Scores for a common standardized college aptitude test are normally distributed with a mean of 498 and a standard deviation of 98. Randomly selected men are given a Preparation Course before taking this test. Assume, for sake of argument, that the Preparation Course has no effect on people's test scores.
If 1 of the men is randomly selected, find the probability that his score is at least 550.3. P(X > 550.3) = ?
This problem was under the chapter central limit theorem but I have no clue how to do this.
I know the central limit theorem is using Z is Z= (xbar-mu)/(sigma/sqrt(n)) I also know the problem mentions standard normal and that is P(x)=1/(sqrt(2pi)) integral e^-((x^2)/2)
Can you guys help? I'm very new to Probability
As was cleared up in the comments this question does not involve the central limit theorem.
The way to do these is to use the fact that if $X$ is $N(\mu,\sigma^2),$ then $$ Z = \frac{X-\mu}{\sigma}$$ is a $N(0,1)$. In your case $\mu=498$ and $\sigma = 98$
You want to compute something of the form $$ P(X> x)$$ where $x=550.3$ in your case. The inequality can be rewritten $$ P\left(\frac{X-\mu}{\sigma} > \frac{x-\mu}{\sigma}\right) = P\left(Z> \frac{x-\mu}{\sigma}\right).$$
Since $Z$ is a standard normal, we have $$ P(Z>z) = \int_{z}^\infty \frac{1}{\sqrt{2\pi}}e^{=t^2/2} dt = 1-\Phi(z)$$ where $\Phi(z)$ is the normal cumulative distribution function.
Thus you want to compute $$ P\left(Z> \frac{x-\mu}{\sigma}\right) = 1 - \Phi\left( \frac{x-\mu}{\sigma}\right).$$