I do my practice and I stuck with this question
Let $X_1, X_2, \cdots, X_9$ represent a random sample from the distribution of $X$. What is the distribution of the sample $\dfrac{\sum_{i = 1}^9 X_i - 18}{15}$? Find $\mathbb P\left[\dfrac{\sum_{i = 1}^9 X_i - 18}{15}< 2\right]$.
and it gives additional information is "Let $X$ is normally distributed with mean $2$ and variance $25$". I try to find out what type of this distribution. it's not chi-square, normal. Can someone help me to solve this question?
Hint: This is very elementary. You should know that
$$\sum_i (a_i\mathcal{N}(\mu_i,\sigma^2_i) + b_i) = \mathcal{N}\left(\sum_i (a_i\mu_i + b_i),\sum_i a^2_i\sigma_i^2\right)$$