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

• Tavia’s class: mean 73, = 8.19

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data = x mean sd normalDensity

Normal Distributions Scatter Plot

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Normal Distributions Scatter Plot

• Distributions are different• Fair comparison not possible …yet

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

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

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

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: 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.

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

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 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.

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

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.

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.