SOLUTION: A year ago, Madison began working at a computer store. Her supervisor asked her to keep a record of the number of sales she made each month. The following data set is a list of her

Question 1205734: A year ago, Madison began working at a computer store. Her supervisor asked her to keep a record of the number of sales she made each month. The following data set is a list of her sales for the last 12 months:
34, 47, 1, 15, 57, 24, 20, 11, 19, 50, 28, 37
a. Determine the standard deviation of the data.
b. Determine the IQR for the above data. What does this mean in the context of the question?
c. Identify any outliers any the data. What measure of central tendency and/or spread does this affect the most? Do you think it would make sense for the supervisor not to consider this value when assessing her job performance? Be specific.
(Please help)
Answer by math_tutor2020(3817) (Show Source): You can put this solution on YOUR website!

Part (a)

Likely your teacher is asking for the sample standard deviation.
If your teacher wants the population version then please let me know.

The first thing we need is the sample mean known as xbar.
Add up the values and divide by the sample size 12.
xbar = (34+47+1+15+57+24+20+11+19+50+28+37)/12 = 28.583333 approximately

Next, we'll subtract the sample mean from each x value. Then square the result.

x	(x-xbar)^2
34	29.340278
47	339.173611
1	760.840278
15	184.506944
57	807.506944
24	21.006944
20	73.673611
11	309.173611
19	91.840278
50	458.673611
28	0.340278
37	70.840278

Each decimal value is approximate.
It is strongly recommended to use a spreadsheet to keep the data organized.
A spreadsheet can make quick work of those computations.

Add up all of the values in the 2nd column. The spreadsheet command SUM is very useful here.
Those values add to 3146.916666 roughly. This is the Sum of the Squared Error (SSE).

Divide the SSE over n-1 = 12-1 = 11 to get the sample variance.
SSE/11 = 3146.916666/11 = 286.083333273 approximately

Lastly, take the square root of the sample variance to get the sample standard deviation.
sqrt(286.083333273) = 16.913998
Like with any other decimal value mentioned, the result is approximate. Round it however your teacher instructs.

Here's another similar problem for more practice
https://www.algebra.com/algebra/homework/Probability-and-statistics/Probability-and-statistics.faq.question.1202737.html

The process discussed above is quite lengthy.
That's why often the simple more preferred route is to use the built-in standard deviation function on the calculator.
If you don't have a calculator or don't have a calculator with such a feature, then many free online tools are available.
Examples are WolframAlpha and GeoGebra.

The standard deviation and variance are measures of spread. The larger the value, the more spread out the data will be.

==================================================================================================================
Part (b)

Original set = {34, 47, 1, 15, 57, 24, 20, 11, 19, 50, 28, 37}

Sorted set = {1, 11, 15, 19, 20, 24, 28, 34, 37, 47, 50, 57}

Since the sample size (n = 12) is an even number, the midpoint is going to be a tie between slots 6 and 7 of the sorted set.
The 6 is from calculating n/2 = 12/2 = 6.

The values at slots 6 and 7 are 24 and 28 respectively.
The midpoint of those values is (24+28)/2 = 26.
The median of the set is 26.

Form two subsets L and U
L = lower set of values smaller than the median
U = upper set of values larger than the median

L = {1, 11, 15, 19, 20, 24}
U = {28, 34, 37, 47, 50, 57}
Both subsets have 6 elements each.

Find the median of set L and set U to get (15+19)/2 = 17 and (37+47)/2 = 42 respectively.
The values 17 and 42 represent Q1 and Q3.

Q1 = first quartile = 17
Q3 = third quartile = 42

Five Number Summary:
min = 1
Q1 = 17
median = 26
Q3 = 42
max = 57

The Q1 and Q3 values help us find the IQR (interquartile range).
IQR = Q3 - Q1
IQR = 42 - 17
IQR = 25

The IQR is another measure of spread. The larger the IQR, the more spread out the values. It's similar to how the standard deviation and variance are also measures of spread.

In the case of the IQR, 50% of the values are between the endpoints Q1 and Q3. We consider this subset the middle 50% of values.

==================================================================================================================
Part (c)

There are two standard ways to determine outliers.

Method 1) Using the standard deviation
Method 2) Using the IQR

------------------------------

Method 1

Refer to part (a).
We calculated the sample mean to be roughly 28.583333
Also, the standard deviation is roughly 16.913998

For a normal distribution, aka bell curve, it turns out that values beyond 3 standard deviations of the mean are outliers.
The vast majority of the data (roughly 99.7%) is within 3 standard deviations of the mean.
Refer to the Empirical Rule.

Let's compute the lower fence (LF) and upper fence (UF)
LF = sampleMean - 3*standardDeviation
LF = 28.583333 - 3*16.913998
LF = -22.158661

UF = sampleMean + 3*standardDeviation
UF = 28.583333 + 3*16.913998
UF = 79.325327

Values between LF and UF are non-outliers.
Values outside these boundaries are outliers.

Looking through the given original data set, we see all of the values are between LF = -22.158661 and UF = 79.325327

Therefore, according to method 1, there aren't any outliers.

------------------------------

Method 2

I'll use Tukey's Rule here.
LF = Q1 - 1.5*IQR
LF = 17 - 1.5*25
LF = -20.5
and
UF = Q3 + 1.5*IQR
UF = 42 + 1.5*25
UF = 79.5

The LF and UF values here are somewhat close to the values from method 1.

Like with the previous method, all values from the original set are between LF and UF.

Therefore, this set has no outliers according to both methods discussed above.

------------------------------

If there were outliers, then the data would be either left-skewed (aka negatively skewed) or right-skewed (positively skewed)

Skewed data sets will be less reliant on the mean and we'll go for the median instead.
In turn it indicates that skewed data sets will not use method 1 as much.
Since there aren't any outliers, we do not have skewed data.

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