A survey of high school seniors from a certain school district who took the SAT has determined that the mean score on the mathematics portion was 650 with a standard deviation of 15.5. Assuming the data can be modeled by a normal probability density function, find a model for these data.
A) $f ( x ) = \frac { 1 } { 15.5 \sqrt { \pi } } e ^ { - ( x - 650 ) / 240.25 }$
B) $f ( x ) = \frac { 1 } { 15.5 \sqrt { 2 \pi } } e ^ { - ( x - 650 ) / 480.5 }$
C) $f ( x ) = \frac { 1 } { 15.5 \sqrt { \pi } } e ^ { - ( x - 650 ) ^ { 2 } / 480.5 }$
D) $f ( x ) = \frac { 1 } { 15.5 \sqrt { 2 \pi } } e ^ { - ( x - 650 ) ^ { 2 } / 480.5 }$
E) $f ( x ) = \frac { 1 } { 15.5 \sqrt { 2 \pi } } e ^ { - ( x - 650 ) ^ { 2 } / 240.25 }$

Question

Quizplus · Accepted Answer

The correct model for these data, based on the normal probability density function, is given by the formula $f(x) = \frac{1}{\sigma \sqrt{2\pi}} e^{-\frac{(x - \mu)^2}{2\sigma^2}}$, where $\mu$ is the mean and $\sigma$ is the standard deviation. Substituting $\mu = 650$ and $\sigma = 15.5$, and noting that $2\sigma^2 = 2(15.5)^2 = 480.5$, the correct model is $f(x) = \frac{1}{15.5\sqrt{2\pi}} e^{-\frac{(x - 650)^2}{480.5}}$.

A Survey of High School Seniors from a Certain School f(x)=115.5πe−(x−650)/240.25f ( x ) = \frac { 1 } { 15.5 \sqrt { \pi } } e ^ { - ( x - 650 ) / 240.25 }f(x)=15.5π​1​e−(x−650)/240.25

A Survey of High School Seniors from a Certain School $f ( x ) = \frac { 1 } { 15.5 \sqrt { \pi } } e ^ { - ( x - 650 ) / 240.25 }$