SOLUTION: IQs are calibrated so that they have a mean of 100 and a standard deviation of 15. a) Find the probability that a randomly selected person has an IQ between 106 and 116. Be sure

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Question 1061625: IQs are calibrated so that they have a mean of 100 and a standard deviation of 15.
a) Find the probability that a randomly selected person has an IQ between 106 and 116. Be sure to draw a sketch.
b) A person with an IQ at or below the 5th percentile is considered �developmentally disabled�. What IQ corresponds to the cutoff for this designation?
Answer by Theo(13342) (Show Source):
You can put this solution on YOUR website!
the easiest way to figure this out is to use the following online statistical z-score calculator.

http://davidmlane.com/hyperstat/z_table.html

the probability that a randomly selected person has an IQ between 106 and 116 is shown below, using this calculator:

the answer shown is .2015

the IQ at or below the 5th percentile is shown below:

the answer shown is 75.322

this calculator allows you to analyze the distribution without having to translate to z-scores.

if you did have to translate to z-scores, you would have done the following, using the formula indicated.

z = (x-m)/s

z is the z-score
x is the raw score
m is the mean
s is the standared deviation.

in the first part of your problem, you would calculate two z-scores.

z1 is the low z-score of the range.
z2 is the high z-score of the range.

you would have gotten:

z1 = (106-100)/15 = .4

z2= (116-100)/15 = 1.067

you would then look up the area on the distribution curve between these z-scores.

the same calculator, using z-scores rather than raw scores, would assist you to find the answer.

that would be shown below:

the4 answer is the same as before, i.e. = .2015

you would also use the same calculator using a mean of 0 and a standard deviation of 1 to finf the z-scpore associated with the fifth percentile of IQs.

that z-score would be -1.645 as shown below:

you would then find the raw score by using the z-score formula as below:

z = (x-m)/s

-1.645 = (x - 100) / 15

solve for x to get x = -1.645 * 15 + 100 which is equal to 75.325

any small difference is more then likely due to rounding.

the z-score is a normalized score.

the mean becomes 0 and the standard deviation becomes 1.

a z-score of 0 is the same as the mean of the raw score.
a standard deviation of 1 is the same as the standard deviation of the raw score.

this means that a z-core is the same as a raw score of 100 in your problem, and a z-score of 1 is the same as a raw score of 115 in your problem, and a z-score of -1 is the same as a z-score of 85 in your problem.