The point here is to calculate the area of the curve between 200 to 800 because it's a distribution of probability and even more, it's a normal distribution then look at it we can already see that the standard variation is 100 and the mean is 500, then
[tex]\begin{gathered} X\sim N(500,100^2) \\ X\operatorname{\sim}N(\mu,\sigma^2) \end{gathered}[/tex]
Then we must find
[tex]P(200\leq X\leq800)[/tex]
In fact, we are doing
[tex]P(\mu-3\sigma\leq X\leq\mu+3\sigma)[/tex]
The normal distribution is symmetrical then
[tex]P(\mu-3\sigma\leq X\leq\mu+3\sigma)=2P(\mu-3\sigma\leq X\leq\mu)=2P(\mu\leq X\leq\mu+3\sigma)[/tex]
Therefore we can just evaluate
[tex]P(\mu-3\sigma\leq X\leq\mu+3\sigma)=2P(\mu\leq X\leq\mu+3\sigma)[/tex]
But why use the standard deviation and the mean? because we already know that value!
[tex]\begin{gathered} P(\mu\leq X\leq\mu+\sigma)=0.3413 \\ \end{gathered}[/tex]
For 2*standard deviation
[tex]P(\mu\leq X\leq\mu+2\sigma)=0.4772[/tex]
And our case, for 3*standard deviation
[tex]P(\mu\leq X\leq\mu+3\sigma)=0.4987[/tex]
Therefore
[tex]\begin{gathered} 2P(\mu\leq X\leq\mu+3\sigma)=2\cdot0.4987 \\ \\ 2P(\mu\leqslant X\leqslant\mu+3\sigma)=0.9974 \end{gathered}[/tex]
Let's go back to our original equation
[tex]P(200\leq X\leq800)=0.9974[/tex]
Then 99.74% of the students score between 200 and 800, now just do 99.8% of 1000
[tex]99.8\%\text{ of 1000 = 998 students}[/tex]
Hence, 998 students scored between 200 and 800
