find-a-the-mean-of-the-distribution-b-the-standard-deviation-of-the-distribution-and-c-the-probabi-6

Question

Find (a) the mean of the distribution, (b) the standard deviation of the distribution, and (c) the probability that the random variable is between the mean and 1 standard deviation above the mean. 
The length of time (in days) required to learn a certain task is a random variable $t$ with probability density function defined by 
$$
f(t)=\frac{e^{-l / 3}}{3} 	ext { for } t 	ext { in }[0, \infty)
$$

Joshua Sieverding · Accepted Answer

Okay, So for this question, we&#x27;re gonna notice that our density function here you to the negative t over three, divided by three is exactly the form of an exponential distribution with Lambda equals 1/3. So if we want to because this is an exponential distribution. We have these formulas down here that we can make use off. So if we want to find the expectation of T, which is the mean we know, it&#x27;s just one over Lambda. So that will be three If we want to find the standard deviation. We know that&#x27;s just the square root of the variants which will be one over lambda again. So three again. And if we want to find the probability that, uh, rt is within one standard deviation of its mean, uh from above, that&#x27;s going to be the probability that T is between three and three plus three. So between three and six and that follows exactly this formula down here. So we know that that&#x27;s going to be 0.23 25 So these are our final answers

Joshua Sieverding · Answer

Okay, So for this question, we&#x27;ve got a distribution. F of t is equal for e to the negative 40. And we can notice that this is what we would refer to as an exponential distribution. So we&#x27;ve got the following formulas to work with. Expected to use one over Lambda Sigma Squared is one over lambda squared and cap FFT is one minus e to the negative. Lambda t. So, for this question, we want to know first off, the mean are expected to you. So that&#x27;s just gonna be won over four. Then we want to know the standard deviation. And the interesting thing about the exponential distribution is that the standard deviation is also the same as, ah, the mean. So we&#x27;re gonna get 1/4, and then the final part is the probability of that. T lies between ah, the means and one standard deviation above the means. So for us, it&#x27;s between the fourth and 1/2. So we&#x27;re gonna do is plug in f of 1/2 and subtract off f of 1/4 from our cumulative distribution function. Ah, and if we plug both these and for here we will get approximately 0.2 3 to 5. So these are our final answers

Joshua Sieverding · Answer

Okay, So for this question, were given a densely function f of tea which is equal to 0.5 e to the negative 0.5 tea. And so we can notice that this follows exactly the format of an exponential distribution that gives us all of these formulas down here to work with. So if we want to find the mean for tea, that&#x27;s just going to be expectation of tea, which we see is one over Lambda. So we&#x27;re gonna get one over 10.5, which would give us 20. Next, we&#x27;re gonna be looking for the standard deviation, which is the same as the square root of the variance. So we&#x27;re again gonna get one over Lambda and again get 20 for the final part. We want to know the probability that, ah t is between its mean and one standard deviation above the mean. So this is between 20 and 40 so we would write this as cap f of 40 minus cap f of 20. So all we have to do is go ahead and plug 40 and 20 into our half of tea so we would get one minus you to the negative Ah, 0.5 times 40 which would be to. And then we&#x27;re gonna subtract off one minus e to the negative one. And if you plug this into a calculator, you will get that. The answer is 0.232 five. So this is our final answer.

Joshua Sieverding · Answer

Okay, So for this question, we&#x27;re giving a density function. F X is equal a 0.1 times e to the negative 0.1 times X for X from zero to infinity. We could notice. This follows exactly The format of the exponential distribution with Lambda equals 2.1. So if we want to find the expectation of X, which is the mean of our random variable, we know that that&#x27;s going to be equal to one over Lambda. So one over 10.1, which is 10. Next. If we want to find the standard deviation, we know that&#x27;s the square root of the variance. So that&#x27;s going to be also one over Lambda for 10 again. Then if we want to find the probability that are variable is between its mean, which is 10 and one standard deviation above its mean, which would be 10 plus 10 for 20 we can look down here and see that that follows exactly the format for this formula, and that will give us point to 32 five. And these are going to be our final answers for this question.

Joshua Sieverding · Answer

Okay, So for this question, we are given in density function ffx that&#x27;s equal to four for X between 2.75 and three. So our first thing we&#x27;re gonna do is notice that this is a uniform rate variable because the density function is a constant. So this means that we can use some of our nice formulas. Ah, the expect the expected X or the mean of our function is just going to BB plus a two, which is the halfway point eso We&#x27;re going to get 2.875 for that. Ah sigma. It&#x27;s going to be a square of sigma squared right here, so we&#x27;ll get B minus a which is going to be 0.25 over route 12 which is approximately 0.0 7 to 1. And finally, for the last part, we want to know the probably that the, uh ren variable is between the mean and one standard deviation above the means such the probability The axe is between 2.875 and 2.875 plus one standard deviation, which we just found to be 0.721 This is the probably the excess between this 0.2 point 9471 Uh, so there&#x27;s plenty of different ways we can go about solving for this. Ah, we can do the integration if we want. Or we can just know that the CDF, the cumulative density function for our uniform and a variable is equal to Ana. So aftereffects is equal to X minus a just 2.75 over B minus a which would be three minus which 275 which would be 2750.25 And so to get this, we can first plug in and take half of 2.9471 minus f of 2.875 and playing both of those ends and subtracting them from each other, we will get 0.288 and so those are our final answers.

Laurent Bergeron · Answer

all right. Moving on to number five. Um, number five is very similar to number tree, actually. So if you interested number tree, don&#x27;t think you should have any problems with number five. We have an exponential gaze again with f of t equals, um, one of the tree times it minus one retreat T. And so if you remember the exponential case, it&#x27;s the finest f of X equals a eight minus 80 with mean equals to one over A and sigma because to one over a So here in problem number five are mean Is one over wanted because a is equal to wonder two. That&#x27;s so here. I can&#x27;t write up of this. Um, all right, this is a and it&#x27;s also here, right? If you compare dysfunction to dysfunction, you can see that a equals wanted so mean so one over a right again Here, Eco&#x27;s won over 100 equals treat bit. This tender deviation is also over a So it&#x27;s also one other wanted also treat awesome. It&#x27;s no in. See, that is a C. We want to know the probability that Oh, by the way, I should read my units here. Um, in this problem. He is the length of time in days. So it mean it&#x27;s true. Days in the standard Division is also treaties if you want to. Um, if you want to know the probability that is between you mean and once tender deviation above mean, which is see, you have to, Jack, let the integral between those two boundaries. So I though a graph of this in the number tree if you want to get more visual explanation, let&#x27;s do it here. That that&#x27;s what we&#x27;re looking for. We want to integrate our probe. You function between our two boundaries that were interested in so news tree. There&#x27;s mu plus signal. Be triple A street hoops want threat a tree here and new place signal six. And then we&#x27;re integrating ft. Which is 1/3 You didn&#x27;t mind this 1/3 to DT. All right, um, wanted here and then in the growth of an exponential is exponential divided by the coefficient, um, key. But we&#x27;re gonna evaluate tea from treat six. So what? We got this 1/3 divided by minus 1/3. That&#x27;s minus one. Um, eating the minus 1/3 at six is heated in minus two and then eating the minus one. Heard that tree is ate My swan. You seen this somewhere? The exact same thing we got for number Tree. And indeed, when you put this in your calculator, you will get the same answer. Isn&#x27;t that very interesting? Well, is a little extra. It&#x27;s visualized why? The probability would be the same even if the value for a changed um so Hoose actually gonna droids here? So here&#x27;s our exponential distribution and let&#x27;s see the mean It&#x27;s right here. Then we know this is a signature because it&#x27;s equal to mean. And so this is sick mice. Well, so here we have, I&#x27;m you sigma and what we just talked late in This problem is the area between those two boundaries. Now, here&#x27;s a trick. It doesn&#x27;t matter what use are what sigma is because even if those numbers changed, the relative of Europe of the curve or how it looks is not gonna change. And since the total area is always gonna be one, the area between the red thing here is not gonna change either. So this is why, even if you change value for a you get the same probability. I think that&#x27;s what they wanted to make you realize here. By comparing gallantry to five and getting the same answer, even with different values. All right, this is that Was it for number five?

Problem

Find (a) the mean of the distribution, (b) the st…

Problem 5 Easy Difficulty

Answer

(a) 3$\qquad$ (b) 3$\qquad$ (c) 0.2325

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Margaret L. Lial, Raymond N. Greenwell, Nathan P. Ritchey

Finite Mathematics and Calculus with Applications

Catherine R.

Caleb E.

Kristen K.

Michael J.

Precalculus Review - Intro

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(a) 3$\qquad$ (b) 3$\qquad$ (c) 0.2325

Finite Mathematics and Calculus with Applications

Catherine R.

Caleb E.

Kristen K.

Michael J.

Precalculus Review - Intro

Functions on the Real Line - Overview

Margaret L. Lial, Raymond N. Greenwell, Nathan P. RitcheyFinite Mathematics and Calculus with Applications

Catherine R.

Caleb E.

Kristen K.

Michael J.

Margaret L. Lial, Raymond N. Greenwell, Nathan P. Ritchey

Finite Mathematics and Calculus with Applications