Solving Problems On Mean Deviation

 

Mean Deviation Practice Questions (Ungrouped Data)

Test your understanding of Mean Absolute Deviation with these 5 Canadian-themed problems. Click "Show Answer & Solution" to check your work!

Q1. A Toronto hockey player scored the following number of goals in 5 consecutive games: 1, 3, 0, 2, 4. What is the mean deviation of their goals?

• A) 1.2 goals
• B) 2.0 goals
• C) 1.5 goals
• D) 0.8 goals

Correct Answer: A) 1.2 goals

Solution:
1. Find the mean: (1 + 3 + 0 + 2 + 4) / 5 = 10 / 5 = 2.
2. Find absolute deviations from mean (2):
|1 - 2| = 1
|3 - 2| = 1
|0 - 2| = 2
|2 - 2| = 0
|4 - 2| = 2
3. Average the deviations: (1 + 1 + 2 + 0 + 2) / 5 = 6 / 5 = 1.2.

Q2. The daily afternoon temperatures recorded in Calgary over 4 days in November were -6°C, -2°C, 1°C, and 3°C. Calculate the mean deviation of the temperature.

• A) 2.5°C
• B) 3.0°C
• C) 3.5°C
• D) 4.0°C

Correct Answer: B) 3.0°C

Solution:
1. Find the mean: (-6 + -2 + 1 + 3) / 4 = -4 / 4 = -1°C.
2. Find absolute deviations from mean (-1):
|-6 - (-1)| = |-5| = 5
|-2 - (-1)| = |-1| = 1
|1 - (-1)| = |2| = 2
|3 - (-1)| = |4| = 4
3. Average the deviations: (5 + 1 + 2 + 4) / 4 = 12 / 4 = 3.0°C.

Q3. Five students in Vancouver recorded their weekly transit commute times: 20, 25, 30, 35, and 40 minutes. What is the mean deviation of their commute times?

• A) 0 minutes
• B) 5 minutes
• C) 6 minutes
• D) 10 minutes

Correct Answer: C) 6 minutes

Solution:
1. Find the mean: (20 + 25 + 30 + 35 + 40) / 5 = 150 / 5 = 30 minutes.
2. Find absolute deviations from mean (30):
|20 - 30| = 10; |25 - 30| = 5; |30 - 30| = 0; |35 - 30| = 5; |40 - 30| = 10.
3. Average the deviations: (10 + 5 + 0 + 5 + 10) / 5 = 30 / 5 = 6 minutes.

Q4. A Tim Hortons manager tracked the number of boxes of Timbits sold per hour over a 4-hour shift: 12, 18, 14, and 16. Calculate the mean deviation.

• A) 1.5 boxes
• B) 2.0 boxes
• C) 2.5 boxes
• D) 3.0 boxes

Correct Answer: B) 2.0 boxes

Solution:
1. Find the mean: (12 + 18 + 14 + 16) / 4 = 60 / 4 = 15.
2. Find absolute deviations from mean (15):
|12 - 15| = 3; |18 - 15| = 3; |14 - 15| = 1; |16 - 15| = 1.
3. Average the deviations: (3 + 3 + 1 + 1) / 4 = 8 / 4 = 2.0 boxes.

Q5. The price of a litre of gas at 6 different stations in Montreal was recorded as: $1.60, $1.65, $1.70, $1.60, $1.75, $1.66. What is the mean deviation of the gas prices?

• A) $0.05
• B) $0.06
• C) $0.04
• D) $0.10

Correct Answer: C) $0.04

Solution:
1. Find the mean: (1.60 + 1.65 + 1.70 + 1.60 + 1.75 + 1.66) / 6 = 9.96 / 6 = $1.66.
2. Find absolute deviations from mean ($1.66):
|1.60 - 1.66| = 0.06; |1.65 - 1.66| = 0.01; |1.70 - 1.66| = 0.04;
|1.60 - 1.66| = 0.06; |1.75 - 1.66| = 0.09; |1.66 - 1.66| = 0.00.
3. Average the deviations: (0.06 + 0.01 + 0.04 + 0.06 + 0.09 + 0.00) / 6 = 0.26 / 6 ≈ $0.043 (rounds to $0.04).