use the normal model n(100,16) describing iq scores to answer the following. a) what percent of peoples iqs…

Explanation:

Step1: Standardize the values

For a normal distribution $N(\mu,\sigma^{2})$, we use the formula $z=\frac{x - \mu}{\sigma}$. Here, $\mu = 100$ and $\sigma=\sqrt{16}=4$.

Step2: Calculate z - scores for part a

For $x = 75$, $z=\frac{75 - 100}{4}=\frac{- 25}{4}=-6.25$. The probability $P(X>75)=1 - P(X\leq75)$. Since the standard normal table usually goes from $z=- 3.49$ to $z = 3.49$ and $P(Z\leq - 6.25)\approx0$, then $P(X>75)\approx1 = 100%$.

Step3: Calculate z - scores for part b

For $x = 85$, $z=\frac{85 - 100}{4}=\frac{-15}{4}=-3.75$. Looking up in the standard - normal table, $P(Z\leq - 3.75)\approx0.0001$. So the percentage of people with IQs under 85 is approximately $0.01%$.

Step4: Calculate z - scores for part c

For $x_1 = 116$, $z_1=\frac{116 - 100}{4}=\frac{16}{4}=4$. For $x_2 = 132$, $z_2=\frac{132 - 100}{4}=\frac{32}{4}=8$. Since $P(Z\leq4)\approx0.99997$ and $P(Z\leq8)\approx1$, then $P(116<X<132)=P(Z\leq8)-P(Z\leq4)\approx1 - 0.99997 = 0.00003$. So the percentage of people with IQs between 116 and 132 is approximately $0.003%$.

Answer:

a. Approximately 100% b. Approximately 0.01% c. Approximately 0.003%