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

[Maybe_Memorie]

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 probabilty?

(ii) I have no idea here


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

 

[I like Serena]

There you are again!

(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?)

(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 probabilty?

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?

(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?

(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?

 

[Maybe_Memorie]

There you are again!



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) ?

 

[I like Serena]

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

Umin = 233g

Good!

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 probabilty?

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!

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)

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?

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.

 

[Maybe_Memorie]

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?

 

[I like Serena]

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

[itex]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[/itex]​

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!

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!

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.

 

[Maybe_Memorie]

You calculated

[itex]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[/itex]​

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

[itex]1 - \binom 5 0 (0.025)^0 (0.975)^5 - \binom 5 1 (0.025)^1 (0.975)^4[/itex]​

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?

 

[I like Serena]

Yes it is.
Just checking if you were still sharp!

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

 

[Maybe_Memorie]

Okay!

Thank you very much for all your help!

 

[I like Serena]

Appreciate the thanks!
And see you next time.