Determine the probability that a random sample of 40 tins has the given combined mass

hndalama

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Sep 26, 2016
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The contents of a consignment of 1200 tins of a product have a mean mass of 0.504 kg and a standard deviation of 92 g. Determine the probability that a random sample of 40 tins drawn from the consignment will have a combined mass of
(a) less than 20.13 kg,
(b) between 20.13 kg and 20.17 kg,
(c) more than 20.17 kg, correct to three
significant figures.

The attached picture shows my attempt and answer for (a).I found a probability of 0.4801 but the book says the probability is 0.0179. How do they get this answer and what is wrong with what i did?
 

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tkhunny

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Well,

1) The Finite Population Correction Factor doesn't do much for you. Either way.
2) Both z-scores are very small in magnitude, -0.05242 and 0.01747 (With the FPCF). Of necessity, this requires their tail probabilities to be rather near 1/2. Your value is far more credible. I get 0.4791 for the left tail, but that's hardly a dispute with your result.
3) The upper z-score is in the neighborhood of the "book's answer". Maybe that's what it means? Hard to say without additional information.
 

hndalama

Junior Member
Joined
Sep 26, 2016
Messages
74
Well,

1) The Finite Population Correction Factor doesn't do much for you. Either way.
2) Both z-scores are very small in magnitude, -0.05242 and 0.01747 (With the FPCF). Of necessity, this requires their tail probabilities to be rather near 1/2. Your value is far more credible. I get 0.4791 for the left tail, but that's hardly a dispute with your result.
3) The upper z-score is in the neighborhood of the "book's answer". Maybe that's what it means? Hard to say without additional information.
Perhaps I should say that I am new to this subject and I am learning by myself from a textbook. I wasn't familiar with the term 'finite population correction factor,' but from what I have found online i think it is the expression sqrt[(NP-N)/(NP-1)]. Please correct me if I'm wrong. The book doesn't use the term but it does explain that this expression is included in the equation to find the 'standard error of the means' when the sample size is 'large'.And then that, the expression is excluded when the population size is very large compared to sample size, infinite or when samples are taken with replacement.
so with that said, -0.5242 is the z score I get with FPCF, and without FPCF I get:
92/ sqrt(40) =14.546
so
z= (503.25 -504)/14.546 = -0.05156

How do you get the z-score value of 0.01747?
 
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