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The Standard Normal Curve Revisited

The Standard Normal Curve Revisited. Can you place where you are on a normal distribution at certain percentiles? 50 th percentile? Z = 0 84 th percentile?

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Page 1: The Standard Normal Curve Revisited. Can you place where you are on a normal distribution at certain percentiles? 50 th percentile? Z = 0 84 th percentile?

The Standard Normal Curve Revisited

Page 2: The Standard Normal Curve Revisited. Can you place where you are on a normal distribution at certain percentiles? 50 th percentile? Z = 0 84 th percentile?

Can you place where you are on a normal distribution at certain percentiles?50th percentile? Z = 084th percentile? Z = 12.5 percentile? Z = -2

Page 3: The Standard Normal Curve Revisited. Can you place where you are on a normal distribution at certain percentiles? 50 th percentile? Z = 0 84 th percentile?

Finding z-scores given the percentileFind the closest percentile on the table to get the z-

score or…Use the invNorm command to find the z-score given

the percentile or area under a normal curve. Press 2nd, DIST, invNorm (percentile as a decimal #). For example, to find the z-score for the 75th percentile, press invNorm (.75).

For the standard normal distribution,What is the median (50%tile)?

invNorm(.50) = 0What is the lower quartile?

invNorm(.25) = -.674What z -score falls at the 95th percentile?

invNorm (.95) = 1.645What z-score has 30% of the data above it?

invNorm(.70) = .524

Page 4: The Standard Normal Curve Revisited. Can you place where you are on a normal distribution at certain percentiles? 50 th percentile? Z = 0 84 th percentile?

SAT ExampleThe approximately normal distribution of SAT scores for the

2009 incoming class at the University of Texas had a mean 1815 and a standard deviation 252.

What did a student at University of Texas get on the SAT if his or her score was 1.6 standard deviations above the average?

Z = 1.6 1.6 = x – 1815

252 403.2 = x – 1815 X = 2218.2 or 2218

What score did a student get who is at the 90th percentile? Z = invNorm(.90) = 1.28 1.28 = x – 1815

252 322.56 = x – 1815 X = 2137.56 or 2138

Page 5: The Standard Normal Curve Revisited. Can you place where you are on a normal distribution at certain percentiles? 50 th percentile? Z = 0 84 th percentile?

Heights Example The heights of 18 to 24 year old males in the US are

approximately normal with mean 70.1 inches and standard deviation 2.7 inches. The heights of 18 to 24 year old females have a mean of 64.8 inches and a standard deviation of 2.5 inches.

• How tall does a US woman between 18 and 24 have to be to be at the 35th percentile?– Z = invNorm(.35) = -.385– -.385 = x – 64.8

2.5– -.9625 = x – 64.8– X = 63.84 inches

• In order to be in the Tall Club of America, you must be in the top 10% height range. What is the cut off for males? For females?– Z = invNorm(.90) = 1.28

1.28 = x – 70.1 1.28 = x – 64.8 2.7 2.5

3.456 = x – 70.1 3.2 = x – 64.8– x = 73.556 inches x = 68 inches

(6’ 1.5” for males) (5’8” for females)

Page 6: The Standard Normal Curve Revisited. Can you place where you are on a normal distribution at certain percentiles? 50 th percentile? Z = 0 84 th percentile?

Born to RunA study of elite distance runners found a mean body weight of 139.1 pounds with a standard deviation of 10.6 pounds. How much would a runner weigh to be in the 75th

percentile? invNorm(.75) = .67 .67 = (x – 139.1)/10.6X = 146.2 pounds

40% of the runners weigh more than pounds. invNorm(.6) = .25 .25 = (x – 139.1)/10.6X = 141.75 pounds

Page 7: The Standard Normal Curve Revisited. Can you place where you are on a normal distribution at certain percentiles? 50 th percentile? Z = 0 84 th percentile?

Using z-scores in a Normal Distribution

Find the percent: find the z-score using the formulathen use the table or normalcdf to find the %

Given the percent: find the z-score using the table or invNorm then solve for the observation using the z-score

formula

Z = observation – mean standard deviation

Page 8: The Standard Normal Curve Revisited. Can you place where you are on a normal distribution at certain percentiles? 50 th percentile? Z = 0 84 th percentile?

RemindersZ – score = standard score = # of standard

deviations above or below the mean

A z- score is NOT a percent!