# Based on the Normal model N(100, 16) describing IQ scores, what percent of people's IQs wouldyou expect to bea) over 80?b) under .90?c) between 112 and 132?

Based on the Normal model N(100, 16) describing IQ scores, what percent of peoples
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1. The distribution of IQ scores is normally distributed with a mean of 100 and a standard deviation of 16.

2. Transform the normal random variable,X, into the standard normal variable Z by using the following formula: $Z=\frac{X-\mu }{\sigma }$ .
a) $P\left(X>80\right)=P\left(\frac{X-100}{16}>\frac{80-100}{16}\right)=P\left(Z>-1.25\right)=1-P\left(Z\le -1.25\right)=1-0.1056=0.8944=89.44$
b) $P\left(X<90\right)=P\left(\frac{X-100}{16}<\frac{90-100}{16}\right)=P\left(Z\le -0.625\right)=0.2659=26.59$

c) $P\left(112

###### Not exactly what you’re looking for?
Jeffrey Jordon

The percentage needed is for the area shown here :

Let us suppose the number corresponding to the needed percentage is X, we get the following expression:

P(X>80)

Knowing that $Z=\frac{X-v}{\sigma }$

So, in the next steps , we will get the Z value from X and calculate its percentage as follows:

P(X >80)

$=P\left(\frac{X-v}{\sigma }>\frac{80-v}{\sigma }\right)$

$=P\left(Z>\frac{80-v}{\sigma }\right)$

$=P\left(Z>\frac{80-100}{16}\right)$

$=P\left(Z>-1.25\right)$

$=1-P\left(Z\le -1.25\right)$

=1-0.1056=0.8944

Therefore, 89.44% of people's IQs would be over 80

b)The percentage to be calculated is for the following colored area:

P(X<90)

$=P\left(Z<\frac{90-100}{16}\right)$

$=P\left(Z<-0.625\right)$

=0.266

Therefore, 26.6% of people's IQs would be under 80

c)The percentage to be calculated represents the data within the covered area:

P(112<X<132)

$=P\left(\frac{112-100}{16}

$=P\left(0.75

=0.2039

Therefore, 20.39% of people's IQs would be between 112 and 132