The Empirical Rule, Standard Scores, and Score Transformations

EDFL 571

The Empirical RuleThere is an “interesting” relationship between the standard deviation, and the manner in which scores are distributed in a frequency distribution or frequency graph.If the distribution of scores is roughly “mound-shaped”, then a surprisingly stable percentage of the scores will fall within one standard deviation +/- of the mean, whatever the actual scores of the distribution may be. Similar recurring values are found for percentages of scores within 2 standard deviations and 3 standard deviations of the mean.These values, representing areas of the distribution, are so stable that they have come to be referred to as the Empirical Rule.Given a mound-shaped score distribution, the following will be true:

68% OF ALL SCORES WILL BE WITHIN ONE STANDARD DEVIATION (1) OF THE MEAN

95% OF ALL SCORES WILL BE WITHIN TWO STANDARD DEVIATIONS (2) OF THE MEAN.

99% OF ALL SCORES WILL BE WITHIN THREE STANDARD DEVIATIONS (3) OF THE MEAN.

By combining these “facts”, we can establish the approximate area in each standard deviation interval

.5% 2% 13.5% 34% 34% 13.5% 2% .5%

A SAMPLE PROBLEM USING THE EMPIRICAL RULE

In a national survey it was found that the average number of hours grade school children spent watching TV per week was 35, with a standard deviation of 10 hours. Use this information, and the Empirical Rule, to answer the following questions.

A SAMPLE PROBLEM USING THE EMPIRICAL RULE

a) What percentage watched fewer than 25 hours per week?

b) What percentage watched between 15hrs. and 45 hours?

c) What is the likelihood that a randomly selected child would be someone who watches less than 5 hours of TV per week?

“SECRET” TO SOLVING EMPIRICAL RULE PROBLEMS- DRAW THE FIGURE

5 15 25 35 45 55 65 raw scores

.5% 2% 13.5% 34% 34% 13.5% 2% .5%

A SAMPLE PROBLEM USING THE EMPIRICAL RULE

a) What percentage watched fewer than 25 hours per week? 16%

b) What percentage watched between 15hrs. and 45 hours? 81.5%

a) What is the likelihood that a randomly selected child would be someone who watches less than 5 hours of TV per week? .5%

Limitations of the Empirical Rule

What if we wanted to know the % of students who watch more than 50 hours of tv per week?Since this is not a standard deviation value, we can’t use the Empirical Rule to answer this question.We need a new tool.

Standard ScoresONE VERY IMPORTANT SKILL IN STATISTICS IS THE ABILITY TO CONVERT SCORES INTO STANDARD SCORE EXPRESSION.A STANDARD SCORE EXPRESSES THE ORIGINAL RAW SCORE IN STANDARD DEVIATION UNITS ABOVE OR BELOW THE MEANTHE MOST COMMON STANDARD SCORE AND THE BASIS FOR ALL OTHERS IS THE Z SCORE.

Finding the Z score

THE Z SCORE IS AN EXPRESSION OF A RAW SCORE IN TERMS OF THE NUMBER OF STANDARD DEVIATIONS ABOVE OR BELOW THE MEAN THAT SCORE IS.

X

z

IF WE KNOW Z AND NEED TO FIND THE RAW SCORE THE FORMULA BELOW WILL ALLOW US TO CALCULATE EASILY.

zX

In the example given a moment ago, the mean was 35, the standard deviation was 10, and we wanted to know about a student who watched 50 hours of tv.That student’s z score would be:

Calculating a z score

z = 50 - 35 z = +15 z = +1.5

10 10

•Meaning, the score is 1.5 standard deviations above the mean.

Other Examples of Standard Scores

By first converting raw scores to z scores, it is possible to further transform them into more useful scales (by multiplying the mean of z, which is always zero, by some constant, and multiplying the standard deviation of z, which is always one, by another constant)Examples include- IQ scale (mean = 100, sd = 15) GRE scale (mean = 500, sd = 100) T scores (mean = 50, sd = 10) NCE scores (mean = 50, sd = 21.06)

Stanine ScoresA different approach to standard scores is to divide the entire range into 9, roughly equal (1/2 sd wide) intervals, called stanines.

Percentile Scores

Yet another approach is to divide the distribution into 100 equal “area” intervals (percentiles), each of which includes exactly one percent of the area/scores Take care in interpreting percentile scores,

width of intervals varies- wider in tails, narrower in middle

Score Transformations

We would like to be able to convert- Raw scores to standard scores, Standard scores to percentile ranksPercentile ranks to standard scoresStandard scores to raw scores

Converting Raw scores to Standard (z) scores

We use the z score formulaSubstitute the known values (raw score, distribution mean, standard deviation) and solve for z

X

z

Converting z Scores to Raw Scores

Substitute what we know (z score, mean, standard deviation), and solve for the raw score

zX

Transforming z Scores to Percentile Ranks

Finding the precise areas in designated areas “under the curve” would be very challenging mathematically, and would require calculusA simpler approach is based on interpreting a table containing known areas “under the curve” that correspond with given z scores (distances from the mean)

Interpreting the Standard Normal Distribution Table

First, our transformation depend on the assumption that the distribution is a “normal distribution” (meaning, a bell curve).With this “standard” distribution shape, it is possible to identify the exact areas corresponding to each z score.We have 2 tables (in Course Documents area) that give area values for each z score- one is the area of the interval from the z score to the mean, the other is the area from the z score to the “tail” in that end of the distribution. We will use both tables, depending on the situation.

Interpreting the Standard Normal Distribution Table- A simple example

In the z column of table one (area between mean and z), find the value 1.00The corresponding area value is .3413 Recall from the Empirical Rule that it said

approximately 34% of the area would be in an interval from the mean to one standard deviation above (or below) the mean.

The standard normal curve table is giving us a more precise value, because it is based on a normal curve (not just a mound-shaped curve)

Another example- Finding the Percentile rank that corresponds to a known value of z

What would be the percentile rank for our z score we calculated earlier, z = +1.5?Remember, the percentile rank indicates the total percentage of scores lower than the observed scoreArea from mean to z score is .4332Now, how much area, in total, would be to the left of our z score value (in other words, what is the percentile rank of our z score)?Total area to left is .5000 (everything below the mean) plus the .4332 above the mean = .9332, or the 93rd percentile rank (approximately 93% of students watch less than 50 hours of tv per week)

Transforming z Scores to Percentile Ranks- Negative z scores

You’ve probably noticed that the table has no negative values for zBut, whenever the original raw score is below the mean, the resulting z score will be negativeThe areas in the table are “symmetrical”- the area corresponding to a negative z score are the same as the areas corresponding to a positive z score, just in the other “half” of the distribution

Transforming z Scores to Percentile Ranks- Negative z scores- An example

Let’s say we have a z score of -.58, and we want to find its corresponding percentile rankIn this case, it will be easier to use table 2 (area beyond z), since we’re interested in knowing how much area is below our z score locationThe area corresponding to our z score of -.58 is .2810Expressed as a percentile rank, this would be the 28th percentile rankIn other words, 28th of all scores would be below the score corresponding to a z score of -.58

Transforming Percentile Ranks to z Scores

If we know the percentile rank, we can transform that back into a z scoreIf the percentile rank were the 75th, what would the corresponding z score be?This time, we begin by “locating” the 75th percentile rank (that’s 50% of the area below the mean, plus an additional 25% above the mean)So, we want to find a z score that corresponds to an area of .2500 between the mean and our target z score valueLooking up .2500, in table 1, the nearest value is .2486, and it’s corresponding z score is +.67Therefore, we can state that when the percentile rank is the 75th, the score it represents is +.67 standard deviations above the mean

Transforming Percentile Ranks to z Scores- When the Percentile Rank is below 50th

If the percentile rank we start with is below the median (less than the 50th %ile ), we can use table 2 to find the corresponding z scoreIf the percentile rank were the 30th, what would the corresponding z score be?Find the area in table 2 nearest to .3000That area is .3015The corresponding z score is .52However, REMEMBER that this z score is BELOW the mean, and therefore is a negative number z = -.52

Several students took a test of scholastic achievement that had a reported mean of 500 and standard deviation of 100. Below are the partial score reports of three students. STUDENT RAW ZSCORE % RANK BOB 640 ____ ______ SUE ____ -1.35 ______ TIM ____ ____ 24th

Putting it all together- A sample problem of score transformations

A graphic, showing the relationship between the z scale, the raw score scale, and the percentile rank scale

Bob

What we know- raw score of 640What we want to know- z score and %ile rankA visual aid to finding those values

Bob- The Solution

First, the z score- Z = (640 - 500)/100 Z = 140/100 Z = +1.4 (visually confirm this with z scale on graph)

Next, the %ile rank Area in table 1, corresponding to z of +1.4 = .4192 Add this to the .5000 (representing the other half

of the distribution) Total area to left of z score is .9192 Percentile rank for Bob’s score is the 91st

percentile rank Confirm this with graph

Sue

What do we know about Sue? Z score = -1.35What do we want to know about Sue? Raw score Percentile Rank

Sue- The Solution

First, let’s solve for her raw score X = (z) + X = (-1.35)(100) + 500 X = (-135) + 500 X = 365

Now, let’s find her percentile rank Area from table 2 associated with z of

(-)1.35 is .0885 Expressed as a %ile rank, this would be the

8th percentile rank Sue’s score exceeded those of 8% of the

test-takers

Tim

What do we know about Tim? %ile rank of 24thWhat do we want to know about Tim? Z score and raw score

Tim- The Solution

First, we need to transform Tim’s %ile rank score into a z score

A percentile rank of 24 is below the median, so use table 2

Find the area closest to Tim’s %ile rank value (.2400) in table 2

That value is .2389 The corresponding z score is .71 Since the value is BELOW the median, this z score is

negative Z = -.71

Next, we can convert the z score into a raw score X = (-.71)(100) + 500 X = (-71) + 500 X - 429

One More Example- %ile Rank above 50

Let’s consider one more student- Jill who had a %ile rank of 82nd

We want to know her raw score and z score

Jill- The Solution

First, we need to transform Jill’s %ile rank score into a z score

A percentile rank of 82 is above the median, so first subtract .5000 from .8200, then use table 1

Find the area closest to .3200 in table 1 That value is .3212 The corresponding z score is .92 Since the value is ABOVE the median, this z score is

positive Z = +.92

Next, we can convert the z score into a raw score

X = (+.92)(100) + 500 X = (92) + 500 X - 592

Wrapping Up- Some Pointers

1. Percentile ranks are always cumulative from left to right.

2. Positive z scores indicate percentile ranks greater than the 50th.

3. Negative z scores indicate percentile ranks less than the 50th.

4. When given a percentile rank greater than 50, subtract 50 from the percentile rank and look up the remainder as the area in table 1.

5. When given a percentile rank less than 50, use table 2, and find the area value closest to the percentile rank value.

6. If z is a positive value, look up the corresponding area in table 1 and add .5000 to get the percentile rank.

7. If z is a negative value, look up the corresponding area in table 2.