The time it takes a person to complete the crossword puzzle in the newspaper is known to be normally distributed, with a mean of 35 minutes, and with a standard deviation of 6 minutes.
Suppose that 9 people are selected at random, and each person completes a crossword puzzle. What is the probability that the average time to complete the puzzles is no more than 34 minutes?
Hi guys, I need some guidance in order to solve this problem.
My first step was to find the $\text P(x\le 34)$ that represents the probability that ONE person resolve the puzzle in less than 34 minutes.
I found it, it is equal to $0.4338$.
However, I have no idea what to do after this.
$0.4338$ is the probability that one person, selected at random, would take no more than $34$ minutes to complete the puzzle. You are interested in the probability that nine people, selected at random, would require a mean time of no more than $34$ minutes to complete the puzzle.
It seems that we should be considering the sampling distribution associated with a sample of nine from the given population for this problem.
Notice that even though the mean of the population is 35 minutes, the mean of a random sample of nine people depends on which nine people are in the sample.
Imagine we generate a new data set by repeatedly sampling nine people from the population and recording the mean of the sample each time.
It turns out that, by the central limit theorem, this new data set will have a normal distribution (for large samples, even if the original population is not normally distributed) with the same mean as the original population.
This new data set will further have a standard deviation $\sigma_{\bar x}$ (which we refer to as a standard error in the context of the original population) that is related to the standard deviation of the original population $\sigma$ and the sample size $n$ via:
$$\sigma_{\bar x}=\frac{\sigma}{\sqrt n}$$
(This makes sense because we would expect a larger sample to have a mean more similar to the population mean than a smaller sample would, on average, and notice that this boils down to $\sigma_{\bar x}=\sigma$ for a sample size of $n=1$, as expected. It is a result worth exploring in more detail, however, with simulations, for example.)
In any case, you are now interested in $\text P(X\le34)$ given that $X\sim \text N\left(35,\frac{6^2}{9}\right)$, or the probability that the sample mean is no more than $34$ given that it is normally distributed with mean $35$ (same as original population mean) and standard deviation $\frac 6 3 = 2$ (smaller than the standard deviation of the original population), that is, $0.3085$.