My First Ever Self Made Calculus - Probability Question

Calculus Level 5

$\large f(x) = \cfrac{x^4}{4} - \cfrac{2x^3}{3} - \cfrac{5x^2}{2} + 6x + 69 \ , \ x \in [-10 , 10]$ For $f(x)$ as defined above, let $A$ be the largest local maximum point and $B$ , the smallest local minimum point. If you choose a random real number $z$ in the interval $[ -10, 10 ]$ , what is the probability that the point $\big(z \ , \ f(z)\big)$ lie on the curve $AB$ ?

The answer is 0.4.

2 solutions

Tom Engelsman
Feb 22, 2017

Let us begin solving this probability problem by first finding the extrema of $f(x)$ above:

$f'(x) = x^3 - 2x^2 - 5x + 6 = (x+2)(x-1)(x-3) = 0 \Rightarrow x = -2, 1, 3$

$f''(x) = 3x^2 - 4x - 5 \Rightarrow f''(-2), f''(3) > 0; f''(1) < 0$

Thus, the local minima are $x = -2, 3$ and the local maximum is $x = 1.$ If we compare these points against the endpoints of the closed interval in question, namely $[-10, 10],$ we find:

$f(-10) = 2925.67; f(-2) = 56.33; f(1) = 72.083; f(3) = 66.75; f(10) = 1712.33$

or $f(-10) > f(10) > f(1) > f(3) > f(-2)$

which means $A = (-10, f(-10))$ and $B = (-2, f(-2))$ . We are interested in the probability that the point $z \in [-10, -2]$ , which can be found by calculating the arc lengths $L_{AB}, L_{total}$ :

$L_{AB} = \displaystyle \int_{-10}^{-2} \sqrt{1 + (f'(x))^{2}}\, dx = \displaystyle \int_{-10}^{-2} \sqrt{1 + (x^3 - 2x^2 - 5x + 6)^{2}}\, dx \approx 2869.8$

$L_{total} = \displaystyle \int_{-10}^{10} \sqrt{1 + (f'(x))^{2}}\, dx = \displaystyle \int_{-10}^{10} \sqrt{1 + (x^3 - 2x^2 - 5x + 6)^{2}}\, dx \approx 4537.3$

and $P(z \in [-10, -2]) = \frac{L_{AB}}{L_{total}} = \frac{2869.8}{4537.3} \approx \boxed{0.6324}.$

My Solution is similar to yours. :) Thanks for uploading your solution. :)

Christian Daang - 4 years, 3 months ago

No prob, Christian! It's a good problem that sparked some great healthy engagement here :)

tom engelsman - 4 years, 3 months ago

How you solve this integrals ???

Kushal Bose - 4 years, 3 months ago

Kushal, I confess I used Wolfram Alpha to evaluate the two arc length integrals due to my schedule today, but the reasoning is there. I welcome any suggestions/tricks on this ugly integrand!

tom engelsman - 4 years, 3 months ago

I also stopped at this point that's why I asked you

Kushal Bose - 4 years, 3 months ago

Isn't this about 1-d geometric probability, why do you need an integral for finding the length of the function curve in the given interval. If we choose any x in the interval [-10, -2] then the point will be on the curve AB and any x in the interval (-2, 10] will not fall on the curve. So the probability is simply (-2 - (-10)) / (10 - (-10)) = 8/20 = 0.4?

Siva Bathula - 4 years, 3 months ago

Siva, the important concept in this problem is to treat AB as a portion of the physical curve f(x), that is all of the points (x, f(x)) between (-10, f(-10)) and (-2, f(-2)). It does not include the line AB connecting these two extreme points. The probability takes into account the length of f(x) between A and B to the total length of f(x) between x = -10 and x = 10.

tom engelsman - 4 years, 3 months ago

I am saying, pick any x between [-10,-2] the point(x,f(x) falls on the curve AB and any x between (-2,10] doesn't fall on the curve AB. So its simply a 1-d geometric probability, 8/20?

Siva Bathula - 4 years, 3 months ago

@Siva Bathula – Actually, $\cfrac{8}{20}$ is not the probability of getting the z value between [-10 , -2] if $\displaystyle f(x) = \cfrac{x^4}{4} - \cfrac{2x^3}{3} - \cfrac{5x^2}{2} + 6x + 69$ . to be simple, i will give an example:

Let $f(x) = \begin{cases} \begin{aligned} \cfrac{1}{9}x^2 & , \ 0 \le x \le 3 \\ 0 & , \ \text{otherwise} \end{aligned} \end{cases}$

$P(1 \le x \le 2) = \displaystyle \int_{1}^{2} \cfrac{1}{9}x^2 = \cfrac{7}{27}$

Hence, if I will use your approach,

the answer will be $\cfrac{2 - 1}{3 - 0} = \cfrac{1}{3}$ which is not equal to $\cfrac{7}{27}$ .

So, you should use integration, not just subtracting it since the pdf is not a continuous uniform ( if i am not mistaken ) .

Then, this problem can be solved by just doing this:

$\cfrac{\int_{-10}^{-2} f(x)}{\int_{-10}^{10} f(x)} \approx 0.62795$

the alternative way is using the formula of the length of an arc, since the problem is talking about point $( z , f(z) )$ , not just the value of z.

Christian Daang - 4 years, 3 months ago

@Christian Daang – Why did the line integral even enter the picture? The problem is talking of an x chosen in between [-10, 10] and whether the point (x,f(x)) falls on the curve AB, the set of all points (x,f(x)) for x in [-10, -2]. And I am not convinced with your counter-example, that problem talks of probability distribution, which is different from the current problem. I suggest you take a look at the wording of the problem, and not jump to using line integrals.

Siva Bathula - 4 years, 3 months ago

Due to the phrasing of the problem, we're looking for the probability that $z \in [-10, 2 ]$ when it is randomly chosen from $[ -10, 10 ]$ . The length of the line segment is immaterial.

Calvin Lin Staff - 4 years, 3 months ago

Christian Daang
Feb 24, 2017

Alternative Solution:

$\cfrac{\int_{-10}^{-2} f(x)}{\int_{-10}^{10} f(x)} \approx 0.62795 \approx 0.630$

The upper bound and the lower bound of the integral of the numerator has been discussed by Mr. @Tom Engelsman . :)

Now, this solution is slightly wrong (fortunately, it is so closed to the answer. XD ) because the question is talking about the point $(z , f(z) )$ not just the probability of z that it lies under the graph of $f(x)$ or z is in the interval of [-10 , -2] (the lower bound and upper bound of the integral of the numerator. ) .

So, the much appropriate way is to solve this using the length of an arc. :)

The function is continuous and well-defined for [-10,10]. So for any value of x, f(x) is well-defined and falls somewhere on the curve of the graph of f(x) between the points (-10,f(-10)) and (10,f(10)). So the probability of a point (x,f(x)) falling on AB simply has to do with x lying between -10 and -2? So isn't this a simple case of 1-d geometric probability?

Siva Bathula - 4 years, 3 months ago

I don't mean the continuity, i mean the probability distribution. :) the probability distribution is not just a continuous uniform, that's why, you can't use that. The formula that you used can be used if the distribution type is a continuous - uniform/uniform distribution type. ( But as you notice, the pdf of this one is not a continuous uniform. :) )

Christian Daang - 4 years, 3 months ago

Due to the phrasing of the problem, we're looking for the probability that $z \in [-10, 2 ]$ when it is randomly chosen from $[ -10, 10 ]$ . The length of the line segment is immaterial.

Calvin Lin Staff - 4 years, 3 months ago

My First Ever Self Made Calculus - Probability Question

The answer is 0.4.

2 solutions

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