Grueling Statistics problem. PLEASE HELP. I dont know whats wrong

[StimpackAddict]

I am in no way shape or form asking someone to do my homework, I have already attempted the problem numerous times.
Here is the problem:

2% of hair dryers produced in a certain plant are defective. Estimate the probablility that of 10,000 randomly selected hair dryers, exactly 225 are defective.

Answers: A) 0.0034 B) 0.0057 C) 0.0065 D) 0.0051

Now in order to solve this, You need to find out what the Mean (MU) is, and what the Variance is (sigma)

Q (chance of defective/failure) is = to .02,

because 1-P = Q, and P = .98 (given)

Mean (Mu) = NxP = (225)(.98)= 220.5

Variance (sigma) = Sq rt. of( (n)(p)(q) ) = sq rt of( (10,000) (.98) (.02) ) = 14

Therfore, p=.98 q=.02 Mu=220.5 and Variance=14

Now you must find the Z scores for the sample size of 225, and because it asks for EXACTLY 225 that are defective, you must take .5 from the number for each Z equation.

Therfore:

Z(1) = Avg Mean - Mu / Variance = 224.5 - 220.5 / 14 = .29 (rounded)

Z(2) = Avg Mean - Mu / Variance = 225.5 - 220.5 / 14 = .36 (rounded)

On my Z Score chart, Z(1)'s score of .29 is = to .6141

and Z(2)'s score of .36 is = to .6406

because .29 < Z < .36 ,

you do this operation: .6406 - .6141

but the problem is, I get the answer .0265, which is no where close to the correct answer, which is .0057

I have triple checked my math, and tried it without rounding even, and I still can't get the answer. What the hell am I doing wrong? I even paralleled this problem to an example problem the teacher did in class.

Any input is appreciated. Thanks.

[Silent A]

[ QUOTE ]
Mean (Mu) = NxP = (225)(.98)= 220.5

[/ QUOTE ]

I haven't had to look at the normal distribution approximation of the binomial theorem in a long time, but are you sure the above is correct? I'm pretty sure it should be:

Mu = 10000*0.02 = 200

[T50_Omaha8]

Sorry but I don't feel like reading your answer.

mean = 200
variance = .02*.98*10000
standard deviation = sqrt(.02*.98*10000)

P(X=225) = phi((225.5-200)/sqrt(.02*.98*10000)) - phi((224.5-200)/sqrt(.02*.98*10000))
where phi(x) is the cumulative distribution function of the standard normal vairable.

Or you can go poisson and the answer is

e^(-200)*200^225/225!

I assume the teacher wants the central limit theorem answer, though (ie the first one i gave...its called the something laplace theorem also). It's kinda weird to approximate the probability it's exactly a certain number.

[Silent A]

[ QUOTE ]
Or you can go poisson and the answer is

e^(-200)*200^225/225!

[/ QUOTE ]

Do you know of an easy, reliable way to calculate this monstrosity?

Edited to add: using Excel I got 0.05925, and using my value for Mu and the normal distribution, I got 0.005788.

[StimpackAddict]

never mind, i figured it out, the problem was that I looked at the 1.81 z score instead of the 1.82 z score, and i had p and Q backwards. Thanks anyways.

[T50_Omaha8]

[ QUOTE ]
Do you know of an easy, reliable way to calculate this monstrosity?

[/ QUOTE ] No. I can hardly use a calculator.

Poisson isn't really meant for means that large (or actually samples that large with success rates that low). It IS, however, meant solely for discrete probability calculations.

[ajizzle]

ok. so i may be way off, but i don't think you need to use a z-score to figure this problem out, since you aren't looking for the probability that something falls in a range, but instead the probability that you get an exact outcome.

so, this is how i would do it.

there are 225 defective, so there are 9775 good ones, with a probability that .02 are defective, .98 are effective.

so its a simple binomial expansion:

(10000 C 225)((.02)^225)((.98)^9775).

i hope i didnt get this wrong. i just graduated with a degree in math, and that would be just plain embarrassing.

[Silent A]

[ QUOTE ]
so its a simple binomial expansion:

(10000 C 225)((.02)^225)((.98)^9775).

i hope i didnt get this wrong. i just graduated with a degree in math, and that would be just plain embarrassing.

[/ QUOTE ]

I'd expect a math guy to realize that (10000 C 225) is a very difficult expression to evaluate. Which is why you're better off going with the normal approximation.

[BruceZ]

[ QUOTE ]
[ QUOTE ]
Or you can go poisson and the answer is

e^(-200)*200^225/225!

[/ QUOTE ]

Do you know of an easy, reliable way to calculate this monstrosity?

Edited to add: using Excel I got 0.05925, and using my value for Mu and the normal distribution, I got 0.005788.

[/ QUOTE ]

I also got 0.005788 from the normal distribution, but from Excel I got 0.005805 by computing C(10000,225)*(0.02^225)*(0.98^9775) as follows:

A1 = 10000, B1 = 225, C1 = 0.02*A1/B1
A2 = 9999, B2 = 224, C2 = C1*0.02*A2/B2
A3 = 9998, B3 = 223, C3 = C2*0.02*A3/B3
...
A225 = 9776, B225 = 1, C225 = C224*0.02*A225/B225

answer = (0.98^9775)*C225

[Silent A]

[ QUOTE ]
[ QUOTE ]
[ QUOTE ]
Or you can go poisson and the answer is

e^(-200)*200^225/225!

[/ QUOTE ]

Do you know of an easy, reliable way to calculate this monstrosity?

Edited to add: using Excel I got 0.05925, and using my value for Mu and the normal distribution, I got 0.005788.

[/ QUOTE ]

I also got 0.005788 from the normal distribution, but from Excel I got 0.005805 by computing C(10000,225)*(0.02^225)*(0.98^9775) as follows:

A1 = 10000, B1 = 225, C1 = 0.02*A1/B1
A2 = 9999, B2 = 224, C2 = C1*0.02*A2/B2
A3 = 9998, B3 = 223, C3 = C2*0.02*A3/B3
...
A225 = 9776, B225 = 1, C225 = C224*0.02*A225/B225

answer = (0.98^9775)*C225

[/ QUOTE ]

I tried doing the same calcs by adding logarithms and got the same answer (0.005805).

And there was a typo in my reported Poisson answer, it should have been (perhaps obviously) 0.005925.

I'm a little surprised the three answers aren't closer.