Type of Distribution

There is a package containing 20 nuts and 20 bolts. A sample of eight is taken from the package at random. Let X be the number of nuts in the sample.

What is the name of the distribution for X?

I'm having trouble identifying distributions, can someone tell me the name of the distribution for X and why?

Please and thank you!

bry32321 said:

There is a package containing 20 nuts and 20 bolts. A sample of eight is taken from the package at random. Let X be the number of nuts in the sample.

What is the name of the distribution for X?

I'm having trouble identifying distributions, can someone tell me the name of the distribution for X and why?

Click to expand...

What types of distributions have you been taught so far? What is your understanding of their definitions?

Thank you!

bry32321 said:

There is a package containing 20 nuts and 20 bolts. A sample of eight is taken from the package at random. Let X be the number of nuts in the sample.

What is the name of the distribution for X?

I'm having trouble identifying distributions, can someone tell me the name of the distribution for X and why?

Please and thank you!

Click to expand...

I don't know a NAME for this distribution - all I know is that it will involve Combinations, because we don't care what order the nuts and bolts are selected in.

How many ways can you select X nuts out of 20? How many ways can you select (8 - X) bolts out of 20? Since these two selections are independent of each other, the total is the product of the two numbers.

If you want probability, divide that product by the number of ways you can select 8 from a total of 40. Words that describe the distribution are symmetric, discrete, mean = mode = median = 4. By my calculation, P(4) = 0.305.

You are saying there are two possibilities: an item from the package is a "nut" or it is a "bolt". The probability any one item is 1/2. Looks like a "binomial distribution" to me.

I started to say this was a binomial distribution. That would be correct if there were a huge number of nuts and bolts. Since there are only a limited number of nuts and bolts (20 of each) the probability changes- the probability of one nut in one selection is 20/40= 1/2, the probability of two nuts in two selections is (20/40)(19/39).= 19/78.

HallsofIvy said:

I started to say this was a binomial distribution. That would be correct if there were a huge number of nuts and bolts. Since there are only a limited number of nuts and bolts (20 of each) the probability changes- the probability of one nut in one selection is 20/40= 1/2, the probability of two nuts in two selections is (20/40)(19/39).= 19/78.

Click to expand...

Right. Perhaps part of the original question stems from how to tell if the distribution is binomial.

The three essential criteria for a Binomial Distribution are:
1) it has to be a "yes/no" question (i.e., binary) - TRUE, each selection is either a nut or not a nut
2) there has to be a fixed number of trials - TRUE, n=8
3) there has to be a universal probability of a "yes" event - FALSE, probability changes with each selection.

The Binomial Distribution is not appropriate when the total population cannot be assumed to be infinite

Just for fun, suppose we try the binomial formula with p=0.5, and see what we would get for P(4)

\(\displaystyle \displaystyle P(4\ |\ \text{binomial}) = (0.5)^4\ (0.5)^4\ \binom 8 4 = 0.273\),

about 10% below the true value, P(4) = 0.305. The binomial distribution is too broad and too short for this problem.

bry32321 said:

There is a package containing 20 nuts and 20 bolts. A sample of eight is taken from the package at random. Let X be the number of nuts in the sample.

What is the name of the distribution for X?

I'm having trouble identifying distributions, can someone tell me the name of the distribution for X and why?

Please and thank you!

Click to expand...

DrPhil said:

I don't know a NAME for this distribution - all I know is that it will involve Combinations, because we don't care what order the nuts and bolts are selected in.

How many ways can you select X nuts out of 20? How many ways can you select (8 - X) bolts out of 20? Since these two selections are independent of each other, the total is the product of the two numbers.

If you want probability, divide that product by the number of ways you can select 8 from a total of 40. Words that describe the distribution are symmetric, discrete, mean = mode = median = 4. By my calculation, P(4) = 0.305.

Click to expand...

FOUND THE NAME OF THIS DISTRIBUTION!! It is a HYPERGEOMETRIC DISTRIBUTION.

Having used "brute force" to come up with the product of two Combinations divided by a third Combination, I compared to formulas in the "Handbook of Probability and Statistics" to see if any matched. And found this:

\(\displaystyle \displaystyle f(x) = \dfrac{\binom k x \binom {N-k} {n-x}}{\binom N n} \),...\(\displaystyle x = 0,1,2,. . ., min(n,k)\)

where N = total number of items in a finite population = 40,
.........n = number of items drawn without replacement = 8
.........k = number of successes (nuts) in finite population = 20
.........N - k = number of failures (not nuts) = 20
.........x = number of successes (nuts) in sample = 0, 1, 2, . . . , 8 (smaller of n or k)
.........n - x = number of failures (not nuts) in sample

Thank you so very much!!!!! I was starting to think it might be hypergeometric.