From Wackerly's *Mathematical Statistics with Applications*, problem 4.30, question c

Question

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 $\int^{\infty}_{- \infty} f(y) \, dy = \int^{t}_{0} 2y \, dy \neq 1$ ...

Answer 1

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.

Answer 2

The first part of this problem (from Wackerly, 7th Edition, problem 4.30(a), page 173) asks for E(Y) and V(Y). These are determined as follows:

$$E(Y) = \int_0^{1}2y^2dy = \frac{2}{3}y^3\rvert^1_0 = 2/3$$

$$E(Y^2) = \int_0^{1}2y^3dy = \frac{2}{4}y^4\rvert^1_0 = 1/2$$

$$V(Y) = E(Y^2) - [E(Y)]^2 = 1/2 - 4/9 = 1/18$$

Given that $X = 200Y - 60$, the mean and variance of X (part b) can be determined:

$$E(X) = 200(2/3) - 60=73.\bar3 $$

$$E(X^2) = E((200Y-60)^2) = E(40000Y^2 - 24000Y + 3600) $$

$$\implies E(X^2) = 40000E(Y^2) - 24000E(Y) + 3600 = 40000(1/2) - 24000(2/3) + 3600 = 7600$$

$$\implies V(X) = E(X^2) - [E(X)]^2 = 7600 - (73.\bar3)^2 = 2222.\bar2$$

Now that $\mu$ and $\sigma$ are known for X and Y, Tchebysheff's inequality with k = 2 yields:

$$P(|Y-\mu| \le k\sigma) \ge 1 - \frac{1}{k^2} $$

$$P(|Y-(2/3)| \le 2\sqrt{1/18}) \ge 0.75 $$

$$\implies -2\sqrt{1/18} \le Y-(2/3) \le 2\sqrt{1/18} $$

$$\implies 2/3 -2\sqrt{1/18} \le Y \le 2/3 + 2\sqrt{1/18}$$

$$0.195262 \le Y \le 1.13807$$

Since y is defined on $0 \le y \le 1$:

$$0.195262 \le Y \le 1$$

This interval signifies that for a given week, there is at least a 75% chance that the robots are operating about 19.53% to 100% of the time. Since the proportion of time in operation was the only variable in the expression for profit, this result implies an interval for the profit.

$$\implies 200(0.195262)-60 \le X \le 200(1)-60$$

$$\implies -20.947 \le X \le 140$$

If one defines profit as being a strictly non-negative quantity, the lower boundary would presumably be zero.

Notably, using Tchebysheff's inequality directly on X instead of Y yields an identical interval, except that the upper boundary of $Y \le 1$ is unaccounted for.