Statistics Problem: Thickness of gears and spacers

[LakeMountD]

A five-gear assembly is put together with spacers between the gears. The mean thicnkess of the gears is 5.03 cm with a standard ev. of .008. the mean thickness of the spacers is .140 cm with a standard dev. of .005. Find the mean and standard deviation of the thickness of the assembled units consisting of five randomly selected gears and four randomly selected spacers.

I am kind of confused on where to even start since I don't truly understand what the problem is asking.

 

[HallsofIvy]

Visualize the assembly! You have 5 gears with 4 spacers between them.
If you assume each gear is exactly 5.03 cm thick and each spacer is exactly 0.14 cm thick how thick would the whole assembly be?
Do you see how that's connected to the mean value of the thickness of the assembly?

To be more technical, if the probability density for the thickness of the gears is f(x) and the probability density for the thickness of the spacers is g(x) then the mean thickness of a gear is [itex]\mu_1=\int xf(x)dx[/itex] and the mean thickness of a spacer is [itex]\mu_2=\int xg(x)dx[/itex].
The mean thickness of a gear and a spacer together is [itex]\int x(f(x)+g(x))dx= \mu_1+ \mu_2[/itex].
The standard deviation of the thickness of the assembly is a little harder. Remember that standard deviation is defined in terms of "square root of a sum of squares".

 

[Architeuthis Dux]

As a scientist, it is important to fully understand the problem before attempting to solve it. In this case, the problem is asking for the mean and standard deviation of the thickness of the assembled units consisting of five randomly selected gears and four randomly selected spacers. This means that we are interested in the thickness of the entire assembly, not just individual gears or spacers.

To find the mean and standard deviation of the assembled units, we can use the following formula:

Mean = (n1 * mean1 + n2 * mean2) / (n1 + n2)

Where n1 and n2 represent the number of gears and spacers respectively, and mean1 and mean2 represent the mean thickness of gears and spacers. In this case, n1 = 5, n2 = 4, mean1 = 5.03 cm, and mean2 = 0.140 cm. Plugging these values into the formula, we get:

Mean = (5 * 5.03 + 4 * 0.140) / (5 + 4) = 4.99 cm

To find the standard deviation, we can use the following formula:

Standard Deviation = √(n1 * (standard dev1)^2 + n2 * (standard dev2)^2) / (n1 + n2)

Using the values given in the problem, we get:

Standard Deviation = √(5 * (0.008)^2 + 4 * (0.005)^2) / (5 + 4) = 0.007 cm

Therefore, the mean thickness of the assembled units is 4.99 cm with a standard deviation of 0.007 cm. This means that on average, the assembled units will have a thickness of 4.99 cm and most of the units will fall within a range of 4.983 cm to 5.003 cm.

In conclusion, it is important to carefully read and understand the problem before attempting to solve it. By using the appropriate formulas and given values, we were able to find the mean and standard deviation of the thickness of the assembled units consisting of five randomly selected gears and four randomly selected spacers.