Find the cumulative distribution function for the probability density function f(x)=38(x−1)2f\left(x\right)=\frac38\left(x-1\right)^2 on the interval [1,3]\left\lbrack1,3\right\rbrack. Use the CDF to find the probability that XX is between 22 and 2.52.5.

P ( 2 ≤ X ≤ 2.5 ) = 0.2969 P\left(2\le X\le2.5\right)=0.2969

Find the cumulative distribution function for the probability density function f ( x ) = 3 8 ( x − 1 ) 2 f\left(x\right)=\frac38\left(x-1\right)^2 on the interval [ 1 , 3 ] \left\lbrack1,3\right\rbrack . Use the CDF to find the probability that X X is between 2 2 and 2.5 2.5 .

we're told to find the cumulative distribution function for the probability density function little f of x equals three eighths times x minus one squared on the interval from one to three. Then we're told to use that cdf to find the probability that x is between two and two point five. So let's start with that first part, just finding our CDF. So our CDF, capital f of x, is going to be equal to the integral. Now here, because this is only on the interval from one to three, I can make this lower bound one as opposed to negative infinity because nothing is happening from negative infinity to one. So bounding this from one up to x, I now want to integrate my PDF with t plugged in where x is. So this is three eighths times t minus one squared and then dt. Now in order to take this integral, I need to do a u substitution. So setting u equal to t minus one, that makes du equal to dt. So replacing that t minus one with u and then that dt with du. I can rewrite this as three eighths integral of a u squared du. Now, this ends up giving me here three eighths times one third u cubed, and I can go ahead and replace that u back with that t minus one so that I can bound it with my original bounds. So that's t minus one cubed and then bounded from one to x. Now this three eighths and one third, those threes will end up canceling and making this a one eighth times t minus one cube bounded from one to x. And now I just want to plug in those bounds here. So plugging in my bounds, this becomes a one eighth. I'm going to keep that out front for now times X minus one cubed minus one minus one cubed. One minus one is zero, so this ends up canceling out. And this ends up giving me here one eighth times x minus one cubed. And this then gives me my cumulative distribution function, f of x. But we're not quite done yet because we were asked to use this cumulative distribution function to find the probability that x is between two and two point five. So how exactly can we now find that probability? We're looking for the probability that x is a greater than or equal to two and less than or equal to 2.5. Using our CDF here, that's gonna be F, big F of 2.5 minus big F of two. So we just wanna plug those values into that CDF that we just found and find the difference. So that's going to be a one eighth times 2.5 minus one eight times two minus one cubed. Now doing this algebra here, we end up ultimately getting one eighth times 3.375, that's that 2.5 minus one cubed. And then minus one eighth because two minus one is one, one cubed is one. That leaves us just with that one eighth. Now putting this into our calculator or making these common fractions and getting that final decimal will give us 0.2969 for our final answer here for the probability that x is in between two and two point five having used that CDF. If you have any questions, feel free to let us know. I'll see you in the next one.

Find the cumulative distribution function for the probability density function f ( x ) = 3 8 ( x − 1 ) 2 f\left(x\right)=\frac38\left(x-1\right)^2 on the interval [ 1 , 3 ] \left\lbrack1,3\right\rbrack . Use the CDF to find the probability that X X is between 2 2 and 2.5 2.5 .

P ( 2 ≤ X ≤ 2.5 ) = 0.2969 P\left(2\le X\le2.5\right)=0.2969

P ( 2 ≤ X ≤ 2.5 ) = 0.0625 P\left(2\le X\le2.5\right)=0.0625

P ( 2 ≤ X ≤ 2.5 ) = 0.8906 P\left(2\le X\le2.5\right)=0.8906

P ( 2 ≤ X ≤ 2.5 ) = 0.0234 P\left(2\le X\le2.5\right)=0.0234

16. Probability & Calculus

Continuous Probability Models

Business Calculus

16. Probability & Calculus

Continuous Probability Models

Struggling with Business Calculus?

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Multiple Choice

Find the cumulative distribution function for the probability density function $f (x) = \frac{3}{8} {(x - 1)}^{2}$ on the interval $open bracket 1 comma 3 close bracket$ . Use the CDF to find the probability that $X$ is between $2$ and $2.5$ .

$P of open paren 2 is less than or equal to X is less than or equal to 2.5 close paren equals 0.2969$

$P of open paren 2 is less than or equal to X is less than or equal to 2.5 close paren equals 0.0625$

$P of open paren 2 is less than or equal to X is less than or equal to 2.5 close paren equals 0.8906$

$P of open paren 2 is less than or equal to X is less than or equal to 2.5 close paren equals 0.0234$

Verified step by step guidance

Step 1: Understand the problem. You are tasked with finding the cumulative distribution function (CDF) for the given probability density function (PDF) f(x) = (3/8)(x - 1)^2 on the interval [1, 3]. The CDF is the integral of the PDF from the lower bound of the interval to a variable upper bound x.

Step 2: Set up the integral for the CDF. The CDF F(x) is defined as F(x) = ∫[1,x] f(t) dt, where t is the variable of integration. Substitute the given PDF into the integral: F(x) = ∫[1,x] (3/8)(t - 1)^2 dt.

Step 3: Solve the integral. Expand (t - 1)^2 to get t^2 - 2t + 1. The integral becomes F(x) = (3/8) ∫[1,x] (t^2 - 2t + 1) dt. Break this into separate integrals: F(x) = (3/8) [∫[1,x] t^2 dt - 2∫[1,x] t dt + ∫[1,x] 1 dt].

Step 4: Compute each integral. Use the power rule for integration: ∫ t^n dt = (t^(n+1))/(n+1). For ∫ t^2 dt, the result is (t^3)/3. For ∫ t dt, the result is (t^2)/2. For ∫ 1 dt, the result is t. Evaluate these expressions at the bounds 1 and x.

Step 5: Use the CDF to find the probability that X is between 2 and 2.5. The probability is given by P(2 ≤ X ≤ 2.5) = F(2.5) - F(2). Substitute the values of F(2.5) and F(2) obtained from the CDF into this formula to compute the probability.

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