I will assume $X \sim \mathsf{Norm}(\mu = .15, \sigma = .1)$ is the annual return. And I assume checking once a year means you would see the random
return $X.$ You are asked for $P(X > 0).$
I don't know whether you are supposed to use printed normal tables or software
to solve this. To use printed normal tables, you need to standardize.
The 'standard score' $Z$ corresponding to the 'raw score' $X$ is found as
$X = (X - \mu)/\sigma.$ (Read about this in your text or lecture notes; the
terminology may be different, but the idea has to be the same.)
So you can solve part (A) as follows:
$$P(X > 0) = P\left(\frac{X - \mu}{\sigma} > \frac{0 - .15}{.1} = -1.5\right)
= P(Z > -1.5).$$
You will have to read in your text how to get the answer, using the particular kind of normal
tables in your text or the kind of software used in your course. Below is a
sketch of the standard normal density function. The total area under the
density curve is 1. The curve is symmetrical about 0. Half of the area is on either side of 0. The area you seek is to the right of the vertical red line.
(Just by looking you can see that the answer has to be a probability more than 1/2. This makes
sense because you've been promised an annual gain of $X,$ which has a positive
mean.)
My guess is that your gain in one month would be $Y = X/12.$ So $\mu_Y = 0.15/12$
and $\sigma_Y = 0.1/12.$ By working through what I have shown you above,
I hope you are now sufficiently 'un-confused' to do part (B).
