By the end of this chapter, you will be able to:
Mastering these skills will help you make confident, data-driven decisions in electronics engineering and beyond!
Compiled by: Peter Kiprono
Engineers rarely work with a single measurement. A technician testing a batch of resistors, a coil of cable or a set of battery cells collects many readings, and a measure of central tendency is one number that represents the typical value of the whole set. The three common measures are the mean, the median and the mode. Each answers a slightly different question, and choosing the right one is part of reporting data honestly.
The mean is the sum of all the values divided by the number of values. For n values x₁, x₂, …, xₙ,
x̄ = (x₁ + x₂ + … + xₙ) ÷ n
where x̄ (read "x bar") is the mean. The mean uses every value, so it is the natural choice for symmetrical data, but one very large or very small value pulls it noticeably.
Example 1: Ten readings of a resistor marked 220 Ω were taken with a multimeter: 218, 221, 220, 219, 222, 220, 223, 217, 220 and 221 Ω. Find the mean.
Sum = 218 + 221 + 220 + 219 + 222 + 220 + 223 + 217 + 220 + 221 = 2201
x̄ = 2201 ÷ 10 = 220.1 Ω
Answer: The mean resistance is 220.1 Ω.
The median is the middle value when the data are arranged in order of size. If n is odd, the median is the value in position (n + 1) ÷ 2. If n is even, it is the mean of the two middle values, in positions n ÷ 2 and (n ÷ 2) + 1. The median is not disturbed by extreme values, so it is the better choice when the data contain a faulty reading or are very uneven.
Example 2: Find the median of the readings in Example 1.
Arranged in order: 217, 218, 219, 220, 220, 220, 221, 221, 222, 223
There are 10 values, so the median is the mean of the 5th and 6th values: (220 + 220) ÷ 2 = 220.
Answer: The median is 220 Ω.
Example 3: A workshop records the number of soldered joints a trainee completes in six timed sessions: 12, 15, 9, 14, 10 and 16. Find the mean and the median.
Mean = (12 + 15 + 9 + 14 + 10 + 16) ÷ 6 = 76 ÷ 6 = 12.67
In order: 9, 10, 12, 14, 15, 16. With six values the median is the mean of the 3rd and 4th: (12 + 14) ÷ 2 = 13.
Answer: Mean = 12.67 joints, median = 13 joints.
The mode is the value that occurs most often. A set of data may have one mode, more than one mode (bimodal or multimodal) or none at all if every value occurs once. The mode is the only measure that can be used for data that are not numbers, such as the most common fault type found in repaired power supplies.
In Example 1 the value 220 appears three times and every other value appears once or twice, so the mode is 220 Ω. In the data 4, 4, 5, 7, 7, 9 the values 4 and 7 each occur twice, so the data are bimodal with modes 4 and 7.
Consider five readings of a bench power supply set to 3.3 V: 3.2, 3.4, 3.3, 3.9 and 3.3 V. The mean is 17.1 ÷ 5 = 3.42 V, the median is 3.3 V and the mode is 3.3 V. The single high reading of 3.9 V has moved the mean away from where most readings lie, while the median and mode stay at 3.3 V. If the 3.9 V reading came from a loose probe, the median describes the supply better than the mean.
When there are many readings they are collected into classes with a frequency for each class. The exact values are then unknown, so the measures are estimated. For the mean, each class is represented by its midpoint m, and
x̄ = Σ(f × m) ÷ Σf
where f is the class frequency and Σf = N is the total frequency.
For the median of grouped data, find the class that contains the N ÷ 2 th value (the median class) and interpolate:
Median = L + [(N ÷ 2 − F) ÷ f] × w
where L is the lower boundary of the median class, F is the cumulative frequency before it, f is its frequency and w is its class width.
For the mode, find the modal class (the highest frequency) and use
Mode = L + [d₁ ÷ (d₁ + d₂)] × w
where d₁ is the modal frequency minus the frequency of the class before, d₂ is the modal frequency minus the frequency of the class after, and L and w refer to the modal class.
Example 4: The lifetimes of 70 LED lamps tested in a workshop are recorded as follows.
| Lifetime (hours) | Frequency |
|---|---|
| 1000 to under 1200 | 6 |
| 1200 to under 1400 | 14 |
| 1400 to under 1600 | 25 |
| 1600 to under 1800 | 18 |
| 1800 to under 2000 | 7 |
Estimate the mean, the median and the mode.
Mean: the midpoints are 1100, 1300, 1500, 1700 and 1900, so
Σ(f × m) = 6 × 1100 + 14 × 1300 + 25 × 1500 + 18 × 1700 + 7 × 1900 = 6600 + 18200 + 37500 + 30600 + 13300 = 106200
x̄ = 106200 ÷ 70 = 1517.1 hours
Median: N ÷ 2 = 35. The cumulative frequency up to 1400 is 6 + 14 = 20, and up to 1600 is 45, so the median class is 1400 to under 1600.
Median = 1400 + [(35 − 20) ÷ 25] × 200 = 1400 + 120 = 1520 hours
Mode: the modal class is 1400 to under 1600, with d₁ = 25 − 14 = 11 and d₂ = 25 − 18 = 7.
Mode = 1400 + [11 ÷ (11 + 7)] × 200 = 1400 + 122.2 = 1522.2 hours
Answer: Mean ≈ 1517.1 hours, median = 1520 hours, mode ≈ 1522.2 hours.
The three values are close to each other, which tells us the distribution of lifetimes is nearly symmetrical about 1520 hours.
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