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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.

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

  2. Assess the proportion of drivers who have a longer reaction time than 1.25 s.

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If you know that a bunch of $n$ random variables are normally distributed with mean $\mu$ and standard deviation $\sigma$, then the sample average is also normally distributed, with mean $\mu$ and standard deviation $\sigma/\sqrt{n}$. Do you know how to get the quantiles of a normal distribution? – Chris Taylor May 22 '12 at 13:29
Chris gave you the key to this and Andre made it explicit. It looks like a homework problem. Does Mathematics require homework problems to be labelled as such? At CrossValidate we do and we only give hints to homework rather than complete solutions. – Michael Chernick May 22 '12 at 15:32
Wasn't a homework realy... im studying for an exam and came across this problem and coulnd't solve it. Thus Andre's solution was very welcome. – tomz May 22 '12 at 20:04

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 $P(X \le 1.25)$.

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