A clown is fired from a cannon towards a cream pie, centered a distance of 15 meters from the cannon. The pie has a radius of 1 meter and the horizontal distance R the clown travels is given by: $\ln R=−2.282+2\ln u+\ln(\sin2\theta)+\epsilon$ where $u$ is the initial velocity in meters per second, $q$ is the angle of the cannon in radians, and $\epsilon$ is an error term with normal distribution $N(0,0.1)$.
The initial velocity has three possible settings: $5, 10\ \text{or}\ 15$ m/sec. The angle of the cannon also has three possible settings: $0.349, 0.436$ or $0.524$ radians. What combination of settings will maximize the probability that the clown will land in the pie, and what is that probability?
Can someone please go through this with me step by step? So far I've only been able to draw a picture to represent it!
Repost!
The only thing tricky about this is that we've taken logarithms of all the "real-world" quantities you might have drawn in your picture. But the picture may throw more confusion than light on the problem.
In order to hit the pie, we need $14 \leq R \leq 16.$ That is, we need $\ln(14) \leq \ln R \leq \ln16.$ We are given that $\ln R$ is normally distributed with mean $−2.282+2\ln u+\ln(\sin2\theta).$
We don't need tables of the normal distribution. We don't even need to use the variance of $\ln R,$ although it is given to us. In fact, we don't even need to know that $\ln R$ is normal, only that it is bell-shaped and symmetric around its mean.
So now we have a random variable, let's call it $X$ to make it easier to think about, where $X = \ln R.$ This random variable $X$ has a symmetric bell-shaped distribution about some mean $\mu.$ We want to maximize the probability that $a < X < b,$ where $a$ and $b$ are constants. (In this particular problem, $a = \ln 14$ and $b = \ln 16$.)
It's obvious that the closer the mean $\mu$ is to the center of the interval $[a,b],$ the larger is the probability that $a < X < b.$ Whenever you shift $\mu$ farther away from the center of that interval, the area under the part of the curve that you cut off on one side is always greater than the area you add on the other side.
The center of the interval is at $$ \frac{a + b}{2} = \frac{\ln 14+\ln 16}{2} \approx 2.7058. $$ The mean of $X$ is $−2.282+2\ln u+\ln(\sin2\theta).$ Since we have exactly three choices for $u$ and three choices for $\theta,$ there are at most nine possible values of the mean. Which of those is closest to $2.7058,$ and which values of $u$ and $\theta$ do you need to choose to get it?
It might help to work out the three possible values of $2\ln u$ and the three possible values of $\ln(\sin2\theta)$ before you add things up. You might notice things that will let you guess the correct answer and eliminate all eight other possible answers quickly. But in the worst case you just evaluate the formula nine times, pick the best outcome, and you're done with the first part of the problem (finding the optimal settings).
For the second part of the problem, you have a known distribution of the normal variable $X$ and you need to compute the probability that $\ln(14) \leq X \leq \ln(16).$ That's when you need to use the fact that $X$ is a normal distribution with variance $\epsilon$ (not just any bell-shaped symmetric distribution) and when you need to drag out the table of the cumulative normal distribution (or use a calculator or software that can "look up" those values for you).