You know that someone takes the bus five days per week to their job. We're told that the waiting times until they can board the bus or a random sample from a uniformed distribution on the interval from 0 to 10 minutes in part a were asked to find the probability density function and then the expected value of the largest of the five waiting times. So to find the probability density function first we have that because X, which is the waiting time, is uniformly distributed on the interval from 0 to 10 minutes, we have ffx is equal to 1/10 minus zero, which is 10 so 1/10 and therefore cumulative distribution function for X capital F of X is X over 10 begin to roll this and therefore we have that g five of y. This is the largest of the five waiting times by a formula from this section is equal to five times why over 10 so ever. Why, to the five minus one is fourth power times lower case F of y, which is simply 1/10 which in turn is equal to five times wider, the fourth over 10 to the fifth. Yeah, and this is only for y between zero and 10. This is the probability density function now to find the expected value. The expected value of y five largest of the times is the integral over the possible values of why so from 0 to 10 of why times the probability density function of why? Which is five times wire the fourth over 10 to the fifth de y And evaluating this integral, we get 50 over six, which is approximately 8.33 and the unit is in minutes. Expected waiting time. Largest waiting time is 8.33 minutes. Next in part B were asked to find the expected value of the difference between the largest and smallest times. Well, again from part A. We have that the expected value of why one could be found by finding G one of why, and this could be found using a formula from this section. And so, if you are to carry out this computation, we find that the expected value of Y one is 10 6 or approximately one point 66 minutes, and therefore the expected value of range y five minus y one this is a linear combinations. This is the same as the difference of expected values. Affected value of y five. I asked the expected value of Y one and using the previous exercise we have 56 minus 16th. So we get 46 which is approximately 6.67 And this is also in minutes. This is the expected value of the range in Part C were asked to find the expected value of the sample Median waiting time. Well, because we have five samples, the median waiting time, the five samples is going to be. Why three. And we have again from a formula in this section that the pdf of white three is G three of y, which is given by five factorial over two factorial times. One factorial tends to factorial times capital f of why the CDF. This is going to be squared times a pdf f of why to the first power times one minus the CDF of y. You mean second power which putting in calculations from part A. This gives us 30 times why squared times 10 minus y squared all over 10 to the fifth. And this is also for why lying between zero and 10 And so it follows that using integration or by symmetry, we find it. The expected value of Y three is going to be five course. You could also do this by direct integration in the unit. Here, again is minutes. Finally in part D were asked to find the standard deviation off the largest time. We're looking for the standard deviation of Y five. Well, first, it's find the second moment of Y five. This is the integral from 0 to 10 of why squared times the pdf of y five, which, when there was five times why did the force over 10 to the fifth de y and evaluating this integral we get 500 over seven, therefore by the shortcut formula for variance the variance of y five. Well, this is going to be the second moment of y five minus the first moment of y five squared. And so we have. This is 500 over seven minus. And from the first part, this is 50/6 squared and calculating the obtained 125 over 63 which is approximately 1.984 from which we obtain, but the standard deviation of why five is the square root of this and is equal to approximately 1.409 minutes.