3.5 Applying the Normal Distribution – Z Scores

Example 1Determine the number of standard deviations above or below the mean each piece of data is. (This is known as a z-score)

a) mean = 100, st dev = 20, data = 110

xxz

20100110

z

5.0z

110 is 0.5 standard deviations above the mean.

b) mean = 25, st dev = 5, data = 18

xxz

52518

z

4.1z

25 is 1.4 standard deviations below the mean.

c) X ~ N(40, 4) , data = 50

xxz

24050

z

5z

50 is 5 standard deviations above the mean.

2440 x

Example 2Determine the percent and percentile corresponding to each z-score from question 1 and sketch. You will need page 398/399 of your textbook.

a) mean = 100, st dev = 20, data = 110, z = 0.5

z = 0.5 => 0.6915 = 69.15% the 69th percentile

100 120 140 160806040

11069.15% 30.85%

b) mean = 25, st dev = 5, data = 18, z = -1.4

z = -1.4 => 0.0808 = 8.08% the 8th percentile

25 30 35 40201510

188.08% 91.92%

c) mean = 40, st dev = 2, data = 50, z = 5

z = 5 => too high = 100% 100th percentile (technically slightly less)

40 42 44 46383634

50100% 0%

Example 3The heights of Doyle students are normally distributed. X ~ N (150, 144).How tall does someone need to be to be in the top 20% ?

mean = 150, st dev = 12data = ? , z = ?

To be in the top 20% you must be in at least the ____ percentile.80th

z = ____ corresponds most closely to 80%.0.84

solve for x

xxz

1215084.0

x

1501284.0 x

08.160x

To be in the top 20% you must be at least 160.08 cm

80% 20%