AP Statistics worksheet on measures of center, featuring problems on calculating mean, median, and trimmed mean for car repair costs and supervisor salaries.

Here are the step-by-step solutions for the problems on your worksheet.

1. Repair Costs (Forward Crash)

Step 1: List the data and order it.
First, let's list all the repair costs from the table to make them easier to work with:
$0, 0, 900, 1254, 234, 979, 707, 670, 769, 4194$

To find the median, we must put these numbers in order from smallest to largest:
$0, 0, 234, 670, 707, 769, 900, 979, 1254, 4194$

Step 2: Calculate the Mean.
The mean is the average. Add all the numbers together and divide by the count (10 cars).
Sum $= 0 + 0 + 234 + 670 + 707 + 769 + 900 + 979 + 1254 + 4194 = 9707$
Mean $= 9707 / 10 = \mathbf{970.7}$

Step 3: Calculate the Median.
The median is the middle number. Since there are 10 numbers (an even amount), we take the average of the two middle ones (the 5th and 6th numbers).
5th number: $707$
6th number: $769$
Median $= (707 + 769) / 2 = \mathbf{738}$

Step 4: Comparison.
The mean ($970.7$) is much higher than the median ($738$). This is because the Volvo S80 cost $4194, which is an outlier (a value very far away from the rest). This high number pulls the average up. Most cars cost less than $1000. Therefore, the median is more representative of a "typical" repair cost.

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2. Repair Costs (Backing Up Crash)

Data Set: $0, 1039, 711, 739, 557, 418, 1191, 1042, 322, 390$

a. Difference between Mean and Median
In Exercise 1, there was a huge outlier ($4194$) that made the mean much larger than the median. In this new data set, the numbers are spread out more evenly without one single extreme value pulling everything up. Therefore, the difference between the mean and median will be smaller.

b. Compute Mean and Median
First, order the data:
$0, 322, 390, 418, 557, 711, 739, 1039, 1042, 1191$

* Mean:
Sum $= 0 + 322 + 390 + 418 + 557 + 711 + 739 + 1039 + 1042 + 1191 = 6409$
Mean $= 6409 / 10 = \mathbf{640.9}$

* Median:
The two middle numbers are the 5th ($557$) and 6th ($711$).
Median $= (557 + 711) / 2 = \mathbf{634}$

* Interpretation:
The typical repair cost for backing into a barrier is around $\$634$. The average cost is slightly higher at $\$640.9$, likely due to a few expensive repairs like the Mercedes ($1191$) and BMW ($1042$).

c. Changing the first observation to 500
If we change the $0$ to $500$:
* Mean: The sum increases by $500$. The new mean would be $(6409 + 500) / 10 = 690.9$. The mean increases.
* Median: The ordered list changes slightly, but the two middle numbers ($557$ and $711$) stay exactly the same. The median does not change.

d. Trimmed Mean
We eliminate the smallest ($0$) and the largest ($1191$).
Remaining data: $322, 390, 418, 557, 711, 739, 1039, 1042$
Sum of remaining $= 5218$
Count of remaining $= 8$
Trimmed Mean $= 5218 / 8 = \mathbf{652.25}$

Trimming Percentage:
We removed 1 value from the low end and 1 from the high end out of 10 total values.
$1/10 = 10\%$. So, this is a 10% trimmed mean.

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3. Supervisor Salaries

Data:
Kern: $\$5354$
Monterey: $\$5166$
Santa Cruz: $\$4443$
Santa Barbara: $\$4129$
Placer: $\$2500$
Merced: $\$2220$

Order the salaries:
$\$2220, \$2500, \$4129, \$4443, \$5166, \$5354$

Which measure determines the salary?
The problem states the salary is the "average of the two counties... in the middle."
The two middle counties are Santa Barbara ($\$4129$) and Santa Cruz ($\$4443$).
Average $= (4129 + 4443) / 2 = \mathbf{4286}$

This calculation describes the Median. (When you have an even number of items, the median is the average of the two middle ones).

What is the other measure?
The other measure of center is the Mean (the arithmetic average of all six salaries).
Sum $= 23,812$
Mean $= 23,812 / 6 \approx \mathbf{3968.67}$

The mean is lower than the median because the two low salaries in Placer and Merced pull the average down. Taxpayers might prefer the mean because it suggests supervisors should be paid less.

Final Answer:

1.
* Mean: $970.7$
* Median: $738$
* Representative Measure: The median is more representative. The mean is skewed upward by the outlier (Volvo S80 at $4194), whereas the median reflects the cost of the majority of the cars better.

2.
* a. The difference would be smaller. The data in exercise 2 does not have an extreme outlier like the $4194 in exercise 1, so the mean and median are closer together.
* b. Mean: $640.9$; Median: $634$. Interpretation: The typical repair cost is about $634. The mean is slightly higher due to a few costly repairs.
* c. The mean would increase (to $690.9$), but the median would remain unchanged ($634$).
* d. Trimmed Mean: $652.25$. Trimming Percentage: $10\%$.

3.
* Measure determining salary: The Median. Its value is $4,286.
* Other measure: The Mean. Its value is approximately $3,968.67.