Find $P(-10Let $X_1,X_2,X_3$ represent the times necessary to perform three successive repair tasks at a repair facility. 
Suppose $X_1,X_2,X_3$ are independent normally distributed random variable with $X_1 \sim N(60,15)$, $X_2\sim N(60,15)$, $X_3\sim N(60,15)$. Let $M=X_1-\dfrac{1}{2}X_2-\dfrac{1}{2}X_3$. Compute $P(-10<M<5)$.
I cannot figure out how to tackle this problem. How to find mean and standard deviation of $M$? Any help is much appreciated.
 A: Characteristic for normal distribution are the properties:


*

*If $X\sim\mathsf{Norm}$ and $c\in\mathbb R-\{0\}$ then $cX\sim\mathsf{Norm}$

*If $X,Y\sim\mathsf{Norm}$ and are independent then $X+Y\sim\mathsf{Norm}$


Further if $Z$ has normal distribution then its distribution is completely determined by its mean and variance.
The two mentioned rules give you the assurance that $M$ has normal distribution. So it now remains to find the mean and variance of $M$ which can be done by means of linearity of expectation and secondly the rule that the variance of two independent random variables is the sum of their variances.
More precisely:


*

*$\mathbb EM=\mathbb E[X_1-\frac12X_2=\frac12X_3]=\mathbb EX_1-\frac12\mathbb EX_2-\frac12\mathbb EX_3$

*$\mathsf{Var}M=\mathsf{Var}X_1+\mathsf{Var}(-\frac12X_2)+\mathsf{Var}(-\frac12X_3)$


The second bullet can be worked out further on base of the general rule $\mathsf{Var}(cX)=c^2\mathsf{Var}X$.
Caution: mostly e.g. $\sim\mathsf{Norm}(60,15)$ means that we are dealing with a normal distribution that has mean $60$ and variance $15$. But unfortunaly it also happens that it stands for normal distribution that has mean $60$ and standard deviation $15$.
A: If $X$ follows a normal distribution $\left(X\sim N(\mu, \sigma^2)\right)$ and $c$ is a constant then:
$$c\cdot X \sim N(c\cdot\mu, c^2\cdot\sigma^2)$$
Also, If $X$ and $Y$ are independent random variables that are normally distributed then their sum is also normally distributed and
$$X+Y \sim N(\mu_X + \mu_Y, \sigma_X^2 + \sigma_Y^2)$$
So in this case where $M = X_1 - \dfrac{1}{2}X_2 - \dfrac{1}{2}X_3$, we have
$$M \sim N\left(\mu_1 -\frac{1}{2}\mu_2 - \frac{1}{2}\mu_3, \sigma_1^2 + \frac{1}{4}\sigma_2^2 + \frac{1}{4}\sigma_3^2\right)$$
and $\mu_1=\mu_2=\mu_3 = 60$, $\sigma_1^2 = \sigma_2^2 = \sigma_3^2 = 15$ so
$$M \sim N\left(0, \frac{45}{2} \right)$$
And with that, you are ready to compute $P(−10<M<5)$.
