PMF of X for Airport Metal Detector Activations

[joemama69]

Homework Statement

Of the people passing through an airport metal detector, 0.5% activate it; Let X denote the number among a randomly selected group of 500 who activate it.

1) What is the PMF of X
i) Using th CLT (approximate PMF)
ii) Using the exact distribution of X

2) P(X = 5) using i and ii

3) P(X<=5) using i and ii

Homework Equations

The Attempt at a Solution

So its been a while since I've taking a probability class but I thought this was a Binomial Distribution problem where n = X, p = .005 and n = 500

P(X = x) = (500 Permutation X) * (.005^X) * (1-.005)^(500-X)

but the output I am getting for X = 1,2,3... = .20, .51, 1.29 which are clearly wrong.

My thought for X = 1 would be P(X=1) = (1/500)*.005 = .00001 which seams reasonable to me but I am confusing myself when I go on to P(X = 2)

Any hints would be appreciated.

 

[jz92wjaz]

500 Permutation X is not part of the Binomial Distribution.

Your first term is correct (The .20 one), but the second and third are wrong.

 

[Ray Vickson]

Homework Statement

Of the people passing through an airport metal detector, 0.5% activate it; Let X denote the number among a randomly selected group of 500 who activate it.

1) What is the PMF of X
i) Using th CLT (approximate PMF)
ii) Using the exact distribution of X

2) P(X = 5) using i and ii

3) P(X<=5) using i and ii

Homework Equations

The Attempt at a Solution

So its been a while since I've taking a probability class but I thought this was a Binomial Distribution problem where n = X, p = .005 and n = 500

P(X = x) = (500 Permutation X) * (.005^X) * (1-.005)^(500-X)

but the output I am getting for X = 1,2,3... = .20, .51, 1.29 which are clearly wrong.

My thought for X = 1 would be P(X=1) = (1/500)*.005 = .00001 which seams reasonable to me but I am confusing myself when I go on to P(X = 2)

Any hints would be appreciated.

You should also have a value for x = 0.

Anyway, why do you think that P(X=1) = (1/500)*.005 ? You already gave a formula for P(X=x); what does it give you when x = 1? (It WON'T be what you wrote!)

 

[joemama69]

Oops, should be Combination I am guessing?

Using that I get P(X = 1,2,3,...) = .20, .26, .21, .13, ...

These just don't seem correct to me. Seem way to high??

 

[Ray Vickson]

Oops, should be Combination I am guessing?

Using that I get P(X = 1,2,3,...) = .20, .26, .21, .13, ...

These just don't seem correct to me. Seem way to high??

They are OK, but you also need to look at P(X=0). After all, it is possible that all 500 fail to activate the device. And, of course, P(X = 0) is a definite part of the binomial distribution.

 

[joemama69]

Ok thank you. So I have the PMF function. The problem asks to compute probabilities using...

1) the CLT (approximation)
2) exact distribution

What is the difference. I thought the CLT allowed the use of binomial dist because of the large n. How do I "approximate PMF" using CLT.

 

[Ray Vickson]

Ok thank you. So I have the PMF function. The problem asks to compute probabilities using...

1) the CLT (approximation)
2) exact distribution

What is the difference. I thought the CLT allowed the use of binomial dist because of the large n. How do I "approximate PMF" using CLT.

Whoever is asking you to use the CLT result is asking you to perform a disastrously bad approximation in this case. Your problem does not at all fit the criteria for getting a reasonable approximation via the CLT. The CLT is a LIMIT theorem; the question is whether you can use the limit as a good approximation when you are not taking something to ∞, but merely to a large value (500 in this case). Under certain circumstances the answer is YES, but not in this case.

However, for this problem there is another type of limit result that can be used instead. Google 'limits of binomial distribution'.

Note added in edit: well, maybe 'disastrously bad' is too strong; just plain 'not very good' is a better description.