IQ Scores by Category

Date: 7/28/96 at 9:26:1
From: MIGUEL SPRUILL
Subject: measures of dispersion, the normal distribution

On standard IQ tests, the mean is 100, with standard deviation of 15. 
The results come very close to fitting a normal curve. Suppose an IQ 
test is given to a very large group of people. Find the percent of 
people whose IQ scores fall into the following categories:

1) greater than 115.
2) greater than 145.

If you find both kinds of standard deviation, the sample standard 
deviation and the population standard deviation, which of the two will 
be a larger number for a given set of data?

Find (a) the range, and (b )the standard deviation for each sample 
round fractional answers to the nearest hundredth. 

1) 67, 83, 55, 68, 77, 63, 84, 72, 65

Date: 7/28/96 at 16:6:13
From: Doctor Anthony
Subject: Re: measures of dispersion, the normal distribution

To find the standardised values for use with normal tables you 
calculate z from  z = (x-m)/s where m = mean and s = standard 
deviation.

For greater than 115 we have 

   z = (115.5 - 100)/15  = 15.5/15 = 1.0333.

The tables give an area 0.8493, so 

   tail area = 0.1507.

So 15.07 percent of the population has an IQ above 115.

For greater than 145 we have  

   z = (145.5 - 100)/15 = 45.5/15 = 3.0333.

The tables give an area .99878, so 

   tail area = 0.00122.

So 0.12 percent of the population has an IQ above 145.

If you use a sample to estimate population mean and s.d. you lose one 
'degree of freedom' in that with mean found, then you no longer have n 
independent values for finding the standard deviation.  So in 
calculating variance from 

   SIGMA(x-m)^2/n (sample variance),

you would use 

   SIGMA(x-m)^2/(n-1) if m and s are calculated from the sample as 
estimates for the population.  

So the standard deviation is greater if you use the sample to estimate 
population standard deviation.    

Find range and standard deviation of the sample 67, 83, 55, 68, 77, 
63, 84, 72, 65.

The range = 84 - 55 = 29.

Variance of the sample is calculated from
 
   SIGMA[x^2/n] - mean^2
   mean = 70.44
   standard deviation = 8.995.

-Doctor Anthony,  The Math Forum
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