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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

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    You need to use a set of cumulative normal distribution tables – 2017-02-17
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    I don't think this is actually _using _ the Central Limit Theorem; it's just using the fact that the sum of $n$ variables each with a general normal distribution is another variable with a general normal distribution. You should understand this, and understand how to describe the distribution of the sum (what its mean and variance are). Also, before tackling the CLT you should have learned how to use Z scores (including where to find the tables mentioned above). If you don't study up on these things I think you will have a lot of trouble as you go further along in this chapter. – 2017-02-17
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    @DavidK I didn't think it was either so I was really confused when this problem was under the CLT chapter. – 2017-02-17
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    I think it's trying to force you to recall the things you should have learned _before_ the CLT chapter, so that you'll be able to continue on to the actual CLT. "I have no clue" indicates to me that you need to review the earlier chapters. – 2017-02-17

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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 $$ P\left(Z> \frac{x-\mu}{\sigma}\right) = 1 - \Phi\left( \frac{x-\mu}{\sigma}\right).$$

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    Ohh, I see. Thank you so much for helping and being really nice about it. Now I have a much clearer idea of how to tackle the other problems. – 2017-02-17