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The application of statistics and probability is essential for refrigeration and air conditioning professionals to analyze performance data, predict system reliability, and optimize maintenance schedules. This chapter focuses on fundamental statistical measures: mean, median, mode, and standard deviation, which are crucial for interpreting data collected from equipment operation, energy consumption, and environmental conditions. Mastery of these concepts enables technicians and engineers to make informed decisions based on empirical data, improving system efficiency and reducing downtime.
The mean or average is the sum of all observed values divided by the number of observations. It provides a central value that represents the dataset, useful for assessing average temperatures, pressures, or energy consumption in refrigeration systems. The formula for the mean \( \bar{x} \) is:
$$ \bar{x} = \frac{\Sigma x_i}{n} $$
where \( x_i \) are the data points and \( n \) is the total number of data points.
Example 1: A refrigeration technician records the daily energy consumption (in kWh) of a cold room over 5 days as follows: 45, 48, 46, 44, 47. Find the mean energy consumption.
Given: \( x_1=45, x_2=48, x_3=46, x_4=44, x_5=47 \), \( n=5 \)
$$ \bar{x} = \frac{45 + 48 + 46 + 44 + 47}{5} $$
$$ \bar{x} = \frac{230}{5} $$
$$ \bar{x} = 46 \text{ kWh} $$
Answer: 46 kWh
Example 2: A technician measures the pressure (in kPa) at 6 different points in a refrigeration system: 120, 115, 118, 122, 117, 119. Calculate the mean pressure.
Given: \( x_1=120, x_2=115, x_3=118, x_4=122, x_5=117, x_6=119 \), \( n=6 \)
$$ \bar{x} = \frac{120 + 115 + 118 + 122 + 117 + 119}{6} $$
$$ \bar{x} = \frac{711}{6} $$
$$ \bar{x} = 118.5 \text{ kPa} $$
Answer: 118.5 kPa
Example 3: The ambient temperatures (in °C) recorded hourly in a supermarket refrigeration room are: 2, 3, 4, 5, 3, 2, 6, 4. Find the mean temperature.
Given: \( x_1=2, x_2=3, x_3=4, x_4=5, x_5=3, x_6=2, x_7=6, x_8=4 \), \( n=8 \)
$$ \bar{x} = \frac{2 + 3 + 4 + 5 + 3 + 2 + 6 + 4}{8} $$
$$ \bar{x} = \frac{29}{8} $$
$$ \bar{x} = 3.625 \text{ °C} $$
Answer: 3.625 °C
Example 4: The hourly humidity readings (%) in a cold storage facility are: 85, 87, 82, 88, 90, 86, 84, 83, 89. Calculate the mean humidity.
Given: \( x_1=85, x_2=87, x_3=82, x_4=88, x_5=90, x_6=86, x_7=84, x_8=83, x_9=89 \), \( n=9 \)
$$ \bar{x} = \frac{85 + 87 + 82 + 88 + 90 + 86 + 84 + 83 + 89}{9} $$
$$ \bar{x} = \frac{774}{9} $$
$$ \bar{x} = 86 \% $$
Answer: 86%
Example 5: A technician measures the compressor discharge temperature (°C) at 7 different time intervals: 95, 97, 100, 98, 96, 99, 101. Find the mean discharge temperature.
Given: \( x_1=95, x_2=97, x_3=100, x_4=98, x_5=96, x_6=99, x_7=101 \), \( n=7 \)
$$ \bar{x} = \frac{95 + 97 + 100 + 98 + 96 + 99 + 101}{7} $$
$$ \bar{x} = \frac{686}{7} $$
$$ \bar{x} = 98 \text{ °C} $$
Answer: 98 °C
The median is the middle value in a data set arranged in ascending or descending order. It is useful when data contains outliers or skewed values, common in temperature or pressure measurements affected by transient conditions. The median divides the data into two equal halves.
Example 1: Find the median of the following compressor pressures (kPa): 110, 115, 112, 118, 116.
Given: \( n=5 \), data: 110, 115, 112, 118, 116
Order data: 110, 112, 115, 116, 118
Median is the middle value (3rd value):
Answer: 115 kPa
Example 2: Find the median of the following temperature readings (°C): 4, 6, 2, 7, 3, 5.
Given: \( n=6 \), data: 4, 6, 2, 7, 3, 5
Order data: 2, 3, 4, 5, 6, 7
Median is average of 3rd and 4th values:
$$ \text{Median} = \frac{4 + 5}{2} $$
$$ \text{Median} = \frac{9}{2} $$
$$ \text{Median} = 4.5 \text{ °C} $$
Answer: 4.5 °C
Example 3: A technician records humidity levels (%) as follows: 80, 85, 82, 83, 88, 90, 87. Find the median humidity.
Given: \( n=7 \), data: 80, 85, 82, 83, 88, 90, 87
Order data: 80, 82, 83, 85, 87, 88, 90
Median is the 4th value:
Answer: 85%
Example 4: Find the median of the following energy consumption data (kWh): 50, 52, 47, 49, 55, 53, 48, 51.
Given: \( n=8 \), data: 50, 52, 47, 49, 55, 53, 48, 51
Order data: 47, 48, 49, 50, 51, 52, 53, 55
Median is average of 4th and 5th values:
$$ \text{Median} = \frac{50 + 51}{2} $$
$$ \text{Median} = \frac{101}{2} $$
$$ \text{Median} = 50.5 \text{ kWh} $$
Answer: 50.5 kWh
Example 5: Find the median of the following compressor discharge temperatures (°C): 95, 97, 93, 99, 96, 98, 94.
Given: \( n=7 \), data: 95, 97, 93, 99, 96, 98, 94
Order data: 93, 94, 95, 96, 97, 98, 99
Median is the 4th value:
Answer: 96 °C
The mode is the value that appears most frequently in a data set. It helps identify the most common operating condition or frequently observed measurement in refrigeration systems, such as most common fault codes or temperature settings. A data set may have one mode (unimodal), more than one mode (bimodal or multimodal), or no mode if all values are unique.
Example 1: Find the mode of the following temperature readings (°C): 4, 5, 4, 6, 7, 4, 5.
Given: data: 4, 5, 4, 6, 7, 4, 5
Number of occurrences: 4 appears 3 times, 5 appears 2 times, others less.
Answer: Mode = 4 °C
Example 2: Find the mode of the following energy consumption data (kWh): 50, 52, 53, 50, 55, 52, 52.
Given: data: 50, 52, 53, 50, 55, 52, 52
Occurrences: 52 appears 3 times, 50 twice, others once.
Answer: Mode = 52 kWh
Example 3: Find the mode of the following pressure readings (kPa): 115, 118, 115, 120, 118, 118, 122.
Given: data: 115, 118, 115, 120, 118, 118, 122
Occurrences: 118 appears 3 times, 115 twice, others once.
Answer: Mode = 118 kPa
Example 4: Find the mode of the following humidity levels (%): 85, 87, 85, 90, 88, 87, 87.
Given: data: 85, 87, 85, 90, 88, 87, 87
Occurrences: 87 appears 4 times, 85 twice, others once.
Answer: Mode = 87%
Example 5: Find the mode of the following compressor discharge temperatures (°C): 95, 96, 97, 95, 97, 98, 99.
Given: data: 95, 96, 97, 95, 97, 98, 99
Occurrences: 95 appears twice, 97 twice, others once.
Answer: Modes = 95 °C and 97 °C (Bimodal)
The standard deviation measures the spread or variability of data around the mean. It is crucial in refrigeration and air conditioning to assess consistency in system parameters like temperature or pressure, indicating stability or identifying anomalies. The formula for the sample standard deviation \( s \) is:
$$ s = \sqrt{\frac{\Sigma (x_i - \bar{x})^2}{n - 1}} $$
where \( x_i \) are data points, \( \bar{x} \) is the mean, and \( n \) is the number of data points.
Example 1: Calculate the standard deviation of energy consumption (kWh) over 4 days: 45, 48, 46, 44.
Given: data: 45, 48, 46, 44, \( n=4 \)
Step 1: Calculate mean
$$ \bar{x} = \frac{45 + 48 + 46 + 44}{4} = \frac{183}{4} = 45.75 $$
Step 2: Calculate squared deviations
$$ (45 - 45.75)^2 = (-0.75)^2 = 0.5625 $$
$$ (48 - 45.75)^2 = 2.25^2 = 5.0625 $$
$$ (46 - 45.75)^2 = 0.25^2 = 0.0625 $$
$$ (44 - 45.75)^2 = (-1.75)^2 = 3.0625 $$
Step 3: Sum squared deviations
$$ 0.5625 + 5.0625 + 0.0625 + 3.0625 = 8.75 $$
Step 4: Divide by \( n-1 = 3 \)
$$ \frac{8.75}{3} = 2.9167 $$
Step 5: Take square root
$$ s = \sqrt{2.9167} = 1.708 $$
Answer: Standard deviation = 1.708 kWh
Example 2: Calculate the standard deviation of compressor pressures (kPa): 120, 115, 118, 122, 117.
Given: \( n=5 \), data: 120, 115, 118, 122, 117
Step 1: Mean
$$ \bar{x} = \frac{120 + 115 + 118 + 122 + 117}{5} = \frac{592}{5} = 118.4 $$
Step 2: Squared deviations
$$ (120 - 118.4)^2 = 1.6^2 = 2.56 $$
$$ (115 - 118.4)^2 = (-3.4)^2 = 11.56 $$
$$ (118 - 118.4)^2 = (-0.4)^2 = 0.16 $$
$$ (122 - 118.4)^2 = 3.6^2 = 12.96 $$
$$ (117 - 118.4)^2 = (-1.4)^2 = 1.96 $$
Step 3: Sum squared deviations
$$ 2.56 + 11.56 + 0.16 + 12.96 + 1.96 = 29.2 $$
Step 4: Divide by \( n-1 = 4 \)
$$ \frac{29.2}{4} = 7.3 $$
Step 5: Square root
$$ s = \sqrt{7.3} = 2.702 $$
Answer: Standard deviation = 2.702 kPa
Example 3: Calculate the standard deviation of hourly temperatures (°C): 2, 3, 4, 5, 3, 2.
Given: \( n=6 \), data: 2, 3, 4, 5, 3, 2
Step 1: Mean
$$ \bar{x} = \frac{2 + 3 + 4 + 5 + 3 + 2}{6} = \frac{19}{6} = 3.167 $$
Step 2: Squared deviations
$$ (2 - 3.167)^2 = (-1.167)^2 = 1.362 $$
$$ (3 - 3.167)^2 = (-0.167)^2 = 0.028 $$
$$ (4 - 3.167)^2 = 0.833^2 = 0.694 $$
$$ (5 - 3.167)^2 = 1.833^2 = 3.361 $$
$$ (3 - 3.167)^2 = (-0.167)^2 = 0.028 $$
$$ (2 - 3.167)^2 = (-1.167)^2 = 1.362 $$
Step 3: Sum squared deviations
$$ 1.362 + 0.028 + 0.694 + 3.361 + 0.028 + 1.362 = 6.835 $$
Step 4: Divide by \( n-1 = 5 \)
$$ \frac{6.835}{5} = 1.367 $$
Step 5: Square root
$$ s = \sqrt{1.367} = 1.169 $$
Answer: Standard deviation = 1.169 °C
Example 4: Calculate the standard deviation of humidity readings (%) in a cold storage: 85, 87, 82, 88, 90.
Given: \( n=5 \), data: 85, 87, 82, 88, 90
Step 1: Mean
$$ \bar{x} = \frac{85 + 87 + 82 + 88 + 90}{5} = \frac{432}{5} = 86.4 $$
Step 2: Squared deviations
$$ (85 - 86.4)^2 = (-1.4)^2 = 1.96 $$
$$ (87 - 86.4)^2 = 0.6^2 = 0.36 $$
$$ (82 - 86.4)^2 = (-4.4)^2 = 19.36 $$
$$ (88 - 86.4)^2 = 1.6^2 = 2.56 $$
$$ (90 - 86.4)^2 = 3.6^2 = 12.96 $$
Step 3: Sum squared deviations
$$ 1.96 + 0.36 + 19.36 + 2.56 + 12.96 = 37.2 $$
Step 4: Divide by \( n-1 = 4 \)
$$ \frac{37.2}{4} = 9.3 $$
Step 5: Square root
$$ s = \sqrt{9.3} = 3.05 $$
Answer: Standard deviation = 3.05 %
Example 5: Calculate the standard deviation of compressor discharge temperatures (°C): 95, 97, 100, 98, 96, 99.
Given: \( n=6 \), data: 95, 97, 100, 98, 96, 99
Step 1: Mean
$$ \bar{x} = \frac{95 + 97 + 100 + 98 + 96 + 99}{6} = \frac{585}{6} = 97.5 $$
Step 2: Squared deviations
$$ (95 - 97.5)^2 = (-2.5)^2 = 6.25 $$
$$ (97 - 97.5)^2 = (-0.5)^2 = 0.25 $$
$$ (100 - 97.5)^2 = 2.5^2 = 6.25 $$
$$ (98 - 97.5)^2 = 0.5^2 = 0.25 $$
$$ (96 - 97.5)^2 = (-1.5)^2 = 2.25 $$
$$ (99 - 97.5)^2 = 1.5^2 = 2.25 $$
Step 3: Sum squared deviations
$$ 6.25 + 0.25 + 6.25 + 0.25 + 2.25 + 2.25 = 17.5 $$
Step 4: Divide by \( n-1 = 5 \)
$$ \frac{17.5}{5} = 3.5 $$
Step 5: Square root
$$ s = \sqrt{3.5} = 1.87 $$
Answer: Standard deviation = 1.87 °C
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Create a free accountThis chapter focused on fundamental statistical measures including the mean, median, mode, and standard deviation, explaining how to calculate and interpret each to summarize data sets effectively. The mean provides the average value, the median identifies the middle value in ordered data, and the mode determines the most frequently occurring value. Standard deviation was introduced as a measure of data dispersion, showing how spread out values are from the mean. The chapter also covered various sampling methods used in data collection, emphasizing their importance in obtaining representative and unbiased samples for analysis. Different sampling techniques were explained to ensure accurate data gathering for engineering applications. Together, these concepts equip students with essential tools to analyze and interpret data in practical scenarios. Mastery of these topics supports better decision-making based on statistical evidence.
A refrigeration technician recorded the following temperatures (°C) during a system test: 2, 4, 6, 8, 10. Calculate the mean temperature. (2 marks)
The pressures (kPa) measured at five points in an air conditioning system are: 100, 105, 110, 115, 120. Find the median pressure. (2 marks)
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