Chapter 6
Zscores
Puzzle 1
Milton took the IQ, EQ and creativity tests that Zach took. His scores were IQ = 170, EI = 70 and creativity = 55. Calculate the zscores relative to the population values in Table 6.1 (in the book). How do Milton’s scores compare to Zach’s?
Table 1: Zach’s and Milton’s raw scores on three tests converted to zscores using the population means and standard deviations for each test
Zach and Milton's scores  

Measure  Zach's raw score  Milton's raw score  Population mean ($\mu$)  Population $SD$ ($\sigma$)  Zach's $z$  Milton's $z$ 
IQ  125  170  100  5  1.67  4.67 
Creativity  30  55  15  11  1.36  3.64 
EI  165  70  150  20  0.75  4.00 
For IQ, Milton’s zscore would be:
$$ z = \frac{X\mu}{\sigma} = \frac{170100}{15} = 4.67 $$
For Creativity, Milton’s zscore would be:
$$ z = \frac{X\mu}{\sigma} = \frac{5515}{11} = 3.64 $$
For EI, Milton’s zscore would be:
$$ z = \frac{X\mu}{\sigma} = \frac{70150}{20} = 4.00 $$
The question also asks us to compare Zach and Milton’s zscores. For IQ Zach got a zscore of 1.67, which is higher than the mean of zero indicating that Zach is more intelligent than the average person. However, Milton’s zscore of 4.67 is incredibly high, it is very unusual to see values outside of $ $ and extremely rare to see values outside of $ $; so, Milton is a genius! For creativity Zach got a similar zscore to the one he got for IQ, he scored z = 1.36, which again puts him above average for this trait, but once again Milton has beaten Zach by quite a large margin by scoring z = 3.64, which again is a very high score and also implies that Milton is a genius. However, the story is quite different when comparing Zach and Milton on emotional intelligence; for this trait Zach got a zscore of 0.75, which puts him a bit above average, whereas Milton scored terribly, he got a z –score of $ 4 $, which is about as far below the mean as you can get, suggesting he has basically no emotional intelligence at all!
Puzzle 2
In the file in Dr Genari’s office, Zach also found that Alice had a score of 57 out of 60 on an IQ test (a different one than the one Zach completed). This test had a mean of 32 and a standard deviation of 7. Using this information and the answer to the previous Puzzle, who has a larger IQ: Alice or Milton?
First we need to convert Alice’s IQ score of 57 into a zscore by using the equation below:
$$ z = \frac{X\mu}{\sigma} = \frac{5732}{7} = 3.57 $$
Therefore, for IQ Alice scored z = 3.57 and looking at the previous question we can see that Milton scored z = 4.67. Both zscores are above 3, suggesting that Alice and Milton are both a lot more intelligent than the average person but Milton has the higher IQ.
Puzzle 3
When Zach met Professor Pincus (Chapter 3), she showed him data from 20 women who rated how important they found various characteristics in men. They rated from 0 (not at all important) to 10 (very important) characteristics such as high salary, humour, kindness and ambition. The data are in Table 3.1 (in the book). A student outside Occam’s asked Celia to rate how important these same characteristics were in a partner. She gave high salary a rating of 5. What is this value as a zscore? (Hint: use the values in Table 3.1 to compute the mean and standard deviation of the distribution of high salary ratings.)
First we need to compute the mean
$$ \bar{X} = \frac{\sum_{i = 1}^n x_i}{n} = \frac{114}{20} = 5.7. $$
Then calculate the standard deviation as we have done with other data sets (the scores for high salary are from Table 3.1 in the book):
Calculating the sums of squared errors for ratings of high salary  

High salary $x$ 
Mean $\bar{X}$ 
Deviance $x\bar{X}$ 
Deviance squared $(x\bar{X})^2$ 

4  5.7  1.7  2.89  
5  5.7  0.7  0.49  
9  5.7  3.3  10.89  
4  5.7  1.7  2.89  
5  5.7  0.7  0.49  
3  5.7  2.7  7.29  
10  5.7  4.3  18.49  
10  5.7  4.3  18.49  
6  5.7  0.3  0.09  
1  5.7  4.7  22.09  
3  5.7  2.7  7.29  
8  5.7  2.3  5.29  
8  5.7  2.3  5.29  
2  5.7  3.7  13.69  
6  5.7  0.3  0.09  
4  5.7  1.7  2.89  
8  5.7  2.3  5.29  
4  5.7  1.7  2.89  
7  5.7  1.3  1.69  
7  5.7  1.3  1.69  
Sum  130.20 
To calculate the variance, use the sum of squares from Table 2:
$$ s^2 = \frac{\text{sum of squared error}}{N1} = \frac{130.2}{19} = 6.85. $$
We can now calculate the standard deviation:
$$ s = \sqrt{6.85} = 2.62. $$
Now we can convert Celia’s rating of 5 for the high salary variable into a zscore by plugging the mean, standard deviation and Celia’s score of 5 into the equation for calculating zscores:
$$ z = \frac{X\bar{X}}{s} = \frac{55.7}{2.62} = 0.27 $$
Therefore, Celia’s zscore for high salary is $ –0.27 $, which is a little bit below the mean. This suggests that Celia doesn’t value high salary as a particularly important characteristic in a romantic partner.
Puzzle 4
For kindness, Celia’s rating yielded a zscore of 0.9. What was her raw score? (Hint: use the values in Table 3.1 in the book to compute the mean and standard deviation of the distribution of kindness ratings.)
First, we compute the mean:
$$ \bar{X} = \frac{\sum_{i = 1}^n x_i}{n} = \frac{184}{20} = 9.2. $$
Then compute the standard deviation as we have done before (the scores for kindness are from Table 3.1):
Calculating the sums of squared errors for ratings of kindness  

Kindness $x$ 
Mean $\bar{X}$ 
Deviance $x\bar{X}$ 
Deviance squared $(x\bar{X})^2$ 

10  9.2  0.8  0.64  
10  9.2  0.8  0.64  
8  9.2  1.2  1.44  
9  9.2  0.2  0.04  
10  9.2  0.8  0.64  
10  9.2  0.8  0.64  
10  9.2  0.8  0.64  
9  9.2  0.2  0.04  
8  9.2  1.2  1.44  
7  9.2  2.2  4.84  
8  9.2  1.2  1.44  
9  9.2  0.2  0.04  
9  9.2  0.2  0.04  
10  9.2  0.8  0.64  
9  9.2  0.2  0.04  
9  9.2  0.2  0.04  
9  9.2  0.2  0.04  
10  9.2  0.8  0.64  
10  9.2  0.8  0.64  
10  9.2  0.8  0.64  
Sum  15.20 
To calculate the variance, use the sum of squares from Table 2:
$$ s^2 = \frac{\text{sum of squared error}}{N1} = \frac{15.2}{19} = 0.8. $$
We can now calculate the standard deviation:
$$ s = \sqrt{0.8} = 0.89. $$
Now we can convert Celia’s zscore for the kindness variable back into her raw score by rearranging the equation for zscores, and then plugging in her zscore and the values of the mean and standard deviation that we have just calculated:
$$ \begin{aligned} z &= \frac{X\bar{X}}{s} \\ X &= zs + \bar{X} \\ &= 0.9 \times 0.89+9.2 \\ &= 10. \end{aligned} $$
Therefore, Celia’s raw score for the kindness was 10 out of 10, suggesting that Celia values kindness very highly in a romantic partner.
Puzzle 5
For ambition, Celia gave a rating of 9, which yielded a zscore of 1.56. What was the standard deviation of ambition ratings? (Hint: use the values in Table 3.1 in the book to compute the mean of the distribution of ambition ratings.)
First we need to compute the mean:
$$ \bar{X} = \frac{\sum_{i = 1}^n x_i}{n} = \frac{135}{20} = 6.75. $$
We can calculate the standard deviation (s) of the ambition ratings by rearranging the equation for finding the zscore and then plugging in the values of z, Celia’s rating (X) and the mean ($\bar{X}$) that are provided in the question:
$$ \begin{aligned} z &= \frac{X\bar{X}}{s} \\ s &= \frac{X\bar{X}}{z} \\ &= \frac{96.75}{1.56} \\ &= 1.44. \end{aligned} $$
Therefore, the standard deviation for ambition was 1.44.
Puzzle 6
For humour, Celia also gave a rating of 9, which yielded a zscore of 0.66. The standard deviation of humour scores was 1.43. What was the mean of the humour ratings?
We can calculate the mean ($\bar{X}$) of the humour ratings by rearranging the equation for finding zscores and then plugging in the values of z, Celia’s rating of humour (X) and the standard deviation (s) that are provided in the question:
$$ \begin{aligned} z &= \frac{X\bar{X}}{s} \\ \bar{X} &= X  zs \\ &= 90.66\times1.43 \\ &= 8.06. \end{aligned} $$
Therefore, the mean rating for humour was 8.06 out of 10.
Puzzle 7
Which of the four attributes did Celia value most? (Hint: use the zscores.)
I have put Celia’s zscores for each characteristic into a table so that we can compare them most easily (Table 4 ). If we look at Celia’s raw scores we can see that she gave the highest rating for kindness (10 out of 10) and the lowest rating for high salary (5 out of 10). However, if we look at her zscores we can see her ratings of the four characteristics in relation to the ratings of the 20 other women. When we do this we can see that her highest zscore by far was for ambition, which was 1.56, suggesting that compared to the other women in the study, Celia values ambition very highly. Her lowest zscore was for high salary ($0.27$); this negative zscore suggests that compared to the other women in the study, Celia does not value high salary as a very important characteristic in a partner.
Celia's raw scores and $z$scores for ratings of four characteristics she might value in a mate  

Characteristic  Celia's $z$score  Celia's raw score 
High salary  0.27  5 
Kindness  0.90  10 
Ambition  1.56  9 
Humour  0.66  9 
Puzzle 8
Celia mentioned Zach’s cheekbones a lot. In the data Professor Pincus looked at there were also 20 women’s ratings of how important they thought attractiveness was. The scores were: 4, 10, 9, 8, 7, 8, 10, 8, 7, 3, 9, 10, 8, 10, 7, 9, 9, 9, 8, 7. Convert this distribution to zscores. Celia rated attractiveness as 9. What was the zscore for this raw score?
First, we calculate the mean
$$ \bar{X} = \frac{\sum_{i = 1}^n x_i}{n} = \frac{160}{20} = 8. $$
Now the sum of squared errors
Calculating the sums of squared errors for ratings of attractivenss  

Attractiveness $x$ 
Mean $\bar{X}$ 
Deviance $x\bar{X}$ 
Deviance squared $(x\bar{X})^2$ 

4  8  4  16  
10  8  2  4  
9  8  1  1  
8  8  0  0  
7  8  1  1  
8  8  0  0  
10  8  2  4  
8  8  0  0  
7  8  1  1  
3  8  5  25  
9  8  1  1  
10  8  2  4  
8  8  0  0  
10  8  2  4  
7  8  1  1  
9  8  1  1  
9  8  1  1  
9  8  1  1  
8  8  0  0  
7  8  1  1  
Sum  66.00 
To calculate the variance use the sum of squares:
$$ s^2 = \frac{\text{sum of squared error}}{N1} = \frac{66}{19} = 3.47 $$
We can now calculate the standard deviation:
$$ s = \sqrt{3.47} = 1.86. $$
Now we know what the mean and standard deviation of the distribution are, we can work out the zscores of each of the raw scores by using the equation:
$$ z = \frac{X\bar{X}}{s} $$
The unique scores in the data are 3, 4, 7, 8, 9 and 10, which we can convert to z as follows
$$ \begin{aligned} z_3 &= \frac{38}{1.86} = 2.69 \\ z_4 &= \frac{48}{1.86} = 2.15 \\ z_7 &= \frac{78}{1.86} = 0.54 \\ z_8 &= \frac{88}{1.86} = 0 \\ z_9 &= \frac{98}{1.86} = 0.54 \\ z_{10} &= \frac{108}{1.86} = 1.08 \end{aligned} $$
The distribution, therefore, becomes
Attractiveness ratings as $z$scores  

Attractiveness $x_i$ 
$z$score $z_i$ 
4  2.15 
10  1.08 
9  0.54 
8  0.00 
7  0.54 
8  0.00 
10  1.08 
8  0.00 
7  0.54 
3  2.69 
9  0.54 
10  1.08 
8  0.00 
10  1.08 
7  0.54 
9  0.54 
9  0.54 
9  0.54 
8  0.00 
7  0.54 
The zscore for Celia’s raw score of 9 was 0.54. This positive zscore suggests that Celia rates attractiveness as quite an important characteristic in a partner.
Puzzle 9
Alice also rated these characteristics of a partner, but using a different rating scale. For kindness, this scale had a mean of 15 and a standard deviation of 7. Alice rated kindness as 21. Who values kindness more: Alice or Celia? (Hint: compare the zscores.)
First we need to convert Alice’s raw score into a zscore:
$$ z = \frac{X\bar{X}}{s} = \frac{2115}{7} = 0.86 $$
Therefore, Alice’s zscore for kindness was 0.86, and if we look back at Celia’s zscore for kindness we can see that it was 0.9. These scores are almost identical, indicating that Alice and Celia value kindness very similarly.
Puzzle 10
In the data Professor Pincus looked at there were also 20 women’s ratings of how important they thought being romantic was. The scores were: 7, 10, 7, 8, 9, 8, 10, 7, 7, 4, 5, 8, 7, 8, 7, 9, 3, 7, 10, 7. Convert this distribution to zscores. Celia rated being romantic a rating of 9. what was the zscore for this raw score?
First, we calculate the mean
$$ \bar{X} = \frac{\sum_{i = 1}^n x_i}{n} = \frac{148}{20} = 7.4. $$
Now the sum of squared errors
Calculating the sums of squared errors for ratings of romantic  

Romantic $x$ 
Mean $\bar{X}$ 
Deviance $x\bar{X}$ 
Deviance squared $(x\bar{X})^2$ 

7  7.4  0.4  0.16  
10  7.4  2.6  6.76  
7  7.4  0.4  0.16  
8  7.4  0.6  0.36  
9  7.4  1.6  2.56  
8  7.4  0.6  0.36  
10  7.4  2.6  6.76  
7  7.4  0.4  0.16  
7  7.4  0.4  0.16  
4  7.4  3.4  11.56  
5  7.4  2.4  5.76  
8  7.4  0.6  0.36  
7  7.4  0.4  0.16  
8  7.4  0.6  0.36  
7  7.4  0.4  0.16  
9  7.4  1.6  2.56  
3  7.4  4.4  19.36  
7  7.4  0.4  0.16  
10  7.4  2.6  6.76  
7  7.4  0.4  0.16  
Sum  64.80 
To calculate the variance use the sum of squares:
$$ s^2 = \frac{\text{sum of squared error}}{N1} = \frac{64.8}{19} = 3.41 $$
We can now calculate the standard deviation:
$$ s = \sqrt{3.41} = 1.85. $$
Now we know what the mean and standard deviation of the distribution are, we can work out the zscores of each of the raw scores by using the equation:
$$ z = \frac{X\bar{X}}{s} $$
The unique scores in the data are 3, 4, 5, 7, 8, 9 and 10, which we can convert to z as follows
$$ \begin{aligned} z_3 &= \frac{37.4}{1.85} = 2.38 \\ z_4 &= \frac{47.4}{1.85} = 1.84 \\ z_5 &= \frac{57.4}{1.85} = 1.30 \\ z_7 &= \frac{77.4}{1.85} = 0.22 \\ z_8 &= \frac{87.4}{1.85} = 0.32 \\ z_9 &= \frac{97.4}{1.85} = 0.87 \\ z_{10} &= \frac{107.4}{1.85} = 1.41 \end{aligned} $$
The distribution, therefore, becomes
Ratings of 'romantic' as $z$scores  

Romantic $x_i$ 
$z$score $z_i$ 
7  0.22 
10  1.41 
7  0.22 
8  0.32 
9  0.86 
8  0.32 
10  1.41 
7  0.22 
7  0.22 
4  1.84 
5  1.30 
8  0.32 
7  0.22 
8  0.32 
7  0.22 
9  0.86 
3  2.38 
7  0.22 
10  1.41 
7  0.22 
The zscore of Celia’s raw score of 9 was 0.87, suggesting that she values the characteristic romantic in a partner more than the average person (nearly a standard deviation higher, in fact).