Find the probability that a normal variable takes on values within .9 standard deviations of its mean?

Question

Find the probability that a normal variable takes on values within 0.9 standard deviations of its mean. (Round your decimal to four decimal places.)

Could someone explain to me how to do this one, I am currently struggling with these normal distribution problems, thank you for any tips that you can provide, I am not exactly even sure where to start

Not sure why this is getting down voted I am not asking for the answer, i just would like to know how to do it

Answer 1

Integrate the normal density function. I let you think about why you don't need to specify any mean or standard deviation explicitly and that the result is always:

$$P=\frac{1}{\sqrt{2\pi}}\int_{-0.9}^{0.9}e^{-\frac{x^2}{2}}\approx0.6319$$

Answer 2

It suffices to take a standard normal random variable $Z \sim \mathcal{N}(0, 1)$ (I encourage you to answer the question "Why?" for this). We're asked to compute $P(|Z| \leq 0.9)$, as z-scores tell us the number of standard deviations we are away from the mean.

We can then write, by definition of absolute value,

$$ P(|Z| \leq 0.9) = P(Z \leq 0.9) - P(Z \leq -0.9) $$

as this is how we compute probabilities using cumulative distribution functions (CDFs). These are easily read in any well-written table like this, and thus we obtain the answer to be $0.6318$.

This is the general approach: convert to z-scores and read things from your table (or a computer). I suggest you continue to read your notes and your texts for more information.