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#1
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I want to start this post by apologizing for posting this here as opposed to the math forum, but it is for the reason that the math forum is dead, and replies would take too long, and I need verification within a few minutes. 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. |
#2
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So i only skimmed your work and I havent slept in a long time but why did you do this...
Mean (Mu) = NxP = (225)(.98)= 220.5 shouldnt the mean of the distribution be .02*10k, which would give the avg number of deffective dryers, then if you run your Z test you can find how many stdevs 225 is from that mean and find the probability of that off a chart? |
#3
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OP: it was obviously pirates
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#4
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pookvis, I just realized this also when I read an actual response in the math forum. The problem is that I labeled both the 225 defective sample size, and the 10000 total BOTH as N, and then multiplied the wrong one.. is this my mistake?
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#5
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yes, N=10k.
Also i think you have your P and Q mixed up, it doesnt really matter what you call it but you want the mean number defective if you are going to use 225. You could also use 10k-225 and start talking about it the way you have it set up its just less intuitive (also your N is still wrong no matter what). |
#6
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I have made the changes... it made it worse pookvis, because now the 2 Z scores are WAYYY off the chart. this cant be right. i cant even get the values to subtract them
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#7
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im gonna PM you the rest of this to spare OOT
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#8
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I'm not sure where you're getting the variance based on the information given here. I think you've got the right idea about everything except for finding sigma.
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#9
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in before lock
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#10
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I have a more elegant solution, and a slightly different answer.
The problem is binomial because it is either effective or not so we can use the binomial probability density function. If you have a Ti83 it will work. binompdf(n,p,k) n = sample space p = probability of success (we determine success as a defective hair dryer) k = exact number of successes we are looking for binompdf(10000,.02,225) = .0058 rJ_ edit*** the reason for our different answers is that mine is exact and you used the normal approximation for a binomial distribution by taking off the .5's |
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