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The Standard Normal Curve Revisited
Can you place where you are on a normal distribution at certain percentiles?50th percentile? Z = 084th percentile? Z = 12.5 percentile? Z = -2
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
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
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)
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
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
RemindersZ – score = standard score = # of standard
deviations above or below the mean
A z- score is NOT a percent!