Question 1197494: For the following problem, use the 68-95-99.7 Rule to approximate the probability rather than using technology to find the values more precisely.
Although controversial and the subject of some recent law suits, some human resource departments administer standard IQ tests to all employees. One IQ test's scores are well modeled by a Normal model with mean 100 and standard deviation 13. If the applicant pool is well modeled by this distribution, what cutoff value would separate a randomly selected applicant into the following regions?
a) the highest 50%
b) the highest 16%
c) the lowest 2.5%
d) the middle 68%
Answer by math_tutor2020(3817)   (Show Source):
You can put this solution on YOUR website!
Part (a)
The mean is at the very center.
Therefore, 50% of the scores are above an IQ of 100
Answer: 100
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Part (b)
Approximately 68% of the scores are within 1 standard deviation of the mean.
This is what the 68 refers to in the "68-95-99.7 Rule" (aka Empirical Rule).
This leaves 100-68 = 32% in the combined tails
So 32/2 = 16% is in each tail
If we want the top 16%, then the cutoff value is 100+13 = 113
Notice I started at the center (100) and increased by one standard deviation (13)
Answer: 113
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Part (c)
According to the 68-95-99.7 Rule, about 95% of the normally distributed population is within two standard deviations of the mean.
(100%-95%)/2 = 2.5% of the population is in each tail
Start at 100 and move down two standard deviations
100-2*13 = 74
Answer: 74
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Part (d)
As mentioned earlier, about 68% of the population in a normal distribution is within one standard deviation of the mean.
Start at the center 100. Add and subtract the standard deviation (13) to determine the lower and upper bounds
100 - 13 = 87
100 + 13 = 113
Answer: The region between markers 87 and 113
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