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
![\begin{gathered} X\sim N(500,100^2) \\ X\operatorname{\sim}N(\mu,\sigma^2) \end{gathered}]()
Then we must find
![P(200\leq X\leq800)](https://img.qammunity.org/2023/formulas/mathematics/college/iicmv7wpsgse925yavwpsmv17lw2o4j0mz.png)
In fact, we are doing
![P(\mu-3\sigma\leq X\leq\mu+3\sigma)](https://img.qammunity.org/2023/formulas/mathematics/college/hfyrdq10hce7ecvnf1tveo60b16xtzaye8.png)
The normal distribution is symmetrical then
![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)](https://img.qammunity.org/2023/formulas/mathematics/college/nfwdmrysapthgfa22apitfmy8v70l78h7v.png)
Therefore we can just evaluate
![P(\mu-3\sigma\leq X\leq\mu+3\sigma)=2P(\mu\leq X\leq\mu+3\sigma)](https://img.qammunity.org/2023/formulas/mathematics/college/eu8p02nykj40c6ibbujwem0t053n2x1rg0.png)
But why use the standard deviation and the mean? because we already know that value!
![\begin{gathered} P(\mu\leq X\leq\mu+\sigma)=0.3413 \\ \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/hi2x1s8yiiksrdhyv0qgm1xx1g507oiibc.png)
For 2*standard deviation
![P(\mu\leq X\leq\mu+2\sigma)=0.4772](https://img.qammunity.org/2023/formulas/mathematics/college/duxhhj1nufnq9whp0efadgb1l0y8q0o2wt.png)
And our case, for 3*standard deviation
![P(\mu\leq X\leq\mu+3\sigma)=0.4987](https://img.qammunity.org/2023/formulas/mathematics/college/pshdjf4njpfj784zst2nivot4u26q7rfnu.png)
Therefore
![\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}](https://img.qammunity.org/2023/formulas/mathematics/college/ixqqb61ijmxfwbm9lxssluq25hqzkuvc61.png)
Let's go back to our original equation
![P(200\leq X\leq800)=0.9974](https://img.qammunity.org/2023/formulas/mathematics/college/sohm7dzotqt4tp2yxriec0hi4gkrhc8npl.png)
Then 99.74% of the students score between 200 and 800, now just do 99.8% of 1000
![99.8\%\text{ of 1000 = 998 students}](https://img.qammunity.org/2023/formulas/mathematics/college/jxhfu446km3skrgxrkuub9e8dhceucw1kw.png)
Hence, 998 students scored between 200 and 800