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

• A. $P(2\le X\le 2.5)=0.2969$
• B. $P(2\le X\le 2.5)=0.0625$
• C. $P(2\le X\le 2.5)=0.8906$
• D. $P(2\le X\le 2.5)=0.0234$

Answer: (A)

Verified step by step guidance

1. 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.
2. 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.
3. 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].
4. 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.
5. 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.