Post by: mano91 on August 31, 2017, 08:37:10 pm
A random sample of three items is slected from a batch of 10 items which contains four defectives.
I am stuck on part c) "construction a probability distribution table which summarises the sampling distribution of the sample proportion of defectives in the sample."
my probabilities are not lining up.
thank you in advance!
Post by: VanillaRice on August 31, 2017, 09:47:02 pm
I'm not quite sure what you mean by your probabilities 'lining up'. Would you be able to elaborate?
Your distribution table should have the probability values for all the values of p hat: 0, 1/3, 2/3, 1
Post by: mano91 on September 01, 2017, 04:27:13 pm
I am unable to achieve the corresponding probabilities for each of those "p hat" values.
the solutions are 0.1, 0.4, 0.4, 0.1 but I cant achieve these.
Thanks,
Emmanuel
Post by: VanillaRice on September 01, 2017, 04:51:53 pm
Hi Vanilla Rice,I've had a go at the question and my answer is not the same as the suggested one. I've checked the textbook, and it looks like you're looking at the answer to the wrong question :P (you've given the answers for question 3). Knowing this, are you now able to get the right answer? If not, please see below for a partial worked solution :)
I am unable to achieve the corresponding probabilities for each of those "p hat" values.
the solutions are 0.1, 0.4, 0.4, 0.1 but I cant achieve these.
Thanks,
Emmanuel
Spoiler
Since the sample size is small, we cannot use the binomial distribution, and must use combinations. This is because the probability changes each time you select an item from the population (of 10).
I'll show you how to do the first one (p hat=0), and I'll let you have a go at the rest ;)
}=\frac{\binom{4}{0}\binom{6}{3}}{\binom{10}{3}}=\frac{1}{6})
where we want 0 out of the 4 defectives and 3 out of the 6 non-defectives (to make up a total of 3 items sampled). This is for a sample of 3 out of a population of 10.
I'll show you how to do the first one (p hat=0), and I'll let you have a go at the rest ;)
where we want 0 out of the 4 defectives and 3 out of the 6 non-defectives (to make up a total of 3 items sampled). This is for a sample of 3 out of a population of 10.
Post by: mano91 on September 01, 2017, 05:47:27 pm
ok, I understand that.
What is the rule that governs when to use counting method or binomial theorem?
Post by: VanillaRice on September 01, 2017, 07:22:25 pm
What is the rule that governs when to use counting method or binomial theorem?Recall the requirements for a binomial random variable - more specifically, the probability, p, must remain constant after every selection/trial/event.
In the question above, we only have 10 items to select one. Let's say we take one (without replacing it) - the probability of it being defective is 4/10. What about the next one being defective? 3/9 (since there is one less in total, and one less defective). The probability is clearly changing after every trial. So, we cannot use the binomial distribution.
Now consider, let's say the question was selecting from a population of 100,000 items, of which 40% (40,000) are defective. What's the probability we select a defective item? 40,000/100,000. What about another one (without replacement)? 39,999/99,999. If we compare these probabilities, they are approximately equal (convert to decimals to check for yourself). Here, we are selecting from a 'large' population. The probability does change, but by a relatively insignificant amount. Therefore, we consider p to be (relatively) constant, which means we can use the binomial distribution.