To solve this problem, we need to determine the probability that someone will complete the test in 1 hour or less, given the probability density function:

$f(x) = \frac{-x^2 + 10x}{36}$

with the domain $0 \leq x \leq 3$.

The probability that someone will complete the test in 1 hour or less can be found by integrating the probability density function from $0$ to $1$:

$P(X \leq 1) = \int_0^1 f(x) \, dx = \int_0^1 \frac{-x^2 + 10x}{36} \, dx$

Let's solve this integral step by step.

Step 1: Set Up the Integral

We need to compute:

$\int_0^1 \frac{-x^2 + 10x}{36} \, dx$

This can be simplified by pulling out the constant $\frac{1}{36}$:

$= \frac{1}{36} \int_0^1 (-x^2 + 10x) \, dx$

Step 2: Integrate Each Term

Now, integrate each term separately:

1. The integral of $-x^2$ from 0 to 1: $\int -x^2 \, dx = -\frac{x^3}{3}$

2. The integral of $10x$ from 0 to 1: $\int 10x \, dx = 5x^2$

Thus, we have:

$\frac{1}{36} \left[ -\frac{x^3}{3} + 5x^2 \right]_0^1$

Step 3: Evaluate the Integral at the Bounds

Substitute $x = 1$ and $x = 0$:

For $x = 1$: $-\frac{(1)^3}{3} + 5(1)^2 = -\frac{1}{3} + 5 = \frac{15}{3} - \frac{1}{3} = \frac{14}{3}$

For $x = 0$: $-\frac{(0)^3}{3} + 5(0)^2 = 0$

Therefore, the result is:

$\frac{1}{36} \cdot \frac{14}{3} = \frac{14}{108} = \frac{7}{54}$

Conclusion

The probability that someone will complete the test in 1 hour or less is:

$\frac{7}{54} \approx 0.1296$

Would you like further details on any steps, or do you have other questions?

Here are five related questions that expand on this information:

1. How would we calculate the probability for someone to complete the test in less than 2 hours?
2. How do we verify that $f(x)$ is a valid probability density function over the interval $[0, 3]$?
3. What is the expected time to complete the test based on this probability density function?
4. How would the probability change if the test was extended to 4 hours?
5. How does integration relate to finding cumulative probabilities in continuous distributions?

Tip: When solving problems with probability density functions, ensure the total area under the curve over the defined interval equals 1 to confirm it's a valid distribution.