Applications of standard deviation

In Deviating from the Mean, we learnt how to calculate standard deviation. Standard deviation is a mathematical tool that helped us make the variance scores in a data set meaningful. 

In this chapter, we are going to look at how standard deviation can be used in everyday life.

For exam, say you get a test back and it says the mean was $73$73 and the standard deviation was $5$5 marks. This means that one standard deviation is equivalent to $5$5 marks. So if you got a score $2$2 standard deviations above the mean, you would have got a score $10$10 marks above the mean. In other words, your mark would be $83$83. 

Similarly, if you got a score $1.5$1.5 standard deviations below the mean, you mark would be $7.5$7.5 marks below the mean. In other words, your mark would be $73-7.5$73−7.5, which would be $65.5$65.5.

Examples

Question 1

The following table shows the marks obtained by a student in two subjects.

a) Find the mark in Science that is $2$2 standard deviations below the mean.

Think: If one standard deviation is $15$15 marks, how many marks would equate to $2$2 standard deviations.

Do: $2$2 standard deviations would be $30$30 marks. So we need to find a score $30$30 marks below the mean.

$89-30=59$89−30=59

So a science mark of $59$59 would be $2$2 standard deviations below the mean.

b) Find the mark in English that is $1.5$1.5 standard deviations above the mean.

Think: If one standard deviation is $10$10 marks, how many marks would equate to $1.5$1.5 standard deviations?

Do: $1.5$1.5 standard deviations would be $15$15 marks. So we need to find a score $15$15 marks above the mean.

$72+15=87$72+15=87

So an English mark of $87$87 would be $1.5$1.5 standard deviations above the mean.

Question 2

The percentage of people in each country with internet access is averaged and found to be $30%$30%. In one particular country, the percentage is $67.5%$67.5%, which is $2.5$2.5 standard deviations above the mean. What is the standard deviation among the countries? Leave your answer as a percentage.

Question 3

In an entrance exam, applicants completed two papers. 

On average, students performed better in Paper $1$1, but their marks were less spread out from the mean in Paper $2$2. The standard deviation of Paper $2$2 could be:

A) 13     B) 9     C) 17

Standard deviation is used as a measure of how widely dispersed a set of measurements are about their mean value.

Under certain assumptions, it is used in hypothesis testing and in establishing confidence intervals. For example and roughly speaking, if the mean value of a measurement made on a sample of experimental subjects that have been given a treatment appears to be different from the corresponding mean of the population before the treatment was given, we will be more likely to believe that the difference is due to the treatment if the standard deviation of the sample is small. Otherwise, the difference could quite likely be explained as a random fluctuation due to the particular sample.

This idea is made more precise when the concept of probability distributions is introduced, together with the associated computational methods.

In educational contexts, standard deviation provides a way of comparing test scores from different subject areas. Different subjects will have different mean test scores and the ranges of results obtained by the student will differ from subject to subject. To overcome this difficulty, test scores are often presented in terms of standard deviation units measured from the mean of each subject.

Example 1

A student obtains a score of $78%$78% in a chemistry test when the mean score for the class is $69%$69%. If the standard deviation of all the scores is $9$9, we see that the student attained a score of $1$1 standard deviation above the mean.

Scores expressed this way are often called z-scores. The student in the example achieved a z-score of $1$1 . Scores given as z-scores are considered to be more exceptional (compared with the population) the greater their distance from the central mean z-score value of $0$0.

Just how exceptional a particular z-score is can be quantified in terms of the proportion of students who achieved a score at least that high if more information is known about the overall distribution of the scores.

Example 2

A student reports a z-score of $-0.6$−0.6 for a test. The parents of the student ascertain that the mean score was $63%$63% with a standard deviation of $11$11. They wish to recover the raw score attained by the student.

Since $1$1 standard deviation is $11$11 percentage points in this example, $-0.6$−0.6 of a standard deviation is $-0.6\times11=-6.6$−0.6×11=−6.6 points. So, the raw score was $63-6.6=56.4$63−6.6=56.4%.

Worked Examples

Question 1

QUESTION 2

QUESTION 3