Find what percentage of sample falls between range [A,b] given a normal model with SD = 30 and mean = 140

Question

So I have a normal model with a standard deviation of 30 and a mean of 140 and then I want to find out what percent of the sample fall in the range of $[120, 180]$. I know by the 68-97-99 rule that $68\%$ fall between $[110, 170]$, so I can assume that a bit less than $68\%$ fall between $[120, 180]$ (Because a normal distribution model is densest near the mean so shifting it right 10 will probably reduce the proportion). Is there a way to find the exact percent that falls within that range just given the SD and mean?

Answer 1

I would begin by transofrming my random variable to a standard normal distribution. So your question for

"Proportion of sample in $[120, 180]$ for mean of $140$ and standard deviation of $30$" is equivalent to:

"Proportion of sample in $[-\frac{2}{3}, \frac{4}{3}]$ for mean of $0$ and standard deviation of $1$" (I just subtracted the mean and divided by the standard deviation)

Now you can get your hands on a chart or some piece of software to do the calculation suggested by Henri in the comments. Thus:

$P(X \leq -\frac{2}{3}) = 0.2524925$

$P(X \leq \frac{4}{3}) = 0.9087888$

These calculations where made by the pnorm function in R. You can subtract both values and get your result (about $65\%$)