Statistical Analysis of Normal Distributions and Z-Scores Assignment

This assignment analyzes normal distributions and z-scores using customer purchase data and student exam scores. The initial part examines probabilities related to customer spending, calculating values using the cumulative distribution function and Excel. It then delves into descriptive measures like mean, median, and standard deviation for two classes, calculating z-scores to compare student performances. The assignment highlights the limitations of raw scores and emphasizes the utility of z-scores in comparing performances across different populations and in standardizing scores. It demonstrates how z-scores provide a more accurate assessment of student performance compared to raw scores by considering the distribution of scores within a class and across different classes. The analysis concludes by comparing student performances in different classes and highlighting how the standard of marking impacts z-scores. The document uses references to support its statistical analysis and findings.

Running head: NORMAL DISTRIBUTIONS AND Z-SCORES
Normal distributions and z-scores
Name of Student
Name of University
Author Note

1NORMAL DISTRIBUTIONS AND Z-SCORES
Table of Contents
Analysing Normally Distributed data..............................................................................................2
1...................................................................................................................................................2
a)..............................................................................................................................................2
b)..............................................................................................................................................2
c)..............................................................................................................................................3
2...................................................................................................................................................4
Looking Past the Mean and Standard Deviation.............................................................................4
3...................................................................................................................................................4
4...................................................................................................................................................5
5...................................................................................................................................................6
6...................................................................................................................................................7
7...................................................................................................................................................8
8...................................................................................................................................................8
References......................................................................................................................................10

2NORMAL DISTRIBUTIONS AND Z-SCORES
Analysing Normally Distributed data
It is given that a local business owner seeks to understand data on his customers in order to
act on the insight to benefit business. The purchase amounts of the customers follow
approximately a normal distribution, with mean $75 and standard deviation $20.
1.
a)
Probability that a random customer spends less than some amount “x” is Φ( x−75
20 ),
where X N ( 0,1 ) , andΦ ( . )being the cumulative distribution function of N ( 0,1 )(Rohatgi & Saleh,
2015).
Then, if Y N ( 75,20 ) , X= Y −75
20
 P(Y< $85) = P(X< 85−75
20 ) =0.6915
 P ($65<Y< $85) = P(Y< $85) - P(Y< $65) = P(X< 85−75
20 ) - P(X< 65−75
20 ) =0.3829
 P(Y>$45) = 1-P(Y<$45) = 1- P(X< 45−75
20 ) = 0.9332
The computations have been done in Excel.
b)
A value of Y N ( 75,20 ) is required where P(Y<y) = 0.75.Additionally a Y value where
P(Y>y) =0.8 is also required.
Then P−1 ( 0.75 ) = 88.4898, for mean 75 and standard deviation 20.

3NORMAL DISTRIBUTIONS AND Z-SCORES
Again,
P(Y>y) = 1-P(Y<y) = 0.8
This means, P(Y<y) =0.2
Then, P−1 ( 0.2 ) = 58.16757533, for mean 75 and standard deviation 20.
Therefore 75% of the customers spend at most $88.49 in a single purchase and 80% of
the customer purchases at least $58.17 worth of goods.
The computations have been done in Excel.
c)
Two values of Y is to be determined which are equidistant from the mean and lie on
opposing sides such that 90% of the values lie within the interval.
Then P (-x < Y −75
20 <x) = 0.90
P (-x <X <x) = 0.90
Or, 1- 2.P (X<x) =0.9 (Rohatgi & Saleh, 2015).
Then, P(X<x) = 0.05
Then upper limit is found to be, y2= 20 ×Φ−1(0.05) +75 = 107.896 and,
The lower limit is found to be, y1= 75-20 ×Φ−1(0.05) = 42.104
Thus 90% of the customer purchases lie in the interval (42.104, 107.896).

4NORMAL DISTRIBUTIONS AND Z-SCORES
2.
The findings in the preceding parts reveal that the bulk of the sales per individual may
range approximately between $42 and $110. The shop owner may increase stock priced within
the range $45 to $107 to increase sales as most customers buy products priced within that range.
Looking Past the Mean and Standard Deviation
3.
The descriptive measures, namely, mean, median, mode, standard deviation, maximum
and minimum of the scores of students in Mr. Moonbeam and Mr. Bookbi’s class are given as
follows (Salkind, 2015):
Dr. Bookbi's Class
mean 87.29
median 86.52
mode 77.17
standard deviation 7.118575556
min 71.84
max 97.93
Table 1
Dr. Moonbeam's Class
mean 93.34
median 93.00
mode 93
standard deviation 3.792122214
min 84.62
max 99.43
Table 2

5NORMAL DISTRIBUTIONS AND Z-SCORES
4.
Taking into account the mean ad standard deviation values, calculated in the preceding
part, the corresponding z-scores of the students for each class has been obtained using
STANDARDIZE function as given in the following tables (Salkind, 2015):
Dr. Bookbi's Class
Student Name Exam Grade Z scores
Alfred 77.17 -1.4217888
Amanda 71.84 -2.1705341
Andrew 77.17 -1.4217888
Anthony 81.00 -0.8837598
Corey 74.45 -1.8038877
Edward 84.23 -0.4300174
Jaime 97.93 1.494525
James 88.47 0.1656074
Jane 95.54 1.1587836
Jerry 83.91 -0.4749702
Kerry 86.41 -0.1237763
Laura 94.78 1.0520207
Leisha 94.78 1.0520207
Marques 90.00 0.3805381
Melissa 93.91 0.9298052
Michelle 92.17 0.6853743
Mostra 93.91 0.9298052
Nicola 83.91 -0.4749702
Rae 83.04 -0.5971856
Roger 88.47 0.1656074
Sally 95.54 1.1587836
Samantha 84.45 -0.3991123
Suzie 84.45 -0.3991123
Theodore 94.56 1.0211156
Thomas 94.56 1.0211156
Tom 86.52 -0.1083238
Trecia 83.69 -0.5058752
Table 3
Dr. Moonbeam's Class
Student Name Exam Grade Z-score

6NORMAL DISTRIBUTIONS AND Z-SCORES
Adam 99 1.60518
Alli 93 -0.09044
Beth 93 -0.09044
Cameron 95 0.436968
Diana 96 0.682214
Dorsey 92 -0.35415
Edna 90 -0.88155
Erica 95 0.436968
Gabriel 99 1.491787
Kendall 90 -0.88155
Kali 91 -0.5361
Kimberley 95 0.436968
Kisha 91 -0.61785
Lacey 99 1.491787
Marc 93 -0.09044
Maria 92 -0.3225
Mary 88 -1.40896
Niles 94 0.173264
Rebecca 99 1.491787
Sara 94 0.270834
Sean 94 0.173264
Shawna 91 -0.57038
Sonya 98 1.259727
Terrell 90 -0.88155
Theresa 85 -2.30029
Tiffany 87 -1.6252
Tracey 96 0.700673
Table 4
5.
The performance of a student in a test is measured by the raw scores. However looking at
the raw scores alone, it may prove to be inadequate to determine the performance in comparison
to other students. Z scores allows one to position the student on the basis of the z-scores obtained
from the raw scores on the standard normal distribution and hence compute the probability of the
student obtaining the raw score in light of the distribution of scores for the class (Lindgren
2017).

7NORMAL DISTRIBUTIONS AND Z-SCORES
The students Tiffany and Sonya in Dr. Moonbean’s class show a difference of 11 marks
in their scores with Tiffany at 87 and Sonya at 98. However when considering the z-score, it is
found that there are 84.41% of the students in the class who have scores between 87 and 98 ,
positioning Tiffany at a much lower position of performance as compared to what the raw scores
indicate.
Considering Erica and Dorsey, the difference is of only 3 marks with Erica scoring 95
and Dorsey scoring 92. Despite such proximity in marks, it is observed through z-scores that
30.73% of the students have scores between 95 and 92 which reveals a wider performance gap.
Mary and Tiffany exhibit only a 1 mark difference with their scores being 88 and 87
respectively. The z scores reveal that 2.74% of the students have scores between 88 and 87.
Therefore the performance when being considered through raw scores alone fail to take into
account, the comparative performance of others and hence the competitive aspect of the test.
6.
A second utility of z scores is that it can allow one to compare the raw scores which come
from different populations following a distribution with different parameters by standardizing
and thus mapping the raw scores onto a standard normal distribution (Lindgren, 2017). Then,
one could compare the performance of a two students in the same subject but from different
classes or the performance of the student in different subjects.
Marques from Dr.Bookbi’s class had scored 90 while Edna from Dr. Moonbean’s class
had scored the same. The raw scores indicate that both are equivalent in performance. However
this does not take into account the difficulty level of the tests conducted by the two lecturers. The
students belong to different populations and thus the scores have different distributions making

8NORMAL DISTRIBUTIONS AND Z-SCORES
any comparison absurd. The z scores however could be compared and it is seen that Marques has
a z score of 0.3805 which is greater than Edna’s -0.385, which clearly indicates that Marques has
performed much better than Edna.
Similarly for Tom from Dr .Bookbi’s class and Tiffany from Dr. Moonbean’s class,
Although Tom has scored 86.52 which is lower than Tiffany’s 87.However, looking at the z-
scores it is seen that Tom has a score of -0.10832 which is greater than Tiffany’s -1.6252 making
Tom the better performer.
Again, for Suzie from Dr .Bookbi’s class and Theresa from Dr. Moonbean’s class,
Theresa had scored 95 which is greater than Suzie’s 84.45.However, the z-scores reveal that
Theresa has a score of -2.300 as opposed to Suzie’s -0.39911 making Theresa a poorer performer
than Suzie.
7.
It is seen that Kali from Dr. Moonbean’s class had scored 91 whereas Marques from Dr.
Bookbi’s had scored 90. If the scores are assumed to be measures of knowledge, then by
comparing the scores of the students one could comment on how much knowledge they have.
However in order to do so, the scores of the two classes much be considered on the same scale.
Then upon standardizing the scores and comparing the z-scores it is seen that Marques has a z-
score of 0.3805 which is higher than Kali’s z-score of -0.5361 which is less than average.
Despite the proximity of their raw scores, it is found that Marques knows more about
information systems than Kali.

9NORMAL DISTRIBUTIONS AND Z-SCORES
8.
It has been found that students in Dr.Bookbi’s class a have scored less than those in Dr.
Moonbean’s class. However it is apparent that the standard of marking in Dr.Bookbi’s is much
harsher than that of Dr. Moonbean. This is because even though raw scores have been found to
be equal or lesser than those of Dr. Moonbean’s class, the position of the student with respect to
the other students when their z-scores are compared is much higher in many instances (Albright
& Winston, 2015). Comparisons in the preceding part provides the evidence for this.

10NORMAL DISTRIBUTIONS AND Z-SCORES
References
Albright, S. C., & Winston, W. L. (2015). Business analytics: Data analysis & decision making.
Cengage Learning.
Lindgren, B. (2017). Statistical theory. Routledge.
Rohatgi, V. K., & Saleh, A. M. E. (2015). An introduction to probability and statistics. John
Wiley & Sons.
Salkind, N. J. (2015). Excel statistics: A quick guide. SAGE Publications.