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.