Question 1002529
 the distribution of IQ scores has a mean of 120 and a standard deviation of 8, use the theorem to determine the interval containing at least 8/9 of the IQ scores. 
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Chebyshev says "at least 1-(1/n)^2 % of the data lies within n 
standard deviations of the mean".
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If 1-(1/n)^2 = 8/9, (1/n)^2 = (1/9) and n = 3
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The interval:
Lower bound:: 120-3*8 = 96
Upper bound:: 120+3*8 = 144
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At least what percentage of these scores must lie between 104 and 136 ? 
104 = 120-8n
8n = 16
n = 2
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136 = 120+8n
8n = 16
n = 2
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Ans: % of data within 2 std of the mean = 1-(1/2)^2 = 0.75 = 75%
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Cheers,
Stan H.
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