2.5 Normal Distributions and z -scores
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Transcript of 2.5 Normal Distributions and z -scores
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2.5 Normal Distributions and z-scores
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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?
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• Stephanie’s class: mean 78, = 9.36
• Tavia’s class: mean 73, = 8.19
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da
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50 60 70 80 90 100x
data = x mean sd normalDensity
Normal Distributions Scatter Plot
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data = x mean sd normalDensity
Normal Distributions Scatter Plot
• Distributions are different• Fair comparison not possible …yet
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Standard Normal Distribution• 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– this process is called
standardizing
2N(0,1 )X
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data = x normalDensity
Standard Normal Distribution Scatter Plot
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z-scores• z = The number of standard deviations a
given score x is above or below the mean
x x z
x xz
• z = z-score– Positive: value lies _________ the mean– Negative: value lies _________ the mean
above
below
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Example 1: Calculating z-scores• Consider the distribution• Find the number of standard deviations each
piece of data lies above or below the mean.• A) x = 11 B) x = 21.5
2N(14,4 )X
x xz
x x
z
11 14
4
0.75
21.5 14
4
1.88Note: z-scores are always rounded to 2 decimal places
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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?
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Example 2
• Use z-scores:
• Stephanie: Taviax x
z
x x
z
94 78
9.36
1.71
92 73
8.19
Tavia’s z-score is higher, therefore her result is better.
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0.00 0.01 0.02 0.03 0.04–2.9 0.0019 0.0018 0.0018 0.0017 0.0016–2.8 0.0026 0.0025 0.0024 0.0023 0.0023–2.7 0.0035 0.0034 0.0033 0.0032 0.0031–2.6 0.0047 0.0045 0.0044 0.0043 0.0041–2.5 0.0062 0.006 0.0059 0.0057 0.0055–2.4 0.0082 0.008 0.0078 0.0075 0.0073–2.3 0.0107 0.0104 0.0102 0.0099 0.0096–2.2 0.0139 0.0136 0.0132 0.0129 0.0125
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
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
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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
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Percentiles
• The kth percentile is the data value that is greater than k % of the population
• Example z = 0.40 z = 1.67( 0.40) 0.6554P z
65.54 % of the data are below this data point. It is in the 66th percentile.
( 1.67) 0.9525P z
95.25 % of the data are below this data point. It is in the 96th percentile.
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Example 3: Finding Ranges
• Given , find the percent of data that lies in the following intervals:
A) 3 < x < 6 B) x > 8.6
2N(7,2.2 )X
For x = 3,x x
z
3 7
2.2
1.82( 1.82)P z 0.3264
For x = 6,x x
z
6 7
2.2
0.45( 0.45)P z 0.0344
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Example 3: Finding Ranges
• Given , find the percent of data that lies in the following intervals:
• A) 3 < x < 6 B) x > 8.6
2N(7,2.2 )X
0.3264( 0.45)P z ( 1.82)P z 0.0344
(3 6) ( 1.82 0.45)P x P z 0.3264 0.0344
0.2920So 29.20% of the data fills this interval.
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Example 3: Finding Ranges
• Given , find the percent of data that lies in the following intervals:
• A) 3 < x < 6 B) x > 8.6
2N(7,2.2 )X
For x = 8.6,x x
z
8.6 7
2.2
0.73
( 8.6) ( 0.73)P x P z
0.7673
( 8.6)P x 1 ( 8.6)P x 1 0.7673 0.2327
So 23.27% of the data lies above 8.6.