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Based on the normal model N(100, 16), describing IQ scores. What percent of people’s IQ would you expect to be:

  1. Percentage of population greater than 80.
  2. Percentage of population less than 90.
  3. Percentage of population between 112 – 132.

The question aims to find the percentage of the people’s IQ with the mean of the population to be 100 and a standard deviation of 16.

The question is based on the concepts of probability from a normal distribution using a z-table or z-score. It also depends on the population’s mean and the population’s standard deviation. The z-score is the deviation of a data point from the population’s mean. The formula for z-score is given as:

\[ z = \dfrac{ x\ -\ \mu}{ \sigma } \]

Expert Answer

This question is based on the normal model which is given as:

\[ N(\mu, \sigma) = N(100, 16) \]

We can find the percentage of population for a given limit by using the $z-score$ which is given as follows:

a) The percentage of population greater than $X \gt 80$ can be calculated as:

\[ p = P(X \gt 80) \]

Converting the limit into $z-score$ as:

\[ p = P \big(Z \gt \dfrac{ 80\ -\ 100 }{ 16 } \big) \]

\[ p = P(Z \gt -1.25) \]

\[ p = 1\ -\ P(Z \lt -1.25) \]

Using the $z-$table, we get the $z-score$ of the above probability value to be:

\[ p = 1\ -\ 0.1056 \]

\[ p = 0.8944 \]

The percentage of population with IQ above $80$ is $89.44\%$.

b) The percentage of population greater than $X \lt 90$ can be calculated as:

\[ p = P(X \lt 90) \]

Converting the limit into $z-score$ as:

\[ p = P \big(Z \lt \dfrac{ 90\ -\ 100 }{ 16 } \big) \]

\[ p = P(Z \lt -0.625) \]

Using the $z-$table, we get the $z-score$ of the above probability value to be:

\[ p = 0.2660 \]

The percentage of population with IQ below $90$ is $26.60\%$.

c) The percentage of population between $X \gt 112$ and $X \lt 132$ can be calculated as:

\[ p = P(112 \lt X \lt 132 \]

Converting the limit into $z-score$ as:

\[ p = P \big(\dfrac{ 112\ -\ 100 }{ 16 } \lt Z \lt \dfrac{ 132\ -\ 100 }{ 16 } \big) \]

\[ p = P(Z \lt -2)\ -\ P(Z \lt 0.75) \]

Using the $z-$table, we get the $z-scores$ of the above probability values to be:

\[ p = 0.9772\ -\ 0.7734 \]

\[ p = 0.2038 \]

The percentage of population with IQ between $112$ and $132$ is $20.38\%$.

Numerical Result

a) The percentage of population with IQ above $80$ is $89.44\%$.

b) The percentage of population with IQ below $90$ is $26.60\%$.

c) The percentage of population with IQ between $112$ and $132$ is $20.38\%$.

Example

The normal model $N(55, 10)$ is given of people describing their age. Find the percentage of people with age below $60$.

\[ x = 60 \]

\[ p = P(X \lt 60) \]

\[ p = P \Big(Z \lt \dfrac{ 60\ -\ 55 }{ 10 } \Big) \]

\[ p = P(Z \lt 0.5) \]

\[ p = 0.6915 \]

The percentage of people with age below $60$ is $69.15\%$.

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