1) Derived scores help interpret raw scores and make them comparable by expressing them in terms of standard deviations from the mean. 2) There are two main types of derived scores - standard scores (z-scores) which indicate how many standard deviations a score is from the mean, and percentile ranks which show the percentage of scores in the group that are the same or lower. 3) Z-scores are calculated using the formula z = (x - M) / SD, where x is the raw score, M is the mean, and SD is the standard deviation. This expresses how many standard deviations a score is from the mean.

1. DR. SARITA ANAND
ASSISTANT PROFESSOR
DEPARTMENT OF EDUCATION
VINAYA BHAVANA (INSTITUTE OF EDUCATION)
VISVA-BHARATI, SANTINIKETAN
sarita.anand@visva-bharati.ac.in
DERIVED SCORES

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3. Derived Score
In order to interpret the scores properly or to make
them comparable we convert the raw scores into
derived scores.
The derived scores help us to know the position of an
individual in his group and we can compare the
performance with others. “A derived score is a
numerical description of an individual’s
performance in terms of norms.”
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4. Derived Score
The derived scores have several uses like:
(a) It helps to know the position of an individual in his
group by knowing how many standard deviation
units above or below the mean he falls.
(b) Standard score obtained on two tests may be
directly compared.
(c) It can be converted into other types of scores such
as percentile norm.
Two types-
(A) Standard Score (z-score or o-score).
(B) Percentile Ranks.
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5. (A) Standard Scores or
z-Score (Small z Score)
Standard scores also indicate the relative position of a
pupil in a group by showing how far the raw score is
above or below average.The standard scores
express the performance of pupils in standard
deviation unit.
This gives us a standard score, usually denoted by
a-score, (read as sigma-‘z’) is obtained by the
formula:
z(or, σ-score) = X – M/SD
where X = score of the individual
M = Mean of the group
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6. (A) Standard Scores
The standard scores represent ‘measurements’ from
the mean in S.D. units. The standard score
indicates how far a particular score is removed
from the mean of the distribution in terms of S.D.
of the distribution. Standard scores conform to
the concept of the normal distribution. In case of
standard scores, the difference between score
units are hypothesized to be equal.
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7. Example 1:
In a test the marks obtained byVicky is 55, the class-
mean being 50 and the S.D. being 10.
...Vicky’s z-score = X-M/SD = 55-50/10 = 1/2 or .5
Thus the raw score of 55 is expressed as 1/2z or .5z
(or 1/2σ or .5 σ) in terms of standard score.
In other words,Vicky’s score is at .5σ (i.e. 1/2 sigma
distance) from the mean or, his score is 1/2σ above
the mean.
(A) Standard Scores
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8. Example 2:
Rakesh’s score in a test is 49.The class mean is 55
and the SD is 3.
... Rakesh’s z-score = X-M/SD = 49-55/3 = -2
The raw score of Rakesh i.e. 49 can be expressed
as – 2z or – 2σ.
Rakesh’s score is at 2 sigma distances from the
mean or his score is 2σ below the mean.
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9. Example 3:
In a test marks obtained by three students are as
follows. The mean = 40, SD = 8. Assuming normal
distribution what are their z-score (sigma-score)
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10. Let us discuss what these standard scores
mean. We know what a normal curve is.
These z-scores can be shown on the base line
of that curve, so that we can know their
position in the group (or class) to which they
belong.
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11. From the above diagram we can know the percentage of
students above and below each student.
Below A there are 50 + 34.13 = 84.13% and above A 100 –
84.13 = 15.87% of pupils. We can also say that A is at a
distance of + 1σ above the mean.
Below B, there are 50 + 34.13 + 13.59 = 97.72% and above
B 100 – 97.72 = 2.28% of students. Again B is at a
distance of + 2σ above the mean.
C’s position is just in the middle of the group. So below C
there are 50% and above C 50% of the group. 11

12. Example 4:
From the data on a test of Arithmetic given below,
whose performance is the best?
Now Amit is 1σ above the mean, Kishore is .5a
above the mean and Shyam is 2σs above the mean.
Thus Shyam’s performance in the test of Arithmetic is the
best.
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13. Example 5:
The mean of a normal distribution is 32 and SD is 10.
What percent of cases will be between 22 and 42?
 Z- Score of 22 = 22 – 32/10 = -1σ
 Z- Score of 42 = 42 – 32/10 = +1σ
We know the position of +1σ and -1σ in the normal
curve. Score 22 is at a distance of – 1σ and score 42
at a distance of+1σ from the mean.
So, the required percent = 34.13 + 34.13 = 68.26. In
other words there are 68.26% of cases between 22
and 42.
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14. Example 6:
In a symmetrical distribution, mean = 20 and σ = 5.
What percent of cases lie above 30?
z-score of 30 = 30-20/5 = + 2σ.
So, score 30 is at a distance of +2σ from the mean.
So percent of cases above 30
= 100 – (50 + 34.13 + 13.59)
= 100 – 97.72 = 2.28.
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15. Example 7:
Radhika’s score in a test of science is given below (Section-A).
Express her score in terms of the scores in Section-B i.e., what
will be the equivalent score of Radhika in section-B?

Radhika’s score is la distance above the mean.
As the standard scores are equal, in section-B also Radhika will
secure 1σ2 i.e. 10 more than M2.
Therefore, in section-B Radhika’s score X2 = M2 + 1σ2 = 60 + 10 = 70.
Thus, X1 score of 55 = X2 score of 70. 15

16. The above Question can also be calculated by putting
the values directly in the formula:
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17. Properties of the standard score
or z-score:
 A score becomes significant only when it is
comparable with other scores. Raw scores
become meaningful when they are converted
into derived scores’ or z-scores.
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18. The derived score’s properties:
1. A z-score has a mean of 0 and standard deviation of 1.
2. We can know the relative position of an individual in
the whole group by expressing the raw score in terms
of a distances above or below the mean.
3. Standard score differences are proportional to raw
score differences.
4. Standard scores on different tests are directly
comparable.
5. One type of standard score can be converted into
another type of standard score.
6. From the formula, z-score = raw score –
mean/standard deviation = X-M/SD, 18

19. Conclusion
it can be derived that:
 (i) If the raw score = mean, z-score is Zero;
 (ii) If the raw score > mean, z-score is positive;
 (iii) If the raw score < mean, z-score is negative.
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20. Advantages of z-scores:
(i) They permit us to convert raw scores into a
common scale which has equal units and which
can be readily interpreted.
(ii) They give us an idea of how well a teacher-made
test is. A good teacher- made test designed to
discriminate among students will generally have a
range between 4 and 5 SDs, i.e., 2.0 to 2.5 SDs on
either side of the mean.
Limitations:
 They involve the use of decimals and negative
numbers.
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21. Z-score
The standard scores or z-scores involve decimals and
directional signs. To avoid this the z-value is multiplied
by’ 10 and then 50 is added to it. The new score is called
Z-score.
Thus, Z-score is a standard score on the scale with a mean
of 50 and SD of 10.
 The formula for computing Z-score is:
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22. Example:
In a test the mean is 50 and SD is 4. Convert a score
of 58 to small z-score and capital score.
Z-score
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23. T-score (Mc Call’s score)
Mc Call suggested a scale with a mean of 50 and a SD of 10 to be used
when the distribution is normal.
T-score enjoys advantage over standard scores as in it the negative or
fractional standard scores can be avoided. (T-score is named after
Thorndike andTerman).
T-score = 50 + 10z
When this formula is applied z is read from the table of normal curve.
Suppose a score of 63 surpasses 84% of cases of the group. Referring
to the table of normal curve we find that such a score is at one
sigma distance from the mean i.e. its σ- distance or z = 1.
 So theT-score equivalent of this score, 63
 = 50 + 10z
 = 50 + 10 x 1 = 60
 Here, in theT-scale it is assumed that the distribution is normal.
This is whyT-score is called a “normalized standard score.” 23

24. In this scale the assumption is that nearly all scores will be within
a range of 5 SDs from the mean. Since each SD is divided into
10 units, the T-score is based on a scale of 100 units, thus it
avoids the negative and fractional standard scores.
Generally the Z value is read from the table of area under normal
curve.
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25. Example:
Suppose Deepak’s score 75 surpasses 84% of cases of
the group. Express it in terms of T-score i.e. find out
the equivalentT-score of 75.
Now referring to the area under normal probability
curve, it will be found that at 1 a distance it will
surpass 84% of cases. In other words the score 75 is
at 1σ distance from mean.
Therefore z = 1.
So,T-score of 75 = 50 + 10z
= 50 + 10 x 1
= 60.
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26. H-score (Hull’s scale)
Hull suggested a scale with mean 50 and SD 14. If H
is a score in Hull’s scale, the formula for
comparison of marks will be
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27. Example:
ExpressAmit’s raw score of 55 in terms of H-score.
Score = 55, Mean = 50 and SD = 5.
H-score (Hull’s scale)
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Reference: https://www.yourarticlelibrary.com/statistics-2/classification-of-score-raw-score-and-derived-score/92557

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