z‐score and Unit Normal Table Exercises Exercise 1 ‐ Mr. Jones has given an exam to his math class. Below you will find the scores for 6 of his students who took the exam. The way Mr. Jones gives grades is based on percentiles. Anybody’s scoring above the 85th percentile gets an A, between the 70th and the 85th percentile students get a B, between the 55th and the 70th percentile students get a C, between the 40th and the 55th percentile students get a D, and below the 40th percentile students get an F. Using their exam scores as a starting point, calculate a grade for each of the six students by first calculating a z‐score, then using the z‐score to calculate a percentile, and finally matching their percentile to a letter grade. The mean and standard deviation for the exam are given below the scores. Student Aida John Lenore David Justin Sarah Exam Mean Exam SD Score 54 52 67 78 69 59 z‐score _______ _______ _______ _______ _______ _______ 55 7 Percentile
_______ _______ _______ _______ _______ _______ Grade _______ _______ _______ _______ _______ _______ Exercise 2 ‐ Miss Rodriguez was talking about grading to Mr. Jones, and liked the idea of giving letter grades based on percentiles. Since she has already given out four exams, she does not want to go back and recalculate each student’s score individually. Instead, she will calculate the cut‐off points for each of the four exams, give these values to her TA, and have him go back and change all the grades based on these values. Below you are given the mean and standard aviation for each of Ms. Rodriguez’s four exams. For each exam, calculate the minimum score needed to get a particular letter grade. Use the same criteria as Mr. Jones in the first exercise, above the 85th percentile is an A, between the 70th and the 85th percentiles is a B, between the 55th and the 70th percentiles is a C, between the 40th and the 55th percentiles is a D, and below the 40th percentile is an F. Exam Exam 1 Exam 2 Exam 3 Exam 4 Mean 55 58 62 45 SD 5 7 8 4 Minimum points for letter grade A B C D ______ ______ ______ ______ ______ ______ ______ ______ ______ ______ ______ ______ ______ ______ ______ ______ Exercise 3 ‐ Mr. Smith has 100 students in his class. The last exam he gave was normally distributed with a mean of 82 and a standard deviation of 10. Use this information to answer the following questions. 1. If Sue scored a 90, and John scored a 95, what is the probability of a student scoring more than Sue but less than John? 2. How many students scored above an 80? 3. What is the probability of a student scoring more than 70, but less than 90? 4. If the bottom 25% get an F, what would be the minimum score needed to get at least at D? 5. If only the top 10% get an A, what score would you need to get an A on the exam? 6. If the middle 30% get a C, what would be the minimum and maximum scores a student could get to earn a C in the exam? 7. How many students scored between a 70 and an 85 on the exam? 8. What is the probability that a student scored less than 75 points on the exam? 9. What percentage of students scored between 70 and 80 points on the exam? 10. What proportion of students scored between 85 and 90 points on the exam? Exercise 4 ‐ You are in charge of enrollment at a prestigious university. The university only accepts the top 10% of GRE scores, and uses these scores to organize students into 4 tiers to determine the order in which they can pick their classes. The top 5% get first choice, then the next 5%, and so on. You already know that the GRE has a mean of 500, and a standard deviation of 100. What GRE scores would define these four tiers? Tier First Second Third Fourth Top 2.5% 2.5%‐5.0% 5.0%‐7.5% 7.5%‐10% Minimum Score _________ _________ _________ _________ Answerrs Exercise 1
1 ‐ To answer these questions you will n
need to conveert each score into a z‐sco
ore, and then use that value
e to look in th
he unit normaal table for the percentagee of the curvee than lies below it. Your answers sshould be as ffollows: Student Aida John Lenore David Justin Sarah Score
e 54 52 67 78 69 59 Exam Meaan Exam SD z‐score ‐0.14 ‐0.43 1.71 3.29 2.00 0.57 Percentile 44.43
33.36
95.64
99.95
97.72
71.57
55 7 To answer these you should begin b
by drawing a distributio
on, and mapp
ping each of the six studen
nts onto it like the image o
on the right. e problem can be concepttualized a littlle easier. Forr example, loo
oking Once you have the imaage made, the
would expect tthe z‐score fo
or Aida to be a little bit less at the imaage of the disstribution on tthe left we w
than zero, and the percentile a little
e less than 50
0. Since each half of the distribution acccounts for haalf of it, then a little less than half should be a little bitt less than a percentile score
e of 50. problem and yyou Now that you have draawn out the p
hat values you should be have a good idea of wh
e the precise expectingg, you can start to calculate
values. z=
X −μ
σ
=
54 − 55 − 1
=
= −0.14 7
7
Once you have a z‐score, you can lo
ook in the tab
ble to find the correspondingg value for the
e area you are interested
d in. You shou
uld look for th
he value in th
he C column off the table for a correspon
nding z‐score value of 0
0.14. Remember that the ttable does no
ot use negative vvalues. The vvalue you get is 0.4443, wh
hich translatess to a percenttile of 44.43. Exercise 2 – The easie
est way to beggin to answerr these probleems ulate the z‐sco
ores that define the percentiles of interrest. is to calcu
In other w
words, we want to know w
what z‐scores define the toop 15%, the ttop 30%, the top 45%, and
d the top 60%
%. So let’s skeetch out a distribution and identify the aareas we are iinterested in.. Then we ccan calculate what z‐score
es would defin
ne those areaas. Grade A B C D Percentile 85 70 55 40 z‐score 1.04 0.52 0.13 ‐0.25
5 To calculaate the first z‐‐score we nee
mal ed to look in tthe units norm
table for tthe value for z. Begin by ccomparing your sketch of tthe problem ((top right imaage) to the im
mages at the start of the ta ble. It should b
be easy to see that the pro
oblem we are
e trying to sollve looks like the image on
n the top left of the top of the unit norm
mal table. We
e want to find the value of z when C equ
uals 15% or .1500. Once the z‐scores are iidentified, the
e table can be
e completed by calculatingg the corresp
ponding scorees for he four examss. each of th
For Exam 1, the answe
ers would be ccalculated as follows: X = (z*σ) +
+ µ = (1.04 * 5
5) + 55 = 5.20
0 + 55 = 60.20
0 X = (z*σ) +
+ µ = (0.52 * 5
5) + 55 = 2.60
0 + 55 = 57.60
0 X = (z*σ) +
+ µ = (0.13 * 5
5) + 55 = 0.65
5 + 55 = 55.65
5 X = (z*σ) +
+ µ = (‐0.25 * 5) + 55 = ‐1.2
25 + 55 = 53.7
75 Exam Exam 1 Exam 2 Exam 3 Exam 4 Mean 55 58 62 45 SD 5 7 8 4 Minimum poin ts for letter ggrade B A D C 60.20 57.60 55.65 53.75 65.28 61.64 58.91 56.25 70.32 66.16 63.04 60.00 49.16 47.08 45.52 44.00 Exercise 3
3 – Begin by ssketching out the question
n. Rememberr the mean was 82, and the standard deviation was 10. ohn scored a 95, what is thhe 1. If Sue scored aand 90, and Jo
probability of aa student sco
oring more than Sue but leess han John? th
he picture on
n the right illustrates the question. We’’re Th
in
nterested in the area shade
ed in gray, wh
hich is the areea between a sco
ore of 90 and a score of 95. Fiirst we need tto calculate z‐scores for sccores of 90 annd 95
5. Since the u
unit normal table only hass z‐scores, wee need to have zz‐scores to be
e able to use iit. X −μ
90 − 82 8
=
= 0.80 σ
10
10
0
X − μ 95
9 − 82 13
z=
=
= 1.30 =
σ
10
10
z=
=
Once you know
O
w the value fo
or the z‐score
es that co
orrespond to 90 and 95, yo
ou can start tto look for thee area. This p
particular pro
oblem is a littlle bit trricky in that th
he area we w
want does nott have the meean or the taill as one of itss boundaries. Th
here are actu
ually several w
ways that we can solve thiss problem using our tools.. Basically, w
we kn
now that the whole distrib
bution accoun
nts for 100% oof itself or 1.00000. If we ccan figure outt the white areas, th
w
hen the gray aarea must be one minus thhe white areaa. X + C = 1.0000
0 then… X == 1.0000 – B –– C if… B + X
We can calcula
W
ate the white area in two p
pieces, one thhe piece to th
he left of the gray area, an
nd the otther the piece
e to the rightt. B = .7881; C = .0968 therefo
ore X = 1.0000
0 ‐ .7881 ‐ .09968 = .1151 Answer ‐ There
e is a .1151 probability of aa student gettting a score b
between 90 aand 95. 2. How many students scored above an 80? z=
X −μ
σ
=
80
8 − 82 − 2
= −0.20 =
10
10
0
Th
he proportion
n of the area above a z‐sco
ore of ‐.20 is .5
5793. Since th
here are 100 sstudents, there must be 1000 * .5793 = 57.93 studen
nts who scored ab
bove 80 on th
he exam. (57.93% of 100 students is 577.93) Answer – 58 sttudents. obability of a student scorring more thaan 70, but lesss than 90? W
3. What is the pr
X −μ
70
7 − 82 − 12
1
=
= −1.20
σ
10
10
0
X − μ 90
9 − 82 8
z=
=
= 0.80
=
σ
10
10
z=
=
Th
he combined area is equal to .3849 + .2
2881 = .6730
Answer – The probability th
hat a student will score be tween 70 and
d 90 is .6730 25% get an F,, what would be the minim
mum score neeeded to get aat least a D? 4. If the bottom 2
XX = (z*σ) + µ
µ = (0.67 * 0) + 82 = 6.70
0 + 82 = 88.70
0 10
Answer – Students would n
need to get a score of at le ast 88.70. A, what score would you n eed to get me on the exam
m? 5. If only the top 10% get an A
X = (z*σ) + µ = (1.28 * 10) + 82 = 12.80 + 82 = 94.80
Answer ‐ Stude
ents would ne
eed to get a sscore of at leaast 94.80. 6. If the middle 3
30% get a C, w
what would be the minimuum and maxim
mum scores aa student could get to
o earn a C in tthe exam? 82 = 85.90 X = (z*σ) + µ = (0.39 * 10) + 82 = 3.90 + 8
+ 82 = ‐3.90 +
+ 82 = 78.10
X = (z*σ) + µ = (‐0.39 * 10) +
udent would need a score of at least 788.10 and not m
more than 85
5.90. Answer – A stu
7. How many students scored between a 7
70 and an 85 oon the exam?? X −μ
70
7 − 82 − 12
1
=
= −1.20
σ
10
10
0
X − μ 85
8 − 82 3
z=
=
= 0.30 =
σ
10
10
z=
=
Siince the prop
portion of the distribution between a z‐‐score of ‐1.200 and a z‐score of .30 is (.1
1179 + .3849) .5029, and there are 100 studen
nts, 50.29% oof 100 is 50.299 students. Answer – 50 sttudents. 8. What is the pr
obability thatt a student sccored less thaan 75 points o
W
on the exam?? z=
X −μ
σ
=
75
7 − 82 − 7
=
= −0.70 10
100
Answer ‐ The p
probability that a student w
will score bellow 70 is .25114. 9. What percenta
W
age of studen
nts scored bettween 70 andd 80 points on
n the exam? X −μ
70
7 − 82 − 12
1
=
= −1.20
σ
10
10
0
X − μ 80
8 − 82 − 2
z=
=
= −0.20 =
σ
10
10
0
z=
=
B = .5793; C = .1151 therefo
ore X = 1.0000
0 ‐ . 5793‐ . 1 151 = .3056 o
or 30.56% Answer ‐ 30.56
6% of studentts scored betw
ween 70 and 80 points on
n the exam. W
on of studentts scored between 85 and 90 points on
n the exam? 10. What proporti
X −μ
85
8 − 82 3
= 0.30
=
σ
10
10
X − μ 90
9 − 82 8
z=
=
= 0.80 =
σ
10
10
z=
=
B = .6179; C = .211
19 therefore X
X = 1.0000 ‐ . 6179 ‐ . 21199 = .1702 portion of students who sccored betweeen 85 and 90 Answer – The prop
pointss on the exam
m was.1702. Exercise 4 ‐ X = (z*σ) + µ = (1.96 * 100) + 500 = 196 + 500 = 696 X = (z*σ) + µ = (1.65 * 100) + 500 = 165 + 500 = 665 X = (z*σ) + µ = (1.44 * 100) + 500 = 144 + 500 = 644 X = (z*σ) + µ = (1.28 * 100) + 500 = 128 + 500 = 628 Tier First Second Third Fourth Top 2.5% 2.5%‐5.0% 5.0%‐7.5% 7.5%‐10% z‐score 1.96 1.65 1.44 1.28 Score 696 665 644 628