2.5 Normal Distributions and
z-scores
Comparing marks
• Stephanie and Tavia are both in the running
for the Data Management award. Stephanie
has 94% and Tavia has 92%, from different
classes.
• Who deserves the award?
• What if I told you it was Tavia?
• Why?
• Stephanie’s class:
mean 78,  = 9.36
Scatter Plot
Normal Distributions
0.07
0.07
0.06
0.06
0.05
0.05
0.04
0.04
0.03
0.03
0.02
0.02
0.01
0.01
0.00
0.00
50
60
70
80
90
100
Scatter Plot
Normal Distributions
data
data
• Tavia’s class: mean
73,  = 8.19
50
60
70
x
data = normalDensity x mean sd
80
x
data = normalDensity x mean sd
• Distributions are different
• Fair comparison not possible
…yet
90
100
Standard Normal Distribution
– this process is called
standardizing
X  N(0,12 )
Scatter Plot
Standard Normal Distribution
0.6
0.5
data
• Mean 0, standard deviation
1
• Can translate each element
of a normal distribution to
standard normal distribution
by finding number of  a
given score is away from the
mean
0.4
0.3
0.2
0.1
-8
-6
-4
-2
0
x
data = normalDensity x
2
4
6
8
z-scores
• z = The number of standard deviations a
given score x is above or below the mean
x  x  z
z
xx

• z = z-score
above the mean
– Positive: value lies _________
below
– Negative: value lies _________
the mean
Example 1: Calculating z-scores
• Consider the distribution X  N(14, 4 )
• Find the number of standard deviations each
piece of data lies above or below the mean.
• A) x = 11
B) x = 21.5
2
z
xx

11  14

4
 0.75
z
xx

21.5  14

4
 1.88
Note: z-scores are always rounded to
2 decimal places
Example 2: Comparing data
using z-scores
• Stephanie and Tavia are both in the running
for the Data Management award. Stephanie
has 94% and Tavia has 92%. If Stephanie’s
class has a mean of 78 and  = 9.36, and
Tavia’s class has a mean of 73 and  = 8.19.
Who deserves the award?
Example 2
• Use z-scores:
• Stephanie:
z
xx

94  78

9.36
 1.71
Tavia
z
xx

92  73

8.19
 
Tavia’s z-score is higher, therefore her result is
better.
z-Score Table
• appendix B, pp. 398-399 of text
• Determines percentage of data that has
equal or lesser z-score than a given value
Example:
P(z < -2.34)
= 0.0096
–2.9
–2.8
–2.7
–2.6
–2.5
–2.4
–2.3
–2.2
0.00
0.0019
0.0026
0.0035
0.0047
0.0062
0.0082
0.0107
0.0139
0.01
0.0018
0.0025
0.0034
0.0045
0.006
0.008
0.0104
0.0136
0.02
0.0018
0.0024
0.0033
0.0044
0.0059
0.0078
0.0102
0.0132
0.03
0.0017
0.0023
0.0032
0.0043
0.0057
0.0075
0.0099
0.0129
0.04
0.0016
0.0023
0.0031
0.0041
0.0055
0.0073
0.0096
0.0125
Only 0.96 % of the data has a lower z-score, and
1 – 0.0096 = 99.04% of the data has a higher z-score
Note
• Notice z-score table does not go above 2.99
or below –2.99
• Any value with z-score above 3 or less than
–3 is considered an outlier
– If z > 2.99, P(z < 2.99) = 100%
– If z < -2.99, P(z < -2.99) = 0%
• If z = 0, P(z < 0) = 50%
– The data point is the mean
Percentiles
• The kth percentile is the data value that is
greater than k % of the population
• Example
z = 0.40
P( z  0.40)  0.6554
65.54 % of the data
are below this data
point. It is in the
66th percentile.
z = 1.67
P( z  1.67)  0.9525
95.25 % of the data
are below this data
point. It is in the
96th percentile.
Example 3: Finding Ranges
• Given X  N(7, 2.22 ) , find the percent of
data that lies in the following intervals:
A) 3 < x < 6
B) x > 8.6
For x = 3, z 
xx

For x = 6, z 
xx

37

2.2
67

2.2
 1.82
 0.45
P( z  1.82)  0.0344
P( z  0.45)  0.3264
Example 3: Finding Ranges
• Given X  N(7, 2.22 ) , find the percent of
data that lies in the following intervals:
• A) 3 < x < 6
B) x > 8.6
P( z  0.45)  0.3264
P( z  1.82)  0.0344
P(3  x  6)  P(1.82  z  0.45)
 0.3264  0.0344
 0.2920
So 29.20% of the data
fills this interval.
Example 3: Finding Ranges
• Given X  N(7, 2.22 ) , find the percent of
data that lies in the following intervals:
• A) 3 < x < 6x  x
B) x > 8.6
For x = 8.6, z 

8.6  7

2.2
 0.73
So 23.27%
of the data
P( x  8.6)  P( z  0.73) P ( x  8.6)  1  P( x  8.6) lies above
 1  0.7673 8.6.
 0.7673
 0.2327