 # 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? FizeauV 2021-08-19 Answered
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