- #1
archaic
- 688
- 214
- Homework Statement
- We are filling some liquid in multiple containers. The volume poured is uniformly distributed between 5.1 and 5.7 deciliters. Every centiliter costs 0.01 dollar.
Knowing that any centiliter added beyond 5.5 deciliters represents an extra cost for the producer, what is the mean of this extra cost?
- Relevant Equations
- N/A
I am not sure about how to approach this.
Since the volume is uniformly distributed, the mean volume is ##(5.7+5.1)/2=5.4##, which is less than ##5.5##. From this, I could say that, on average, the producer won't spend any extra dollars.
But then I thought that maybe I should interpret this as the average if we only consider the cases where the volume is more than ##5.5##, so that my random variable is now ##5.5\leq X\leq 5.7##.
If so, my new mean is ##5.6##, which is just one centiliter above the lower-bound volume for there being an extra cost, resulting in a mean of ##0.01##$.
Or maybe ##\int_{5.5}^{5.7}\left[0.01(x-5.5)\right]f(x)dx##, where ##f(x)## is the pdf?
Since the volume is uniformly distributed, the mean volume is ##(5.7+5.1)/2=5.4##, which is less than ##5.5##. From this, I could say that, on average, the producer won't spend any extra dollars.
But then I thought that maybe I should interpret this as the average if we only consider the cases where the volume is more than ##5.5##, so that my random variable is now ##5.5\leq X\leq 5.7##.
If so, my new mean is ##5.6##, which is just one centiliter above the lower-bound volume for there being an extra cost, resulting in a mean of ##0.01##$.
Or maybe ##\int_{5.5}^{5.7}\left[0.01(x-5.5)\right]f(x)dx##, where ##f(x)## is the pdf?