The time taken by a randomly selected applicant for a mortgage to fill out a certain form has a normal distribution with mean value 10 min and standard deviation 2 min. If five individuals fill out a form on one day and six on another, what is the probability that the sample average amount of time taken on each day is at most 11 min?
In an attempt to work out the problem, I started by identifying the known variables:
$X$: the time taken by a randomly selected applicant for a mortgage to fill out a certain form (that is, one form)
$\mu_{X}=10$
$\sigma_{X}=2$
Now, there are obviously more variables but this is where it starts to get a bit confusing for me. In the first part of the problem, $X$ clearly refers to the time it takes to fill out a single form. But in the second part of the problem, the population ($n=5$) fills out a different number of papers on each day, changing the population the question at the end of the problem is addressing. Presumably, there is some other population $Y$ and I'm looking for $P(\overline{Y}\leq11)$. Figuring out what population $Y$ encompasses and how to set up the solution is where I'm stumped. Any nudge in the right direction is greatly appreciated.
Best Answer
The population is the same on both days, but the samples of different sizes.
On Day 1 the average $\bar X_1 \sim \mathsf{Norm}(10, 2/\sqrt{5})$ and $$P(\bar X_1 < 11) = P\left(\frac{\bar X_1 - 10}{2/\sqrt{5}} < \frac{11-10}{2/\sqrt{5}}\right) = P(Z < 1.118) = 0.8682,$$ where $Z$ is standard normal.
Similarly, on Day 2: $\bar X_2 \sim \mathsf{Norm}(10, 2/\sqrt{6})$ and $P(\bar X_2 < 11) = 0.8897.$
I used software to evaluate the probabilities. If you use printed normal tables, on Day 1 you will have to use $P(Z < 1.12)$ instead of $P(Z < 1.118)$ or use interpolation. So you may get a slightly different answer to four places. Similarly for Day 2.
I will leave it to you to do the intermediate steps for Day 2, to show how to use the assigned method of evaluating standard normal probabilities (software or tables) in each case, and to finish the problem.