What Proportion of Pins Produced Will Be Unusable?

[semidevil]

I don't remember the exact story problem, but it goes something like this.

You are suppose to make a pin that is 6 cm long. Unfortunately, the machine will produce pins that 6 + y cm long. where 0 <= y <= 2, and the function f(y) = K(y + y^2).

if the pin is more then 7 cm long, then it is unusable. Find the proportion that is produced that will be unusable.

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ok, so if y = 0, or y = 1, then it is fine, and if y = 2, then it is unusable.

so can I just integrate from 0 to 1 of f(y), and then subtract from 1, to get the answer?

the book says the answer is 23/28...but I don't know how it got that.

 

[quantumdude]

ok, so if y = 0, or y = 1, then it is fine, and if y = 2, then it is unusable.

Not quite. It should read: If 0<=y<=1, then it is fine, and if 1<y<=2, then it is unusable.

so can I just integrate from 0 to 1 of f(y), and then subtract from 1, to get the answer?

Yes, but first, you have to find K. To get it you have to normalize the probability distribution by integrating f(y) over the sample space and setting it equal to 1. This will allow you to solve for K. Then you can do what you have proposed above.

 

[wais]

Yes, you can integrate from 0 to 1 of f(y) and subtract it from 1 to find the proportion that is usable. This is because f(y) represents the probability that the pin will be longer than 7 cm, and we want to find the proportion that is shorter than 7 cm.

To find the proportion that is unusable, we can simply subtract the proportion that is usable from 1. So the equation would be:

1 - ∫(0 to 1) f(y)dy = 1 - ∫(0 to 1) K(y + y^2)dy

= 1 - [Ky + (Ky^2)/2] evaluated at y = 0 and y = 1

= 1 - [K(1) + (K(1)^2)/2] + [K(0) + (K(0)^2)/2]

= 1 - [K + K/2] + 0

= 1 - 3K/2

We know that y can range from 0 to 2, so we need to find the value of K that would make the integral of f(y) from 0 to 2 equal to 1.

∫(0 to 2) K(y + y^2)dy = 1

= 2K + (4K)/3

= (10K)/3

So, (10K)/3 = 1

K = 3/10

Substituting this value of K into our equation for unusable proportion, we get:

1 - 3(3/10)/2

= 1 - 9/20

= 11/20

Therefore, the proportion of pins that will be unusable is 11/20 or 0.55. This is not the same as the book's answer of 23/28, so it's possible that there was a mistake in the book's solution or in the given information.