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A student has a class that is supposed to end at $9 : 00$ a.m. and another that is supposed to begin at$9 : 10$ a.m. Suppose the actual ending time of the 9 a.m. class is a normally distributed rv $X_{1}$ withmean $9 : 02$ and standard deviation 1.5 $\mathrm{min}$ and that the starting time of the next class is also a normally distributed rv $X_{2}$ with mean $9 : 10$ and standard deviation 1 $\mathrm{min}$ . Suppose also that the time necessary to get from one classroom to the other is a normally distributed rv $X_{3}$ with mean 6 min and standard deviation 1 min. What is the probability that the student makes it to the second class before the lecture starts? (Assume independence of $X_{1}, X_{2},$ and $X_{3},$ which is reasonable if the student pays no attention to the finishing time of the first class.)

.8340

Intro Stats / AP Statistics

Chapter 4

Joint Probability Distributions and Their Applications

Section 11

Supplementary Exercises

Probability Topics

The Normal Distribution

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Okay, So we're given that the amount of time it takes for class one to end is normally distributed. The amount of time it takes for the class to to start is normally distributed. And the amount of time it takes to get from class one to class two is also normally distributed. And they have these normal distributions, I have been down here. X one is the time that it takes to get to The time it takes for the first class to end. X two is the time it takes for the second class to begin. An X three is the time It takes to get from Class 1 to class too. So what I'm gonna do is I'm gonna let t equal X one plus X three and X one plus X three is the time it takes for the first class to end, added to the time it takes to get from class one to class too. So this will be equal to This will be equal to do one of these lines is equal to a normal distribution where we add the main values and we add the variances and this is allowed because all three of these are normally distributed and they're all independent of each other. We can do that when they're independent of each other. So if we add 902-6 minutes we get 908 and if we add one squared to 1.5 squared 1.5 squared is 2 to 5. So this is 3 to 5. And now that we have this normal distribution of the time it takes to get to class two added to the time it takes for the first class to end. We can now set this up With this 3rd distribution. So what we're trying to find is we want this time this normal distribution to be less than at the time it takes for the class to to start. So this normal distribution, so we want t To be less than X two. And so what we're gonna do now is I'm gonna rewrite this by just minus ng ti from both sides. So you get zero Is less than X two -T. And now we can set this up as another normal distribution by minus ng the two means and then we still add the variances. So this would be normal Where we have 908. We're sorry, we have 9 10 minus 98 just to two minutes and we have 3.25 Added to one squared up here. So that is 4.25 as our variants. And now what we can do is just set up a Z score. So right now we want to find the probability that X two minus T is greater than zero, but I'm going to change this. So we've found the probability of Z is greater than 0-. Are you divided by our variance? I'm sorry, our standard deviation and this will let us use then the standard normal distribution table to find our probability. So if we let zero minus you, if we set up this, so we actually find zero minus, you will use to so we know zero minus you is negative two and the standard deviation, It's just the square root of our variance, which is 4-5. So we're trying to find the probability that Z is greater than -2 over the square to 4.25. And we can do this by setting this up a different way, we can say that this is equal to one minus the probability that Z is actually less than or equal to -2 over the square to 4.25. And this will make it. So we can actually look at our distribution table And find the value that we need to, which is the value at -2 divided by the square to 4.25. So if we simplify this further, We'll see that -2 divided by the square Of 4-5, goes to about -97. And now if we go and look at our Normal distribution table, we'll see that the value at negative .97 Is just about .16 60 And so now all we have to do is take one and subtract this value of .1660. And we'll get a resulting value of .834, which is then our probability.

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