find-a-the-mean-of-the-distribution-b-the-standard-deviation-of-the-distribution-and-c-the-probabi-5

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 years) that a seedling tree survives is a random variable $t$ with probability density function defined by
$$
f(t)=0.05 e^{-0.05 t} \quad 	ext { for } t 	ext { in }[0, \infty)
$$

Joshua Sieverding · Accepted 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;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, 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;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.

Laurent Bergeron · Answer

hi, everyone. Today we&#x27;ll be covering exercises from Chapter 11 1 tree named the Special Probability Density functions. So hopefully you&#x27;re already comfortable with Chapter 11 12 and 11 Do in Chapter 11 country, where you take a look at true different. Ah, very useful and using everyday life probably the identity functions such as uniform ah, probability, density function. And that&#x27;s the one that I could use for this problem from no one. So let&#x27;s have a quick review of what is a uniform? Um probability density function. So, see, this is a graph here. Um, the if the problem is uniform, it means that it doesn&#x27;t change in value. So here we have a certain So are you X in this problem, it could be the length of belief. And why is the probably I&#x27;m just gonna write product between two values of X. So between two different lengths, which we caught A and B, the probability is equal. So that&#x27;s what it&#x27;s telling us that between can be and so that music as likely to get the length of this length or a leaf of this length, all possibilities are equal. And for this specific case, um the function that describes this. Um, it&#x27;s probability his f of X. It has to be a constant of x ray. There can&#x27;t be any X in here because this is constant. You look in the manual, this is what you see. So as we consider is no ex. Well, one thing we should mention is that X has to be has to be between A and B. So why this constant? This is because the area off your probability density function has to be one. Why? Because the area of a program, the density function, is also probability. It&#x27;s the probability that your experiment will be between the boundaries of the area. So let me rephrase that if you take for example, um the area I&#x27;m drawing ready here, this should be equal to the probability that the leaf is between a and this value here. And so if you think the whole area that probably as to be one has to be one, because belief has to be somewhere. So that didn&#x27;t help you understand why we need this constant here. Why is it won over B minus a. Well, if you would take the integral of this rectangle. You would get the height, times, length and the length here. What the length is equal to be minus a right if you take B minus a times one over B minus A. You get one. All right, so now that&#x27;s that was just a quick review of the uniformed density function. No, it&#x27;s focusing that problem. Problem number one. The length in centimeter of the leaf of a certain plant is continuous. Random variable would probably density function defined by of picks because 5/7. And that tells you x between, um, four and 5.4. So what you wish you should The information you should expect from this is that in our special case of the uniform on, bro the density function, this is a and this is being, and we&#x27;ve just covered that it&#x27;s extra vehicle to one over B minus a right. Um, if you take 5.4 minus four and you do one over to result, you get 5/7. So this is all consistent with the definition that we had previously. Now, for this exercise, we want a you know that mean the distribution. Well, since we have a special case that we&#x27;ve still you already. We can look in the manual and find an equation for the mean of a uniform density function, and the mean is simply a plus. B divided bite you. So that will be four was 5.4, divided by two, and this is equal to 4.7. Um, those are centimeters because we&#x27;re talking about the length of leaf. Now be we want to know the standard effusion, and this one is equal to one over its credit of 12 B minus C. Again, that&#x27;s an equation that you find in the in this chapter for the special case. This is only for the uniform, probably the density function. It&#x27;s only gonna work when you have square box like that that is defined between a and B and that the eyes the height is one over, you might say. So it&#x27;s calculated this If we put um, 5.4 minus four, that&#x27;s 1.4. So 1.4 divided by square root of 12 that is approximately zero point 404 centimeters right and not see, we want to know the probability that they&#x27;re random. Variable is between the mean and one standard deviation above me. All right, I&#x27;m gonna You knew that for this so that we can visualize this better. So are a function is like that, right? And then we&#x27;ve got hey here and would be here on We can even write that. This is for this is 5.4. We found the mean here at 4.7. This makes sense because it&#x27;s right in the mill. So on average, that&#x27;s the lanky yet no and B. We also found that the standard deviation, which is how likely you are to deviate from the mean to be 0.4. So that news this length here, zero point four for us. Now we want to know the probability that the leaf is between the mean and one standard deviation way from demeanor are above the mean. So we wanted to know what&#x27;s the probably That belief falls into this red square here that it is between four points and centimeters and five point Wong or four centimeters. Um, all right, so the equation we need for that is, and we need to find the area of this Red Square to find the area the best way to do it is to use an interval. So 4.7 times and the indigo is from 4.7 24 27 plus zero point 44 And what we&#x27;re gonna integrate is if affects that probably function. That&#x27;s how you get probably that something is between two different guns. All right, so, um, let&#x27;s keep going. If X is, um five, we&#x27;re seven. And so we can be right. This is for integral from 4.72 5.14 of five or seven X. And if you saw this, you get you can take the constant way, and then it&#x27;s simply into North Constant. It is X, but we have to be evaluated from 4.7 two 5.24. So, um, we&#x27;re gonna replace X here with this number and that minus X with this number, and that gives this. I see your point two. He ate six. So that&#x27;s a pro, the probably decimal, with one being 100%. So if you want to convert this into person, did you can always too times 100% and then you would get 28 point 28 points. 86 percent. All right, so that&#x27;s it for that problem. You can always refer back to this problem when you have. Ah, when you want to know more about the uniform. Uh oh. The special case of the uniform probability density function, cause I&#x27;m not gonna do the review everything, um but yeah, that&#x27;s a problem for you in the next one.

Joshua Sieverding · Answer

Okay. So for this problem, we have a distribution. Affects is equal to 5/7 for X between three and 4.4. So whenever we have any distribution at all, we confined things, using integration in other techniques. But with something like this, we can notice that it&#x27;s a uniform random variable. That means that the probability the X is any of thes points between here eyes exactly the same. Right? So if we&#x27;re asked anything about this distribution, we can actually know some things just from memory or from looking them up easily. For example, the mean of this distribution expected X is just the middle point of our X value. So for this one, we&#x27;re going to be looking at 3.7. Ah, the medians. Well, okay, we don&#x27;t need that one. The standard deviation Ah, which we can denote. Sigma is going to be the square root off the variants on the variance is just going to be 4.4 minus three squared over 12. And the whole thing is under the square here to be clear, so this would then be equal to just one point for over route 12 and the probability that the invariable is between the median and the mean. We&#x27;ll figure that out. We would first need to figure out what the median is. Um, And for a uniform random variable, the median is actually the same as the mean, its 3.7. So the probability that X is between 3.7 and 3.7. Since we&#x27;re talking about a continuous random variable here, this has to be zero. So our final answers for this question will be these three here.

Problem

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

Problem 4 Easy Difficulty

Answer

$\begin{array}{lll}{\text { (a) } 20} & {\text { (b) } 20} & {\text { (c) } 0.2325}\end{array}$

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Finite Mathematics and Calculus with Applications

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$\begin{array}{lll}{\text { (a) } 20} & {\text { (b) } 20} & {\text { (c) } 0.2325}\end{array}$

Finite Mathematics and Calculus with Applications

Heather Z.

Kayleah T.

Caleb E.

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

Heather Z.

Kayleah T.

Caleb E.

Michael J.

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

Finite Mathematics and Calculus with Applications