z-score given the percentile

January 31, 2018 | Author: Anonymous | Category: Math, Statistics And Probability, Normal Distribution
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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 = 0  84th percentile? Z = 1  2.5 percentile? Z = -2

Finding z-scores given the percentile  Find 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) = 0

 What is the lower quartile? 

invNorm(.25) = -.674

 What z -score falls at the 95th percentile? 

invNorm (.95) = 1.645

 What z-score has 30% of the data above it? 

invNorm(.70) = .524

SAT Example  The 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 Run  A 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)/10.6  X = 146.2 pounds

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

pounds.

Using z-scores in a Normal Distribution Z = observation – mean standard deviation

 Find the percent:  find the z-score using the formula  then 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

Reminders  Z – score = standard score = # of standard deviations

above or below the mean  A z- score is NOT a percent!

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