Math worksheet focusing on calculating mean absolute deviation for various data sets, including sports scores and student prizes.

Here are the step-by-step solutions for each problem on the worksheet.

Problem 1

Step 1: Find the Mean Absolute Deviation (MAD) for Data Set A
Data Set A: $23, 49, 58, 72, 81, 89, 90$

1. Find the Mean (Average):
Add all the numbers together and divide by the count (7).
$$23 + 49 + 58 + 72 + 81 + 89 + 90 = 462$$
$$462 \div 7 = 66$$
The mean is 66.

2. Find the Absolute Deviations:
Find the distance of each number from the mean ($|x - 66|$).
* $|23 - 66| = 43$
* $|49 - 66| = 17$
* $|58 - 66| = 8$
* $|72 - 66| = 6$
* $|81 - 66| = 15$
* $|89 - 66| = 23$
* $|90 - 66| = 24$

3. Find the Mean of these Deviations:
Add the deviations and divide by 7.
$$43 + 17 + 8 + 6 + 15 + 23 + 24 = 136$$
$$136 \div 7 \approx 19.428...$$
Rounded to two decimal places: 19.43

Step 2: Find the Mean Absolute Deviation (MAD) for Data Set B
Data Set B: $25, 46, 54, 61, 70, 79, 85$

1. Find the Mean:
$$25 + 46 + 54 + 61 + 70 + 79 + 85 = 420$$
$$420 \div 7 = 60$$
The mean is 60.

2. Find the Absolute Deviations:
Find the distance of each number from the mean ($|x - 60|$).
* $|25 - 60| = 35$
* $|46 - 60| = 14$
* $|54 - 60| = 6$
* $|61 - 60| = 1$
* $|70 - 60| = 10$
* $|79 - 60| = 19$
* $|85 - 60| = 25$

3. Find the Mean of these Deviations:
$$35 + 14 + 6 + 1 + 10 + 19 + 25 = 110$$
$$110 \div 7 \approx 15.714...$$
Rounded to two decimal places: 15.71

Comparison:
$19.43 > 15.71$, so the MAD for A is greater than the MAD for B.

***

Problem 2

Step 1: Find the MAD for Team A
Team A Points: $42, 54, 54, 50, 57, 58$

1. Find the Mean:
$$42 + 54 + 54 + 50 + 57 + 58 = 315$$
$$315 \div 6 = 52.5$$
The mean is 52.5.

2. Find the Absolute Deviations:
* $|42 - 52.5| = 10.5$
* $|54 - 52.5| = 1.5$
* $|54 - 52.5| = 1.5$
* $|50 - 52.5| = 2.5$
* $|57 - 52.5| = 4.5$
* $|58 - 52.5| = 5.5$

3. Find the Mean of these Deviations:
$$10.5 + 1.5 + 1.5 + 2.5 + 4.5 + 5.5 = 26$$
$$26 \div 6 \approx 4.333...$$
Rounded to two decimal places: 4.33

Step 2: Find the MAD for Team B
Team B Points: $55, 15, 82, 65, 44, 99$

1. Find the Mean:
$$55 + 15 + 82 + 65 + 44 + 99 = 360$$
$$360 \div 6 = 60$$
The mean is 60.

2. Find the Absolute Deviations:
* $|55 - 60| = 5$
* $|15 - 60| = 45$
* $|82 - 60| = 22$
* $|65 - 60| = 5$
* $|44 - 60| = 16$
* $|99 - 60| = 39$

3. Find the Mean of these Deviations:
$$5 + 45 + 22 + 5 + 16 + 39 = 132$$
$$132 \div 6 = 22$$
Rounded to two decimal places: 22.00

Comparison:
$4.33 < 22.00$, so the MAD for Team A is less than the MAD for Team B.

***

Problem 3

Step 1: Find the MAD with '65'
Data Set: $15, 22, 12, 73, 44, 56, 65$

1. Find the Mean:
$$15 + 22 + 12 + 73 + 44 + 56 + 65 = 287$$
$$287 \div 7 = 41$$
The mean is 41.

2. Find the Absolute Deviations:
* $|15 - 41| = 26$
* $|22 - 41| = 19$
* $|12 - 41| = 29$
* $|73 - 41| = 32$
* $|44 - 41| = 3$
* $|56 - 41| = 15$
* $|65 - 41| = 24$

3. Find the Mean of these Deviations:
$$26 + 19 + 29 + 32 + 3 + 15 + 24 = 148$$
$$148 \div 7 \approx 21.142...$$
Rounded to two decimal places: 21.14

Step 2: Find the MAD without '65'
New Data Set: $15, 22, 12, 73, 44, 56$ (6 numbers)

1. Find the New Mean:
$$15 + 22 + 12 + 73 + 44 + 56 = 222$$
$$222 \div 6 = 37$$
The new mean is 37.

2. Find the New Absolute Deviations:
* $|15 - 37| = 22$
* $|22 - 37| = 15$
* $|12 - 37| = 25$
* $|73 - 37| = 36$
* $|44 - 37| = 7$
* $|56 - 37| = 19$

3. Find the Mean of these Deviations:
$$22 + 15 + 25 + 36 + 7 + 19 = 124$$
$$124 \div 6 \approx 20.666...$$
Rounded to two decimal places: 20.67

Comparison:
$21.14 > 20.67$, so the MAD with '65' is greater than the MAD without '65'.

──────────────────────────────────────

Final Answer:

1)
Mean Absolute Deviation for A = 19.43
Mean Absolute Deviation for B = 15.71
Mean Absolute Deviation for A > Mean Absolute Deviation for B

2)
Mean Absolute Deviation for Team A = 4.33
Mean Absolute Deviation for Team B = 22.00
Mean Absolute Deviation for Team A < Mean Absolute Deviation for Team B

3)
Mean Absolute Deviation with ‘65’ = 21.14
Mean Absolute Deviation without ‘65’ = 20.67
Mean Absolute Deviation with ‘65’ > Mean Absolute Deviation without ‘65’