Normal Distribution Application Given: $\mu=80$, $\sigma=15$, $500$ respondents
a. Find $P(74\lt x \lt 101)$
b. Find number of respondents with score $\lt 98$

For a., my answer is $0.5746$ (using the formula $z=\cfrac{x-\mu}{\sigma}$); for b., $442$ respondents (using the formula $z=\cfrac{x-\mu}{\sigma}$ and multiplying the answer by $500$ respondents).

This question is flashed in our automated quiz where I got a score of $0$. I wanted to insist that I got the correct answers when we get back to class. May I verify them? Thank you.

For part a. I get $P(74\lt x \lt 101) = P\left(\cfrac{74-80}{15} \lt z \lt \cfrac{101-80}{15}\right)=0.574665082$.
For part b. I get $P(x \lt 98)\times 500 = P\left(-99 \lt z \lt \cfrac{98-80}{15}\right) \times 500\approx 0.885\times500=442.5$. For this I just used the fact that the at the end of the left tail of the Gaussian it takes large and negative $z$. So I chose $-99$ (ideally this should be $-\infty$ but this is undefined).
