# Measures of Central Tendencies and Dispersions in the FE Electrical Exam

You might wonder, “Why do I need to know to master Measures of Central Tendencies and Dispersions in FE Electrical Exam preparation?” I’m an electrical engineer, not a statistician.” Well, let me tell you, understanding these concepts is crucial for success in both the FE Electrical Exam and the electrical engineering field.

Measures of Central Tendencies, such as mean, median, and mode, allow us to summarize extensive data sets into a single value representing the “center” of the data. On the other hand, Measures of Dispersion, such as variance and standard deviation, give us an idea of how spread out the data is.

These concepts are used in various applications in electrical engineering, such as analyzing signal quality, designing control systems, and optimizing circuit performance. You’ll be lost in electrical engineering without thoroughly understanding Measures of Central Tendencies and Dispersions.

So buckle up and get ready to learn! This blog post will delve deeper into these concepts and equip you with the proper knowledge concerning Measures of central tendencies and dispersions in the FE Electrical Exam.

## Importance of Central Tendencies and Dispersions in Electrical Engineering

Measures of Central Tendencies and Dispersions play a significant role in Electrical Engineering. The following key points highlight their significance and real-world applications:

**Signal Processing**– Electrical engineers work with often noisy and unpredictable signals. Understanding a signal’s mean, variance, and standard deviation can help filter out the noise and identify the underlying patterns. For example, in audio processing, the mean and variance of a signal can be used to remove background noise and enhance speech clarity.**Control Systems**– Electrical engineers control various physical processes using feedback systems. The control system’s performance depends on the measurements’ accuracy and the feedback signals’ precision. Measures of Central Tendencies and Dispersions are used to analyze the system’s behavior and make improvements. For instance, in an industrial control system, the mean and standard deviation of a sensor’s readings can adjust the system’s parameters to maintain optimal performance.**Circuit Design**– In electrical circuit design, ensuring that the circuit components’ values fall within specific tolerances is crucial to avoid unexpected behavior. Measures of Central Tendencies and Dispersions can be used to calculate the expected values and tolerances for circuit components. For example, in a power supply circuit, the mean and standard deviation of the input voltage can be used to design the voltage regulation circuitry.**Reliability Analysis**– Electrical engineers are concerned with the reliability and durability of electrical systems. Measures of Central Tendencies and Dispersions are used to analyze systems’ performance and identify improvement areas. For instance, in a solar power system, the mean and variance of the solar panel output can be used to predict the system’s overall performance and reliability.**Quality Control**– Electrical engineers work with manufacturing processes that require precise measurements and control. Measures of Central Tendencies and Dispersions are used to monitor the manufacturing process and ensure that the products meet quality standards. For example, in producing electronic components, the mean and standard deviation of the component’s dimensions can be used to monitor the manufacturing process and identify any variations that might affect the component’s performance.

## Measures of Central Tendencies

In statistics, there are three standard measures of central tendency: mean, median, and mode. Let’s define each of them first before diving into their relationship.

**Mean**– The mean is the sum of all the values in a data set divided by the number of values in the set. It is also known as the average. For example, the mean of the numbers 2, 4, 6, and 8 is (2+4+6+8)/4 = 5.**Median**– The median is the middle value in a sorted data set. The median is the central value if the data set has an odd number of values. If the data set has an even number of values, the median is the average of the two middle values. For example, the median of 1, 2, 3, 4, and 5 is 3.**Mode**– The mode is a data set’s most frequently occurring value. For example, the mode of the numbers 1, 2, 2, 3, and 4 is 2.

Consider an electrical engineer designing a new device’s power supply. The engineer measures the voltage output of the power supply over several days and records the following data: 5V, 5.2V, 5.3V, 5.1V, 5.2V, 5V, 5.2V, 5.3V.

To analyze this data, the electrical engineering experts calculate the mean: (5+5.2+5.3+5.1+5.2+5+5.2+5.3)/8 = 5.175V. The mean gives us an idea of the central tendency of the data set, but it doesn’t tell us anything about the variability of the data.

Then, the experts calculate the median: first, they sort the data set in ascending order, so the values are 5V, 5V, 5.1V, 5.2V, 5.2V, 5.2V, 5.3V, and 5.3V. The median is the average of the two middle values, which are 5.2V and 5.2V.

So the median is (5.2V+5.2V)/2 = 5.2V. The median gives us an idea of the central tendency of the data set, but it doesn’t consider the frequency of each value.

Finally, they calculate the mode: the mode is 5.2V, as it is the most frequently occurring value in the data set. The mode gives us an idea of the most common value in the data set, but it doesn’t tell us anything about the variability or distribution of the data.

By using all three measures of central tendency, the engineer gains a more complete understanding of the data set’s characteristics. In this case, the mean voltage output is 5.175V, the median voltage output is 5.2V, and the mode voltage output is 5.2V.

These values suggest that the power supply delivers a fairly consistent voltage output, with occasional variations above or below the mean value.

### Calculating Central Tendencies for Group Data

#### Mean

The following table shows the class boundaries and frequencies of resistances measured in a sample of electrical components:

Class Boundaries (Ohms) | Frequency |

5 – 10 | 4 |

10 – 15 | 8 |

15 – 20 | 12 |

20 – 25 | 6 |

25 – 30 | 2 |

To calculate the mean for this group data, we need first to find the midpoint of each class interval. We can do this by adding the lower and upper-class boundaries and dividing the sum by 2. Then, we multiply each midpoint by its frequency, sum these products, and divide by the total frequency of the data.

The table with the required columns for the solution is:

Class Boundaries | Midpoint | Frequency | fx |

5 – 10 | 7.5 | 4 | 30 |

10 – 15 | 12.5 | 8 | 100 |

15 – 20 | 17.5 | 12 | 210 |

20 – 25 | 22.5 | 6 | 135 |

25 – 30 | 27.5 | 2 | 55 |

Total | 32 | 530 |

To find the mean, we sum the fx column and divide by the total frequency (n):

Mean = fx / n = 530 / 32 = 16.56

Therefore, the mean resistance for this sample of electrical components is 16.56 ohms.

**Mode**

To find the mode of a grouped data set, we need to determine the class interval with the highest frequency. In our example, the class interval with the highest frequency is 15-20 ohms, with a frequency of 12. Therefore, the mode of this sample of electrical components is 15-20 ohms.

It’s worth noting that there can be more than one mode in a data set if there are two or more class intervals with the same highest frequency. However, in this case, there is only one mode.

To calculate the mode of a grouped data set using a formula, we can use the following equation:

Mode = L + [(Fm – f1)/(2Fm – f1 – f2)] x h

where:

- L is the lower boundary of the modal class interval
- Fm is the frequency of the modal class interval
- f1 is the frequency of the class interval (immediately) preceding the modal class interval
- f2 is the frequency of the class interval (immediately) leading to the modal class interval
- h is the class width (i.e., the size of each class interval)

Using the same table from our previous example, we can find the mode as follows:

Class Boundaries | Frequency | Parameters (from formulae) |

5 – 10 | 4 | |

10 – 15 | 8 | F1 |

15 – 20 | 12 | Fm |

20 – 25 | 6 | F2 |

25 – 30 | 2 |

Here, the modal class interval is 15-20 ohms, with the highest frequency 12.

The lower boundary of the modal class interval is 15, the frequency of the interval immediately preceding it (10-15 ohms) is 8, and the frequency of the leading class (20-25 ohms) is 6. The class width (h) is 5 ohms (the same for all classes).

Plugging these values into the formula, we get:

Mode = 15 + ((12 – 8) / (2*12 – 8 – 6)) * 5

= 15 + (4 / 10) * 5

= 15 + 2

= 17

Therefore, the mode of this sample of electrical components is 17 ohms.

#### Median

To calculate the median of a grouped data set, we first need to find the cumulative frequency for each class interval. To do this, we can add up the frequencies for each class interval from the start of the data set up to and including that interval.

Using the same table from our previous examples, we can add a column for cumulative frequency:

Class Boundaries | Frequency | Cumulative Frequency |

5 – 10 | 4 | 4 |

10 – 15 | 8 | 12 |

15 – 20 | 12 | 24 |

20 – 25 | 6 | 30 |

25 – 30 | 2 | 32 |

The total number of observations (n) in this sample is 32. We can calculate the position of the median using the formula (n/2).

In the given case, we have n = 32, so the position of the median class can be found by:

(n/2) = (32/2) = 16

This means that the median is the value that falls near the 16th observation in the data set. It falls within the 15-20 ohms class interval (Hint: Look for the most minor and nearest cumulative frequency for 16) as the cumulative frequency for the 15-20 ohms interval is 24.

Next, we need to find the median for this class interval using the formula:

Median = L + [(n/2 – Cf) / f] x h

Where:

- L is the lower boundary of the class interval containing the median
- n is the total number of observations
- Cf is the cumulative frequency up to the end of the class interval containing the median
- f is the frequency of the class interval containing the median
- h is the class width (i.e., the size of each class interval)

For the class interval 15-20 ohms, we have:

- L = 15
- n = 32
- Cf = 12
- f = 12
- h = 5

Plugging these values into the formula, we get:

Median = 15 + [(16 – 12) / 12] x 5

= 15 + (4 / 12) * 5

= 15 + (5/3)

= 16.67 ohms (rounded to two decimal places)

Therefore, the median of this sample of electrical components is 16.67 ohms.

## Measures of Dispersions

Variance and standard deviation are two measures of dispersion commonly used in statistics to describe the spread or variability of a data set.

### Variance

The Variance measures how much the data points in a data set vary from the data set’s mean. Mathematically, variance is defined as the average of the squared differences between each data point and the data set’s standard. The formula for variance is:

**Variance = (Σ(xi – x̄)²) / (n – 1)**

where xi is the ith observation, x̄ is the mean of the data set, and n is the sample size.

### Standard Deviation

Standard deviation is the square root of the variance and measures the variation or dispersion of data values. The standard deviation is expressed in the same units as the data, and it tells us how much the individual data points are likely to differ from the mean. The formula for standard deviation is:

**Standard deviation = sqrt(Variance)**

### Relation between Variance and Standard Deviation

Now, let’s consider an example to explain the relationship between variance and standard deviation. Suppose we have a data set that contains the resistance values of 10 electrical components, measured in ohms:

12, 18, 22, 25, 29, 32, 36, 40, 45, and 50

To calculate the variance, we first need to find the mean of the data set:

x̄ = (12 + 18 + 22 + 25 + 29 + 32 + 36 + 40 + 45 + 50) / 10

= 30.2 ohms

Next, we can calculate the variance using the formula:

Variance = [(12-30.2)² + (18-30.2)² + (22-30.2)² + (25-30.2)² + (29-30.2)²

+ (32-30.2)² + (36-30.2)² + (40-30.2)² + (45-30.2)² + (50-30.2)²] / (10 – 1)

= 218.49

Therefore, the variance of this data set is 218.49 ohms^2.

To calculate the standard deviation, we can take the square root of the variance:

Standard deviation = sqrt(218.49) = 14.78 ohms

This tells us that the resistance values in this data set are fairly spread out, with a standard deviation of 14.78 ohms. The larger the standard deviation, the more spread out the data is.

In electrical engineering, the standard deviation of a data set can be used to determine the tolerance level of electronic components. If the standard deviation is too large, it may indicate that the components are not within the required tolerance level (having too much fluctuation) and must be replaced.

### Conclusion

Understanding measures of central tendencies and dispersions are crucial for success in the field of electrical engineering, as well as for passing the FE Electrical Exam. These measures provide valuable insights into data sets, allowing engineers to make informed decisions based on statistical analysis.

By mastering these statistical concepts, electrical engineers can improve the accuracy and reliability of their electrical setups and better understand the performance of electronic devices and control systems.

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