}}\text{.}[/tex]4. The interquartile range (IQR) is given by:
[tex]\text{IQR}=Q_3-Q_1\text{.}[/tex]Where Q_1 and Q_3 are the first and third quartiles. The lowest quartile (Q1) covers the smallest quarter of values in your dataset.
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Using the definitions above, we compute the mean, the median and the standard deviation for the samples taken by Manuel and Gretchen.
Manuel's sample
• Sample = {3, 6, 8, 11, 12, 8, 6, 3, 10, 5, 14, 9, 7, 10, 8}
,• Count = 15
1. Mean
Using the formula above, we get:
[tex]\mu=\frac{120}{5}=8.[/tex]2. Median
We order the data set:
[tex]3,3,5,6,6,7,8,(8),8,9,10,10,11,12,14.[/tex]From the ordered data set, we see that the central number 8 divides the data set into two equal parts.
So the median of this sample is:
[tex]\bar{x}=8.[/tex]3. Standard deviation
Using the formula above, we get:
[tex]\sigma=\sqrt[]{\frac{138}{15-1}}\cong3.14.[/tex]4. Interquartile range
Dividing the data sample into quartiles, we have:
[tex]3,3,5,6|6,7,8|8|8,9,10|10,11,12,14.[/tex]We have:
• Q_1 = 6,
,• Q_3 = 10.
So the interquartile range is:
[tex]\text{IQR }=Q_3-Q_1=10-6=4.[/tex]Gretchen's sample
• Sample = {22, 4, 7, 8, 12, 15, 10, 7, 9, 6, 13, 3, 8, 10, 10}
,• Count = 15
1. Mean
[tex]\mu=\frac{144}{15}=9.6.[/tex]2. Median
We order the data set:
[tex]3,4,6,7,7,8,8,(9),10,10,10,12,13,15,22.[/tex]From the ordered data set, we see that the central number 8 divides the data set into two equal parts.
So the median of this sample is:
[tex]\bar{x}=9.[/tex]3. Standard deviation
[tex]\sigma=\sqrt[]{\frac{307.6}{15-1}}\cong4.69.[/tex]4. Interquartile range
Dividing the data sample into quartiles, we have:
[tex]3,4,6,7|7,8,8|9|10,10,10|12,13,15,22.[/tex]We have:
• Q_1 = 7,
,• Q_3 = 12.
So the interquartile range is:
[tex]\text{IQR }=Q_3-Q_1=12-7=5.[/tex]Answers
Manuel's sample
0. Mean = 8
,1. Median = 8
,2. Standard deviation ≅ 3.14
,3. Interquartile range = 4
Gretchen's sample
0. Mean = 9.6
,1. Median = 9
,2. Standard deviation ≅ 4.69
,3. Interquartile range = 5
