In statistics, the normal distribution is an example of a continuous probability distribution. With the normal distribution, many things can be concluded or predicted. Furthermore, the normal distribution also plays a role in communication science. Examples of applications in the field of communication science can be found in analyzing audience behavior, media consumption, survey data, and various performance metrics.
Case 1: The Media Consumption
A survey about media consumption habits has been distributed online. Suppose the time it takes respondents to complete the survey is normally distributed with a mean of 8 minutes and a standard deviation of 2 minutes. What is the probability that a randomly selected respondent will complete the survey in: a) less than 10 minutes? b) more than 10 minutes?
To answer the questions in Case 1 part a), type the following in a cell of an Excel worksheet: =NORM.DIST(10;8;2;1) [See Figure 1.]
Figure 1
After the Enter key is pressed, the value 0.841 will appear. So, the probability that a respondent will complete the survey in less than 10 minutes is 84.1%. In the code, the values 8 and 2, respectively, represent the mean and standard deviation of the normal distribution. The value 1 is practically always used in calculating the probability value using the normal distribution.
To answer part b), type =1-NORM.DIST(10;8;2;1). After the Enter key is pressed, the value 0.159 will appear. So, the probability that a respondent will complete the survey in more than 10 minutes is 15.9%. [See Figure 2.]
Figure 2
Let us discuss the answer to part a) further. The value of 0.841, which Excel displayed, is none other than the area under a curve called the normal curve. Look at Figure 3 below.
Figure 3
The blue area above has an area of 0.841. The normal curve represents one of the most important probability distributions in statistics, called the normal distribution. The curve is bell-shaped and has the equation 
for every real number x. There are two parameters in this distribution, namely 
which is the mean, and 
which is the standard deviation. The curve is symmetric about the vertical axis through the mean.
Figure 4
In Figure 4, the blue curve is the normal curve. The red vertical axis is the line passing through the mean (μ). This axis is the axis of symmetry of the curve. If the curve is folded along this line, the left and right sides will overlap.
The total area under the curve is 1 or 100%. See Figure 5.
Figure 5
Since the total area under the normal curve is 1, the area of the unshaded region in Figure 3 is 1 – 0.841 = 0.159. That is why in answering question b) in Excel, type =1-NORM.DIST(10;8;2;1).
Figure 6
The value of 0.159 in Figure 6 shows the probability that a respondent will complete the survey in more than 10 minutes. The answers to parts a) and b) can be explained by Figure 7 as follows.
Figure 7
In Fig. 7, the part in pink represents the answer to part a), while the blue part represents the answer to part b). Note that 0.841 + 0.159 = 1 = 100%. The probability that a respondent will complete the survey in less than 10 minutes is 0.841. The probability that a respondent will complete the survey in more than 10 minutes is 0.159. Now, consider the following case.
Case 2: Performance Metric
The readability index measures how easy or difficult a written text is to read and understand. By calculating the readability index, we can evaluate the effectiveness of written communication in reaching and capturing the attention of the intended audience. A communication researcher analyzes the readability score of news articles using the standard readability index. It is known that the readability scores are normally distributed with a mean of 60 and a standard deviation of 10. What percentage of news articles have a readability score between 55 and 70?
The question in this case can be represented by the area of the shaded region in Figure 8.
Figure 8
To get the area of the region, in one of the cells of the Excel sheet type: =NORM.DIST(70;60;10;1)-NORM.DIST(55;60;10;1). After pressing the Enter key, the value 0.533 will be obtained. So, about 53.3% of news articles have a readability score between 55 and 70. In the code, the value 60 indicates the average and 10 indicates the standard deviation. The values 70 and 55 are the boundaries of the area in question. The first term of the code, namely NORM.DIST(70;60;10;1), is the area of the region in Figure 9a.
Figure 9a
The second term of the code, namely NORM.DIST(55;60;10;1), represents the area of the region in Figure 9b.
Figure 9b
Therefore, the answer to the question (i.e. 0.533) is the subtraction of two areas, namely 0.841 and 0.309. The difference in the calculation of 0.001 is due to rounding.
PROBLEMS
Problem 1
A social media analyst studies the number of likes a company’s posts receive. The average number of likes per post is 500, with a standard deviation of 100. Assume the number of likes is normally distributed. What is the probability that a randomly selected post will receive fewer than 450 likes?
Problem 2
A study is conducted to analyze the length of student presentations in a communication theory class. Previous research has shown that the average presentation is 15 minutes long, with a standard deviation of 3 minutes. Assume that the length of presentations is normally distributed. What is the probability that a presentation will last longer than 20 minutes?
Problem 3
A political campaign uses sentiment analysis software to measure public opinion of its candidates based on social media posts. The software generates a sentiment score ranging from 0 to 100, where 50 represents neutral sentiment, a score above 50 represents positive sentiment and a score below 50 represents negative sentiment. After analyzing a large sample of posts, the campaign found that the average sentiment score was 58, with a standard deviation of 8. Assuming the sentiment scores are normally distributed. What percentage of social media posts express positive sentiment (scores above 60)?
