Question

The average IQ score is 100. The standard deviation is 15. If we assume IQ scores arenormally distributed: DRAW THE BELL CURVE!

Answer 1

Given a normal distribution with a mean of 100 and a standard deviation of 15:

`\begin{gathered} \mu=100 \\ \sigma=15 \end{gathered}`

The graph of the Bell curve is:

a)

To find the percentage of the population with an IQ between 70 and 130, we need to find the z-scores, using the equation:

`Z=(X-\mu)/(\sigma)=(X-100)/(15)`

Then, the z-scores of 70 and 130 are:

`\begin{gathered} Z_(70)=(70-100)/(15)=-2 \\ Z_(130)=(130-100)/(15)=2 \end{gathered}`

Then, the proportion can be calculated using the probability:

`P(-2Looking at z-distribution tables, this probability is (to two decimal places):[tex]P(-2<strong>Then, 95% of the population has an IQ between 70 and 130.</strong><p>b)</p><p>Unusually high scores are often calculated as those values above 3 standard deviations above the mean, so if in this case the mean is 100, and the standard deviation is 15, then the unusually high scores are:</p>[tex]\begin{gathered} \mu+3\sigma=100+3\cdot15=100+45=145 \\ \Rightarrow IQ\ge145 \end{gathered}`

c)

The range of IQ within 3 standard deviations is:

`\begin{gathered} \mu\pm3\sigma=100\pm45 \\ \Rightarrow\lbrack55,145\rbrack \end{gathered}`

Then, the corresponding z-scores are:

`\begin{gathered} Z_(55)=(55-100)/(15)=-3 \\ Z_(145)=(145-100)/(15)=3 \end{gathered}`

The proportion can be calculated using the probability:

[tex]P(-3Using a z-distribution table, this probability is (to 3 decimal places):[tex]P(-3Then, 99.7% of the population will have an IQ score within 3 standard deviations of the average IQ score.