Z-Scores
Z-Score:Itisameasureofatheposi4onspecificvalueina
datasetrela4vetomeanintermsofthestandarddevia4on
units.Itissome4mescalledtheStandardScore.
Valueof92is1.45
standarddevia4onunits
abovethemean
Valueof72is0.35
standarddevia4onunits
belowthemean
Sec4on2.5,Page45
1
ZScoreProblems
2.50
Shelia and Joan competed in two events at a
track meet. In the long jump, Shelia’s jump was
0.5 meters above the mean of the group and
Joan’s jump was 0.3 meters beyond the mean.
The standard deviation was 0.25 meters.
In the 100 meter run Shelia’s time was 5
seconds faster than the mean and Joan’s time
was 8 seconds faster than the mean. The
standard deviation was 3 seconds.
Who had the best combined performance?
Why?
Problems,Page52
2
NormalModel
Valuesonthex-axisextendfrom-∞to+∞.
Thetotalareaunderthecurveequals1.0.
Theareaaboveanyintervalonthex-axisequalsthe
percentofdatavaluesintheinterval.
Sec4on6.1
3
NormalModels
Standard Normal Curve
µ = 0, σ = 1
−( x− µ )2
f (x) =
2
e
σ 2π
2σ
€
Sec4on6.1
4
Probabili4esandNormalCurves
To demonstrate the process of finding probabilities
related to populations that can be described by a
normal curve, let’s consider IQ scores. IQ scores
are normally distributed with a mean of 100 and a
standard deviation of 16. If a person is picked at
random, what is the probability that her IQ is
between 100 and 115: that is, what is
P(100<x<115)?
Sec4on6.3,Page124
5
FindingAreaswiththeTI-83
PRGM – NORMDIST
1 – ENTER
100 ENTER: Enters Lower Bound
115 ENTER: Enters Upper Bound
100 ENTER: Enters Mean
16 ENTER: Enters Standard Deviation
Displays results:
The area above the interval is .3257.
Therefore the probability of selecting a
person at random with an IQ between 100
and 115 is .3257 or 32.57%.
Sec4on6.3,Page124
6
FindingAreaswiththeTI-83
90 100
σ=16
Find P(x > 90)
PRGM – NORMDIST – ENTER
1 – ENTER
90 ENTER
2ND EE99 – ENTER: On the TI 83, 2nd E99 = ∞
100 – ENTER
16 – ENTER
Display:
P(x > 90) = .7340
Sec4on6.3,Page125
7
Problems
Problems,Page133
8
Problems
Problems,Page133
9
StandardNormalCurve
Z-Curve
z-axis
µ=0 σ=1
The standard normal curve or distribution is
simply a normal distribution with a mean of 0 and
a standard deviation of 1. The values on the
horizontal axis are z-values.
Before computers, all problems had to be
converted to a standard normal distribution and
then solved with tables in the book that relate the
z-values to the areas above the intervals.
Example: Find the area above the z-axis
bounded by -1.5<z<2.1
NORMDIST 1
LOWER BOUND = -1. 5
UPPER BOUND = 2.1
Mean = 0
STANDARD DEVIATION = 1
Answer: AREA = .9153
Sec4on6.2,Page122
10
Problems
Problems,Page132
11
DetermineDataValueswithTI-83
Suppose that in a large class, your instructor tells you
that you need to be in the top 10% of the class to get
an A. For a particular exam, the mean is 72 and the
standard deviation is 13. (Assume a normal
distribution) What grade is required for an A?
Top 10%
Area from the
left = 1-.10=.90
Area = .10
µ = 72, σ = 13
Score = ?
PRGM – NORMDIST – 2
.90 – ENTER
72 – ENTER
13 – ENTER
Display:
A test score of 88.66
or above is required to
be in the top 10%
Sec4on6.3,Page125
12
Problems
6.51 IQ scores are normally distributed with a
mean of 100 and a standard deviation of 16. Find
the following:
a.  The 66th percentile.
b.  The 80th percentile.
c.  The minimum score required to be in the top
10%.
d.  The minimum score to be in the top 25%.
6.52 Find the two z-scores that bound the middle
30% of the standard normal distribution.
Problems,Page133
13