Math Probability: Need to find the Variance

[CryptoMath]

Homework Statement

You buy 15 boxes each containing 4 microprocessors. All microprocessors come from a very large lot in which 9.5 % of microprocessors are defective. Y = the number of boxes in which there is less than 2 defective microprocessors.

QUESTION:
Find the Variance of the variable Y

Homework Equations

The Attempt at a Solution

I am thinking this is Binomial.

We are looking of a number of boxes that has less than 2 defective, which means (0 or 1) is acceptable.

We know there is 15 boxes with 4 microprocessors each.
So 15*4 = 60 total microprocessors
9.5% of microprocessors are defective, so 60*9.5%=5,7

n=15
p=??

Am I thinking this right ? Pretty much stuck here.

Thanks

 

[Ray Vickson]

Homework Statement

You buy 15 boxes each containing 4 microprocessors. All microprocessors come from a very large lot in which 9.5 % of microprocessors are defective. Y = the number of boxes in which there is less than 2 defective microprocessors.

QUESTION:
Find the Variance of the variable Y

Homework Equations

The Attempt at a Solution

I am thinking this is Binomial.

We are looking of a number of boxes that has less than 2 defective, which means (0 or 1) is acceptable.

We know there is 15 boxes with 4 microprocessors each.
So 15*4 = 60 total microprocessors
9.5% of microprocessors are defective, so 60*9.5%=5,7

n=15
p=??

Am I thinking this right ? Pretty much stuck here.

Thanks

What is the probability (call it $p$) that a (randomly-chosen) box has fewer than 2 defectives? That computation is elementary.

Once you have figured out the value of $p$, you can compute the probability distribution of Y = number of acceptable boxes out of 15 (where an acceptable box has < 2 defectives). That computation ought to be familiar.

 

[andrewkirk]

Crypto, this problem has the same general form as your earlier one about Binomial+Poisson. In that one you had to first work out the probability of a page having no errors, and then use that as a binomial parameter. This is the same except that instead of working out the prob of a page having no errors, you are working out the prob of a box containing fewer than two defective units. This doesn't really add anything to what Ray said, but I thought it might help to point out the similarity to the other problem, so you can get the hang of breaking down these puzzles.

 

[CryptoMath]

What is the probability (call it $p$) that a (randomly-chosen) box has fewer than 2 defectives? That computation is elementary.

Once you have figured out the value of $p$, you can compute the probability distribution of Y = number of acceptable boxes out of 15 (where an acceptable box has < 2 defectives). That computation ought to be familiar.

Crypto, this problem has the same general form as your earlier one about Binomial+Poisson. In that one you had to first work out the probability of a page having no errors, and then use that as a binomial parameter. This is the same except that instead of working out the prob of a page having no errors, you are working out the prob of a box containing fewer than two defective units. This doesn't really add anything to what Ray said, but I thought it might help to point out the similarity to the other problem, so you can get the hang of breaking down these puzzles.

I did that:

Binomcdf(15,0.095,0,1)
=0.576018

Variance formula for Binomial
= npq
with n = 4 and p = 0.576018 and q = (1-0.576018)

= 0.976885

Does that make sense ?
Thank you for your help and time.

 

[Ray Vickson]

I did that:

Binomcdf(15,0.095,0,1)
=0.576018

Variance formula for Binomial
= npq
with n = 4 and p = 0.576018 and q = (1-0.576018)

= 0.976885

Does that make sense ?
Thank you for your help and time.

You are not buying 4 boxes; you are buying 15 boxes.

 

[andrewkirk]

There are two steps:
(1) work out probability that a given box will have less than two defective units
(2) then work out the variance of the number of boxes, out of the 15, that have less than two defective units.
Which of the two steps is the following?

I did that:

Binomcdf(15,0.095,0,1)
=0.576018

If it is step 1 then it should not be using 15, as there are four units per box.
If it is step 2 then it should not be using 0.095, as that is not the probability of a box having less than two defective units.

If you label your steps and state what you are doing, you will be less likely to get confused between the two steps.

 

[CryptoMath]

There are two steps:
(1) work out probability that a given box will have less than two defective units
(2) then work out the variance of the number of boxes, out of the 15, that have less than two defective units.
Which of the two steps is the following?

If it is step 1 then it should not be using 15, as there are four units per box.
If it is step 2 then it should not be using 0.095, as that is not the probability of a box having less than two defective units.

If you label your steps and state what you are doing, you will be less likely to get confused between the two steps.

Okay;

Step 1:
15*4=60
Binomcdf(60,0.095,0,1)
=0.018287

Step2:
Variance formula for Binomial
= npq
with n = 15 and p = 0.018287 and q = (1-0.018287)

= 0.269287

 

[andrewkirk]

Step 1:
15*4=60
Binomcdf(60,0.095,0,1)
=0.018287

Why is the number 15 in there? For step 1, given a randomly selected box containing four units, where each unit has a 9.5% chance of being defective, and defectiveness of any two units is independent of each other, we are trying to work out the probability that less than two of the units are defective. The number 15 is not relevant to the calculation.
That's why I think it's so important to get into the habit of writing the explanatory words around the formulas, rather than just writing the formulas cold as you are doing. If you had prefaced your formula with an explanatory sentence, you would have noticed that the number 15 does not occur in the sentence, and hence probably should not be in the formula either.

 

[CryptoMath]

Why is the number 15 in there? For step 1, given a randomly selected box containing four units, where each unit has a 9.5% chance of being defective, and defectiveness of any two units is independent of each other, we are trying to work out the probability that less than two of the units are defective. The number 15 is not relevant to the calculation.
That's why I think it's so important to get into the habit of writing the explanatory words around the formulas, rather than just writing the formulas cold as you are doing. If you had prefaced your formula with an explanatory sentence, you would have noticed that the number 15 does not occur in the sentence, and hence probably should not be in the formula either.

I just got the correct answer:
I had 60 because I was thinking we needed to total amount of microprocessor

Step 1:

Binomcdf(4,0.095,0,1)
=0.9524

Step2:
Variance formula for Binomial
= npq
with n = 15 and p = 0.9524 and q = (1-0.9524)
=0.679136

 

[andrewkirk]

Well done