Example 2.36 from Statistics, 2.7 Measures of the Spread of the Data

Measures of the Spread of the Data: example 2.36

Two students, John and Ali, from different high schools, wanted to find out who had the highest GPA when compared to his school. Which student had the highest GPA when compared to his school?

Student	GPA	School Mean GPA	School Standard Deviation
John	2.85	3.0	.7
Ali	77	80	10

Table 2.37

Work it out on paper first. Then open the solution one step at a time, and stop as soon as you can finish on your own.

Solution
For each student, determine how many standard deviations (#ofSTDEVs) his GPA is away from the average, for his school. Pay careful attention to signs when comparing and interpreting the answer.

$z = # of STDEVs = \frac{value - mean}{standard deviation} = \frac{x + μ}{σ}$
For John, $z = # o f S T D E V s = \frac{2.85 - 3.0}{.7} = - 0.21$
For Ali, $z = # o f S T D E V s = \frac{77 - 80}{10} = - 0.3$
John has the better GPA when compared to his school because his GPA is 0.21 standard deviations below his school's mean, while Ali's GPA is .3 standard deviations below his school's mean.
John's z-score of –.21 is higher than Ali's z-score of –.3. For GPA, higher values are better, so we conclude that John has the better GPA when compared to his school. The z-score representing John's score does not fall as far below the mean as the z-score representing Ali's score.

How did it go?

Student

GPA

School Mean GPA

School Standard Deviation

John

2.85

3.0

Ali

Solution

Solution