Solving Normal Distribution Homework: Qn, TNE, u, SD, NS

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Homework Help Overview

The discussion revolves around a problem involving normal distribution in the context of quality control for a company's packaging process. The company specifies that the average weight of packages should not be less than a nominal quantity (Qn) and that a certain percentage of packages can deviate from this quantity. Participants are exploring the implications of standard deviation, probabilities related to the weights of packages, and the conditions under which packages are considered non-standard.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning, Problem interpretation, Assumption checking

Approaches and Questions Raised

  • Participants discuss the calculation of standard deviation needed to meet the non-standard package criteria and explore the implications of different values for the mean fill weight. Questions arise about the correct application of z-scores and the binomial distribution for calculating probabilities related to the number of non-standard packages.

Discussion Status

There is an ongoing exploration of various methods to calculate probabilities related to the weights of packages. Some participants have provided guidance on using binomial distribution concepts, while others are questioning the assumptions and calculations made by their peers. Multiple interpretations of the problem are being considered, and there is no explicit consensus yet.

Contextual Notes

Participants are working under the constraints of specific nominal quantities and tolerances for weight deviations. The discussion includes assumptions about the independence of the filling process and the normal distribution of weights, which are critical to the calculations being discussed.

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Homework Statement



A company specifies that the actual contents of packages shall not be less, on average, than the Nominal Quantity (Qn) and that not more than 2.5% of packages may be Non-Standard (NS), i.e. deviate from Qn by more than "total negative error" (TNE). Thus, if Qn=230g and TNE=9g a package is NS if it weighs less than 221g.
Assume a normal distribution and containers are filled independently.

(a) Let Qn=230g and TNE=9g and suppose the company sets the average fill to be u=230g.

(i) What must the standard deviation of the filling operation be so that only
2.5% of containers are NS?

(ii) Suppose the actual standard deviation is 6g, what minimum value should u be so that
the NS regulation is met? Round to whole units.


(b) Suppose u is the value found in (a)(ii) and tha five containers are selected at random
and weighed.

(i) What is the probability that two or more of them are NS?

(ii) What is the probability that the combined weight of the five congtainers is less than
1200g?


(c) What is the probability that an inspector will need to weigh exactly 6 containers
from the production line to discover a NS container. That is, the first five are standard
and the sixth is NS.


Homework Equations





The Attempt at a Solution



I don't have my actual workings with me, but here are my methods.


(a) (i) 2.5% is NS, so I found the z-value, Z, which gives 0.025.
Z = (221-230)/SD

From this I found the standard deviation.

(ii) Z = (221 - Umin)/6
Solved for Umin


(b) (i) P(2 or more are NS) = 1 - P(1 or 0 is not NS)
How do I calculate this probability?

(ii) I have no idea here


(c) Do I have to do a binomial distribution?
 
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There you are again! :smile:

Maybe_Memorie said:
(a) (i) 2.5% is NS, so I found the z-value, Z, which gives 0.025.
Z = (221-230)/SD

From this I found the standard deviation.

Your method is good, your z-value is wrong.
(A typo I presume?)


Maybe_Memorie said:
(b) Suppose u is the value found in (a)(ii) and tha five containers are selected at random
and weighed.

(i) What is the probability that two or more of them are NS?

(b) (i) P(2 or more are NS) = 1 - P(1 or 0 is not NS)
How do I calculate this probability?

What is the chance that exactly 0 containers are NS?
What is the change that only the 1st container is NS?
What is the change that only the 2nd container is NS?

Can these chances be added up?


Alternatively, what do you know about the binomial distribution?
And in particular the cumulative binomial distribution?
Can you think of parameters that would be applicable?


Maybe_Memorie said:
(ii) What is the probability that the combined weight of the five congtainers is less than
1200g?

(ii) I have no idea here

If you have 2 normal distributions and you add them up.
What is the resulting mean?
And what is the resulting variance?

What if you add 5 normal distributions?


Maybe_Memorie said:
(c) What is the probability that an inspector will need to weigh exactly 6 containers
from the production line to discover a NS container. That is, the first five are standard
and the sixth is NS.

(c) Do I have to do a binomial distribution?

No, not in this case.

This would be:
P(first ok AND second ok AND third ok AND fourth ok AND fifth ok AND sixth is NS)

Do you know how to calculate this?
 
I like Serena said:
There you are again! :smile:



Your method is good, your z-value is wrong.
(A typo I presume?)

Sorry, that was a typo. Should have been z = -1.96
Standard deviation = 4.5918

Umin = 233g

What is the chance that exactly 0 containers are NS?
What is the change that only the 1st container is NS?
What is the change that only the 2nd container is NS?

Can these chances be added up?


Alternatively, what do you know about the binomial distribution?
And in particular the cumulative binomial distribution?
Can you think of parameters that would be applicable?

P(chosen bottle is NS) = 0.025

So the desired probability is
1 - 5C0.(0.025)^0.(0.975)^5 - 5C1.(0.025)^1.(0.975)^4 - 5C2.(0.025)^2.(0.975)^3
= 0.00015045

If you have 2 normal distributions and you add them up.
What is the resulting mean?
And what is the resulting variance?

What if you add 5 normal distributions?

I'm not really sure here..

No, not in this case.

This would be:
P(first ok AND second ok AND third ok AND fourth ok AND fifth ok AND sixth is NS)

Do you know how to calculate this?

Is it (0.975)^4.(0.025) ?
 
Maybe_Memorie said:
Sorry, that was a typo. Should have been z = -1.96
Standard deviation = 4.5918

Umin = 233g

Good! :smile:


Maybe_Memorie said:
P(chosen bottle is NS) = 0.025

So the desired probability is
1 - 5C0.(0.025)^0.(0.975)^5 - 5C1.(0.025)^1.(0.975)^4 - 5C2.(0.025)^2.(0.975)^3
= 0.00015045

(b) (i) P(2 or more are NS) = 1 - P(1 or 0 is not NS)
How do I calculate this probability?

You're in the neighbourhood.
However, you switched your chances around, since the chance on not NS is 0.975.
And you should leave out the last term, since that is for 2 bottles! :wink:

Btw, you're effectively looking at a binominal distribution here, with parameters n=5 and p=0.975.
So you could also calculate it with the cumulative binomial distribution.
That is: P(2 or more are NS) = 1 - P(1 or 0 is not NS) = 1 - binomcdf(n=5, p=0.975, k=1)


Maybe_Memorie said:
If you have 2 normal distributions and you add them up.
What is the resulting mean?
And what is the resulting variance?

What if you add 5 normal distributions?

I'm not really sure here..

If you add up 2 independent normal distributions, you'll get a new normal distribution, with:
1. a mean that is the sum of the two means
2. a variance, that is the sum of the two variances
(See wikipedia: http://en.wikipedia.org/wiki/Sum_of...andom_variables#Independent_random_variables").

Can you continue this?


Maybe_Memorie said:
This would be:
P(first ok AND second ok AND third ok AND fourth ok AND fifth ok AND sixth is NS)

Do you know how to calculate this?

Is it (0.975)^4.(0.025) ?

Almost, your reasoning is sound, but I think you miscounted.
The first power should be 5, since their are 5 bottles ok.
 
Last edited by a moderator:
I like Serena said:
You're in the neighbourhood.
However, you switched your chances around, since the chance on not NS is 0.975.

When I did it I was putting getting a NS as a success and a standard as a failure. I understand your way, but not why mine is wrong.

If you add up 2 independent normal distributions, you'll get a new normal distribution, with:
1. a mean that is the sum of the two means
2. a variance, that is the sum of the two variances
(See wikipedia: http://en.wikipedia.org/wiki/Sum_of...andom_variables#Independent_random_variables").

Can you continue this?

For 5 items the mean would be the sum of the 5 means.
The variance would be the sum of the 5 variances.

Each item has U=233, SD=6, Variance=36
So for the sum, U=1165, SD=13.4164

Normalize 1200, and get the corresponding probability.
Is this right?

Almost, your reasoning is sound, but I think you miscounted.
The first power should be 5, since their are 5 bottles ok.
[/QUOTE]

Sorry, that was a type. (0.975)^5.(0.025)
Do I have to do anything else to this? Isn't this the same as first 4 are ok, 5th is NS, 6th is ok?
 
Last edited by a moderator:
Maybe_Memorie said:
When I did it I was putting getting a NS as a success and a standard as a failure. I understand your way, but not why mine is wrong.

P(chosen bottle is NS) = 0.025

So the desired probability is
1 - 5C0.(0.025)^0.(0.975)^5 - 5C1.(0.025)^1.(0.975)^4 - 5C2.(0.025)^2.(0.975)^3
= 0.00015045

You calculated
1 - \binom 5 0 (0.025)^0 (0.975)^5 - \binom 5 1 (0.025)^1 (0.975)^4 - \binom 5 2 (0.025)^2 (0.975)^3​

which is:
1 - P(0 bottles NS) - P(1 bottle NS) - P(2 bottles NS) = P(3, 4 or 5 bottles NS)​

but that was not the question!

Maybe_Memorie said:
For 5 items the mean would be the sum of the 5 means.
The variance would be the sum of the 5 variances.

Each item has U=233, SD=6, Variance=36
So for the sum, U=1165, SD=13.4164

Normalize 1200, and get the corresponding probability.
Is this right?

Yep! :smile:


Maybe_Memorie said:
Sorry, that was a type. (0.975)^5.(0.025)
Do I have to do anything else to this? Isn't this the same as first 4 are ok, 5th is NS, 6th is ok?

That is correct!
And yes it is the same.
 
I like Serena said:
You calculated
1 - \binom 5 0 (0.025)^0 (0.975)^5 - \binom 5 1 (0.025)^1 (0.975)^4 - \binom 5 2 (0.025)^2 (0.975)^3​

which is:
1 - P(0 bottles NS) - P(1 bottle NS) - P(2 bottles NS) = P(3, 4 or 5 bottles NS)​

but that was not the question!

I had a typo, I should have written
1 - \binom 5 0 (0.025)^0 (0.975)^5 - \binom 5 1 (0.025)^1 (0.975)^4​

This gives
1 - P(0 bottles NS) - P(1 bottle NS) = P(2, 3, 4 or 5 bottles NS)​

The question asked "what is the probability at least two of them are NS, so isn't this correct?
 
Yes it is.
Just checking if you were still sharp! :smile:

(It seems I switched the chances around myself. :shy:)
 
Okay! :smile:

Thank you very much for all your help! :smile:
 
  • #10
Appreciate the thanks!
And see you next time. :cool:
 

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