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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Sadie Eaton
Answered 2021-08-20 Author has 104 answers

1. The distribution of IQ scores is normally distributed with a mean of 100 and a standard deviation of 16.
X N(100,16)
2. Transform the normal random variable,X, into the standard normal variable Z by using the following formula: Z=Xμσ .
a) P(X>80)=P(X10016>8010016)=P(Z>1.25)=1P(Z1.25)=10.1056=0.8944=89.44
b) P(X<90)=P(X10016<9010016)=P(Z0.625)=0.2659=26.59

c) P(112<X<132)=P(11210016<X1001613210016)=P(0.75<Z<2)=P(Z<0.75)P(Z<2)=0.22660.0228=0.2038=20.38

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Jeffrey Jordon
Answered 2021-10-06 Author has 2047 answers

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=Xvσ

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

P(X >80)

=P(Xvσ>80vσ)

=P(Z>80vσ)

=P(Z>8010016)

=P(Z>1.25)

=1P(Z1.25)

=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(Z<9010016)

=P(Z<0.625)

=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(11210016<Z<13210016)

=P(0.75<Z<2)

=0.2039

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

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