A statistics and probability problem Assume that the average reaction time of drivers is normally distributed with average of 1.1 s and a standard deviation of 0.3 s.


*

*Compute the probability that in the selected simple random sample of
50 drivers we compute sample average, that is less than 1 s.

*Assess the proportion of drivers who have a longer reaction time than 1.25 s.
 A: A large start: Let $X_1$, $X_2$, and so on up to $X_{50}$ be the reaction times of the drivers. We are interested in the random variable $Y$, where
$$Y=\frac{1}{50}(X_1+X_2+\cdots+X_{50}).$$
Because the $X_i$ are independent normal with mean $1.1$ and variance $(0.3)^2$, the random variable $Y$ is normally distributed with mean $\frac{1}{50}(50)(1.1)=1.1$ and variance $\frac{1}{50^2}((50)(0/3)^2)=\frac{(0.3)^2}{50}$. 
To put it in another way, the standard deviation of $Y$ is $\frac{0.3}{\sqrt{50}}$. (You may have just been told the formula for the standard deviation of the sample mean, the expression $\frac{\sigma}{\sqrt{n}}$ may be familiar. But I could not resist sketching in some of the theory.)
Now you want to find the probability that the normal $Y$ with mean $1.1$ and standard deviation $0.3/\sqrt{50}$ is less than $1$. 
Note that $0.3/\sqrt{50}\approx 0.0424264$. So we want the probability that 
$$Z<\frac{1-1.1}{0.0424264},$$
where $Z$ is standard normal. So you will be calculating, roughly, the probability that $Z<-2.35702$. I assume you know how to use tables of the standard normal to do that. But in case the negative number causes difficulty, by symmetry our probability is the same as the probability that $Z>2.35702$, and tables easily give you $P(Z \le 2.35702)$.
Question $2$ is much faster to solve. If $X$ is normal mean $1.1$, standard deviation $0.3$, all we want is $P(X>1.25)$. The usual "table" way is to first compute $P(X \le 1.25)$. 
