binomial probability distribution

Thank you, misterT, MrAl.

misterT said that cumulative binomial distribution should be used for that problem but now I have checked the book again and there is no mention of cumulative binomial distribution (CBD) anywhere. All the problems in that exercise, except the one we are discussing, are solved using BD formula. Now I wonder if the book doesn't even discuss CBD then why it would include any problem for which CBD is to be used.

It would nice of you if you could help me with the query in this post too? It seems Steve is busy and John has missed the post.

Regards
PG

PG1995 said:

Thank you, misterT, MrAl.

misterT said that cumulative binomial distribution should be used for that problem but now I have checked the book again and there is no mention of cumulative binomial distribution (CBD) anywhere. All the problems in that exercise, except the one we are discussing, are solved using BD formula. Now I wonder if the book doesn't even discuss CBD then why it would include any problem for which CBD is to be used.

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I can just put my opinion in here. I do believe the CBD is useful, but the problem stated "do not use tables". Hence, I would just use the BD and calculate everything manually. It's not really that hard. I just did it out in about 5 minutes. Note that the probability that 4 or more meet the criterion equals one minus the probability that 3 or less meet it. Obviously, the latter is easier to calculate than the former, if you do your own manual summing using the BD.

There is no need to calculate the value of p because the problem only asks about consistency between claims. Hence, just assume p=0.4 and check to see if the numbers match up.

PG, don't make a problem harder than it is. Just provide the simple answers they ask. They tell you the assistants think p=0.4 (and so q=0.6). It doesn't matter how or why they think this. The newpaper determined it's own numbers and you can check to see if the BD with p=0.4 and n=15 will yield their numbers.

PG1995 said:

It would nice of you if you could help me with the query in this post too? It seems Steve is busy and John has missed the post.

Regards
PG

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I had looked at this when you first posted it, but I found the terminology confusing also. I'm not sure of the answer but probably business majors would laugh at us and say the definition is well-known.

steveB said:

I can just put my opinion in here. I do believe the CBD is useful, but the problem stated "do not use tables". Hence, I would just use the BD and calculate everything manually. It's not really that hard. I just did it out in about 5 minutes. Note that the probability that 4 or more meet the criterion equals one minus the probability that 3 or less meet it. Obviously, the latter is easier to calculate than the former, if you do your own manual summing using the BD.

There is no need to calculate the value of p because the problem only asks about consistency between claims. Hence, just assume p=0.4 and check to see if the numbers match up.

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Got it! I was looking at the problem from a totally different and wrong angle and this led to so many questions. Thank you. I wish you had put your opinion earlier!

Regards
PG

steveB said:

the problem stated "do not use tables".

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It stated that only in problem 5-23 not for problem 5-24. I still think that the problem was intended to be solved using tables. That is pretty common method in statistics. I have one full book just for tables..

Hi

Could you please help me with this query? Thank you.

Regards
PG

misterT said:

It stated that only in problem 5-23 not for problem 5-24. I still think that the problem was intended to be solved using tables. That is pretty common method in statistics. I have one full book just for tables..

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You may be right. It's not entirely clear to me which is preferable. Problem 5.24 is an extension of 5.23, but they don't explicitly restrict us in 5.24. However, it's a 5 minute problem either way. If the table is handy, we might as well use it, but otherwise we can sum up a few terms just as fast as we can find the table and look up the numbers.

Certainly, there is nothing wrong with using CBD tables, as they let us solve faster and we will be less prone to making errors, in general. I think PG should make sure he can solve the problem either way, and he should make sure he understands the relationship between the BD and CBD.

PG1995 said:

Hi

Could you please help me with this query? Thank you.

Regards
PG

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Expected loss must consider the probability and the probability is a crucial factor in making these types of business decisions. Imagine if insurance companies did not consider probability of occurrence when they determined your insurance premiums.

Of course, Mr Poppin will consider the probability when making his decision, but the expected loss is a relevant factor. It seems worth spending $200 to prevent an expected loss of $528, especially when he will be "disgraced". Also, 10% chance of losing $5000 is a scary thought.

What if the expected loss was only $100. Well, then many business owners might consider a 10% risk worth the expected savings. Others might consider the possible loss of $5000 unacceptable. However, Mr. Poppin seems to have a lot of pride, so maybe even then he would consider %10 risk too much of a gamble to risk being disgraced. However, even our prideful Mr. Poppin might reconsider in a case where the cost to hire is $200,000, the expected loss is $300,000 and the probability is only 0.1%.

Thank you, Steve.

steveB said:

Expected loss must consider the probability and the probability is a crucial factor in making these types of business decisions. Imagine if insurance companies did not consider probability of occurrence when they determined your insurance premiums.

Of course, Mr Poppin will consider the probability when making his decision, but the expected loss is a relevant factor. It seems worth spending $200 to prevent an expected loss of $528, especially when he will be "disgraced". Also, 10% chance of losing $5000 is a scary thought.

What if the expected loss was only $100. Well, then many business owners might consider a 10% risk work the expected savings. Others might consider the possible loss of $5000 unacceptable. However, Mr. Poppin seems to have a lot of pride, so maybe even then he would consider %10 risk too much of a gamble to risk being disgraced. However, even our prideful Mr. Poppin might reconsider in a case where the cost to hire is $200,000, the expected loss is $300,000 and the probability is only 0.1%.

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Just to let you know that I'm not against Mr. Poppin's hiring of temporary employees, especially when he could be disgraced and his very pride could be hurt!

But that 'expected loss' doesn't seem a 'real' thing; it's more like an indicator. Let me try to say it differently. If there are going to be more than 72,000 people then the loss will be $5000 and that's all. The loss won't be $528. The loss is going to be either $0 (when number of people doesn't exceed 72,000) or $5000. Here, the figure "$528" is not something real; it's just a product of monetary loss and probability of loss, the larger the value the more is it of concern. Suppose, what if the loss could be $20000 and probability of number of people exceeding 72000 is just 3%. Even then the product of loss and probability is 600. I would say "600" is a risk factor. I understand that my wording is not very clear but right now that's the best I could do. Thanks.

Regards
PG

PG1995 said:

Thank you, Steve.



Just to let you know that I'm not against Mr. Poppin's hiring of temporary employees, especially when he could be disgraced and his very pride could be hurt!

But that 'expected loss' doesn't seem a 'real' thing; it's more like an indicator. Let me try to say it differently. If there are going to be more than 72,000 people then the loss will be $5000 and that's all. The loss won't be $528. The loss is going to be either $0 (when number of people doesn't exceed 72,000) or $5000. Here, the figure "$528" is not something real; it's just a product of monetary loss and probability of loss, the larger the value the more is it of concern. Suppose, what if the loss could be $20000 and probability of number of people exceeding 72000 is just 3%. Even then the product of loss and probability is 600. I would say "600" is a risk factor. I understand that my wording is not very clear but right now that's the best I could do. Thanks.

Regards
PG

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I agree with what you are saying here, and I was trying to make similar points. Yes, the $528 is not something real, and it is an indicator that helps to make decisions.

In your counterexample, a $600 expected loss, with 3% risk and $20000 maximum cost is different than $600 expected loss with 20% risk and $3000 maximum cost. Two people might very well make different decisions in these two cases. Also, even in one case, one person might consider 3% risk too low to worry about and another person might consider $20000 too much to risk.

I guess I'm not understanding what your issue is. Expected loss is just a useful indicator and many people will make decisions considering the indicator as one aspect of the decision, and others will make decisions solely based on the indicator. What is your concern about that?

EDIT: Let me just expand on this a little more. What if you are a business man making these decisions very often? Then the expected value becomes a very "real" number for you. On average, you will see your expected losses due the the "law of large numbers". This is how the business of insurance is run. The expected values are born out in a large scale business like insurance, and, provided the insurance company determines expected values correctly, they will make money by making decisions in this way.