Mathematics, 14.06.2021 21:10 anaiyamills

Dannette and Alphonso work for a computer repair company. They must include the time it takes to complete each repair in their repair log book. The dot plots show the number of hours each of their last 12 repairs took. Part a. Calculate the median, mean, IQR, and standard deviation of each data set. Part b. Which measure of central tendency and spread should you use to compare the two data sets? Explain your reasoning. Part c. Determine whether there are any outliers in either data set. Dannette's Repair Times х х X X X X Х Х + 9 + 1 0 Relations 2 3 4 8 10 12 5 6 7 Repair Time (hours) Geometry Alphonso's Repair Times Groups X Trigonometry X Х X X X х X х Statistics 7 X + 3 10 9 0 4 12 Series 8 1 2 5 7 Repair Time (hours) Greek PLZ HELP

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Answer from: Arianahinton9856

(a):

Dannette Alphonso

$\bar x_D = 4.33$ $\bar x_A = 5.17$

M_D = 2.5 M_A = 5

$\sigma_D = 3.350$ $\sigma_A = 1.951$

IQR_D = 7 IQR_A = 1.5

(b):

Measure of center: Median

Measure of spread: Interquartile range

(c):

There are no outliers in Dannette's dataset

There are outliers in Alphonso's dataset

Step-by-step explanation:

Given

See attachment for the appropriate data presentation

Solving (a): Mean, Median, Standard deviation and IQR of each

From the attached plots, we have:

IQR_A = 1.5 ---- Dannette

$A = \{3,4,4,4,4,5,5,5,5,6,6,11\}$ ---- Alphonso

n = 12 --- number of dataset

Mean

The mean is calculated

$\bar x = \frac{\sum x}{n}$

So, we have:

$\bar x_D = \frac{1+1+1+1+2+2+3+7+8+8+9+9}{12}$

$\bar x_D = \frac{52}{12}$

$\bar x_D = 4.33$ --- Dannette

$\bar x_A = \frac{3+4+4+4+4+5+5+5+5+6+6+11}{12}$

$\bar x_A = \frac{62}{12}$

$\bar x_A = 5.17$ --- Alphonso

Median

The median is calculated as:

$M = \frac{n + 1}{2}th$

$M = \frac{12 + 1}{2}th$

$M = \frac{13}{2}th$

M = 6.5th

This implies that the median is the mean of the 6th and the 7th item.

So, we have:

$M_D = \frac{2+3}{2}$

$M_D = \frac{5}{2}$

M_D = 2.5 ---- Dannette

$M_A = \frac{5+5}{2}$

$M_A = \frac{10}{2}$

M_A = 5 ---- Alphonso

Standard Deviation

This is calculated as:

$\sigma = \sqrt{\frac{\sum(x - \bar x)^2}{n}}$

So, we have:

$\sigma_D = \sqrt{\frac{(1 - 4.33)^2 +.............+(9- 4.33)^2}{12}}$

$\sigma_D = \sqrt{\frac{134.6668}{12}}$

$\sigma_D = 3.350$ ---- Dannette

$\sigma_A = \sqrt{\frac{(3-5.17)^2+............+(11-5.17)^2}{12}}$

$\sigma_A = \sqrt{\frac{45.6668}{12}}$

$\sigma_A = 1.951$ --- Alphonso

The Interquartile Range (IQR)

This is calculated as:

IQR =Q_3 - Q_1

Where

$Q_3 \to$ Upper Quartile and $Q_1 \to$ Lower Quartile

Q_3 is calculated as:

$Q_3 = \frac{3}{4}*({n + 1})th$

$Q_3 = \frac{3}{4}*(12 + 1})th$

$Q_3 = \frac{3}{4}*13th$

Q_3 = 9.75th

This means that Q_3 is the mean of the 9th and 7th item. So, we have:

$Q_3 = \frac{1}{2} * (8+8) = \frac{1}{2} * 16$ $Q_3 = \frac{1}{2} * (5+6) = \frac{1}{2} * 11$

Q_3 = 8 ---- Dannette Q_3 = 5.5 --- Alphonso

Q_1 is calculated as:

$Q_1 = \frac{1}{4}*({n + 1})th$

$Q_1 = \frac{1}{4}*({12 + 1})th$

$Q_1 = \frac{1}{4}*13th$

Q_1 = 3.25th

This means that Q_1 is the mean of the 3rd and 4th item. So, we have:

$Q_1 = \frac{1}{2}(1+1) = \frac{1}{2} * 2$ $Q_1 = \frac{1}{2}(4+4) = \frac{1}{2} * 8$

Q_1 = 1 --- Dannette Q_1 = 4 ---- Alphonso

So, the IQR is:

IQR = Q_3 - Q_1

IQR_D = 8 - 1 IQR_A = 5.5 - 4

IQR_D = 7 --- Dannette IQR_A = 1.5 --- Alphonso

Solving (b): The measures to compare

Measure of center

By observation, we can see that there are outliers is the plot of Alphonso (because 11 is far from the other dataset) while there are no outliers in Dannette plot (as all data are close).

Since, the above is the case; we simply compare the median of both because it is not affected by outliers

Measure of spread

Compare the interquartile range of both, as it is arguably the best measure of spread, because it is also not affected by outliers.

Solving (c): Check for outlier

To check for outlier, we make use of the following formulas:

Lower =Q_1 - 1.5 * IQR

Upper =Q_3 + 1.5 * IQR

For Dannette:

Lower = 1 - 1.5 * 7 = -9.5

Upper = 8 + 1.5 * 7 = 18.5

Since, the dataset are all positive, we change the lower outlier to 0.

So, the valid data range are:

$Valid = 0 \to 18.5$

From the question, the range of Dannette's dataset is: 1 to 9. Hence, there are no outliers in Dannette's dataset

For Alphonso:

Lower = 4 - 1.5 * 1.5 =1.75

Upper = 5.5 + 1.5 * 1.5 =7.75

So, the valid data range are:

$Valid = 1.75\to 7.75$

From the question, the range of Alphonso's dataset is: 3 to 11. Hence, there are outliers in Alphonso's dataset

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