From Wackerly's *Mathematical Statistics with Applications*, problem 4.30, question c I find the text of this exercise ambiguous. The question states:

The proportion of time $Y$ that an industry robot is in operation
during a 40-hour week is a random variable with probability density
function $$ f(y) = 2y, \qquad 0 \leq y \leq 1 $$
For the robot under study, the profit for a week is given by $X = g(Y) = 200Y - 60$.
Find an interval in which the profit should lie for at least 75 % of
the weeks that the robot is in use.

My interpretation is that we are required to find $Y : P(X > 0) = 0.75$.
So, the probability to get $Y \geq \dfrac{60}{200} = 0.3$ should be $0.75$.
$P(Y \leq 0.3)$ is easily computed as $F(0.3) = \int^{\infty}_{- \infty} f(y) \, dy = \int^{0.3}_{0} 2y \, dy = 0.09$, thus $P(Y \geq 0.3) = 0.91 > 0.75$ already.
My guess then is, the problem requires us to restrict the original interval for $Y$ from $[0, 1]$ to $[0, t]$, so that $P(Y \geq 0.3) = 0.75$ holds for the new interval.
However, in case the original $[0, 1]$ interval of definition gets modified, then the integral of $f(y)$ above is not going to be equal to $1$, with the consequence that the cumulative distribution function $P(Y \leq y) = F(y) = \int^{\infty}_{- \infty} f(y) \, dy$ won't be well-defined anymore, given that now $\int^{\infty}_{- \infty} f(y) \, dy = \int^{t}_{0} 2y \, dy \neq 1$ ...
 A: My interpretation of the question is like this:  For each 40-hour week, let $Y_i$ represent the random proportion of time that the robot is in operation, for week $i \in \{1, 2, 3, \ldots \}$.  Let $X_i = g(Y_i)$ represent the profit (or loss) for week $i$.  Find an interval $[L, U]$ such that for sufficiently large $n$, $L \le X_i \le U$ is true for $i \in S$ where $|S| \approx 0.75n$; that is to say, $$\Pr[L \le X \le U] = 0.75. \tag{1}$$
This in turn would imply $$\begin{align}
0.75 &= \Pr[g^{-1}(L) \le Y \le g^{-1}(U)] \\
&= \Pr\left[\frac{60+L}{200} \le Y \le \frac{60+U}{200}\right] \\
&= \left(\frac{60+U}{200}\right)^2 - \left(\frac{60+L}{200}\right)^2 \\
&= \frac{(60+U)^2 - (60+L)^2}{200^2}.
\end{align}$$
Solving $U$ in terms of $L$ yields
$$U = -60 + \sqrt{L^2 + 120L + 33600}, \tag{2}$$ which demonstrates that the answer is not uniquely determined:  We may choose any bounds satisfying $-60 \le L < U \le 140$ that obeys $(2)$; e.g., when $L = -60$, we have $U = 100 \sqrt{3} - 60$, and when $L = 40$, we have $U = 140$.  This is depicted in the following plot, where $L$ is the lower bound in blue and $U$ is the upper in orange.

