Question #fb52f

1 Answer
Nov 15, 2017

A) #~~75#
B) #~~25.3%#
C) #1.33# (two decimals)

Explanation:

To answer these questions, we would have to know if the standard IQ test results fall into a normal distribution. There is some debate online about that (many say it's meant to be a normal distribution, while the actual results are more likely to diverge from a normal distribution in the "tails"), but for this problem I believe it is intended to be treated that way.

Thus, this problem describes a normal distribution with a mean #mu# of 100 and a standard deviation #sigma# of 15. In notation terms, we say this is a #N(100,15^2)# distribution. To answer the questions it is helpful to convert this problem into the equivalent standard normal distribution #N_s(0,1^2)# (mean of 0, std. deviation of 1) and then we will be able to reference standard normal distribution area/probability tables.

Question A

Asking what IQ score has 95% of people above it is precisely the same as asking what IQ score has 5% of people below it, since we know that all IQ scores combined would represent 100% of all scores. For the standard normal distribution, we need to know which z-score has a left-tail area that is 0.05; this would be the point where 5% (ie 0.05) of the area under the standard normal distribution lies to the left of that z-score.

To get this, we can either consult a z-score table or use a z-score calculator. I will opt to use a z-score calculator, which finds that a z-score of -1.64 (to two decimals) is our target.

Now, we use the z-score formula to determine the correlated IQ score #x# according to our #N(100,15^2)# distribution this z-score matches:

#z = (x-mu)/sigma#

#-1.64 = (x-100)/15#

#-24.6 = x-100#

#x = 75.4~~75#

Question B

To find the percentage of people who would score less than 90, we work in an opposite direction than Question A. In this case, we will convert the IQ score of #x = 90# into a standard normal distribution z-score, and then consult a z-score calculator or table to determine the proportion of all z-scores that are lower than that value.

#z = (x-mu)/sigma#

#z = (90 - 100)/15#

#z = -10/15 = -0.67# (two decimals)

A z-score calculator tells us that for this z-score, approximately 25.3% of all z-scores lie to the left of this z-score. Another way of looking at this is saying that 25.3% of people would be expected to have a lower IQ score than 90.

Question C

This question is more straightforward. We can use the z-score formula to convert an IQ score of 120 from a distribution of #N(100,15^2)#:

#z = (x - mu)/sigma#

#z = (120 - 100)/15#

#z = 20/15 = 1.33# (two decimals)