I'm not sure if this is the proper forum for this post, so if it is not, please do move...
The task of finding and removing apples that house worms before they get to the grocery store is a big problem. COnsumers have been shown to react poorly upon finding worms their apples. To combat this problem, Wormfinder Inc. has developed an amazing new nonintrusive test for worms in apples. THis test is, called WormScan and has the incredible false negative rate of exactly 0 (i.e. if an apple is declared by wormscan to be free of worms it is guaranteed to have no worms in it). Unfortunately, such a performance comes with a cost; the false positive rate is 3% (i.e. 3% of all good apples are marked as having a worm inside). Statistically it has been found that 0.2% (1 in 500) apples have worms.
(a) What percentage of apples will test as having worms?
( Give than an apple has tested as having worms, what is the probability that there is a worm inside?
So for (a) I got this for an answer:
P(YX) = P(XY)P(Y) / P(X) => P(XY)P(Y) / [(P(XY)P(Y) + P(Xnot Y)P(not Y)]
=> (0.03)(0.002) / [(0.03)(0.002)+(.97)(.998)] = 0.0000619 or 0.00619%
I don't think this is right, I used Bayes Rule for this, and it just seems like a ridiculously small number to obtain from this question, and as for ( I have no idea where to begin. I tried using Bayes rule for P(XY), basically reversing the logic to obtain the answer, and got the same answer...
Any help with this would be greatly appreciated. I have exhausted my text resources and can not seem to find any other resources or examples.
So basically I am just asking for a guideline of how to solve this or some examples or anything really.
Statistics HW help?
Page 1 of 11 Replies  1363 Views  Last Post: 09 February 2013  07:43 PM
Replies To: Statistics HW help?
#2
Re: Statistics HW help?
Posted 09 February 2013  07:43 PM
IngeniousHax, on 09 February 2013  02:55 PM, said:
Unfortunately, such a performance comes with a cost; the false positive rate is 3% (i.e. 3% of all good apples are marked as having a worm inside). Statistically it has been found that 0.2% (1 in 500) apples have worms.
(a) What percentage of apples will test as having worms?
(/> Give than an apple has tested as having worms, what is the probability that there is a worm inside?
So for (a) I got this for an answer:
P(YX) = P(XY)P(Y) / P(X) => P(XY)P(Y) / [(P(XY)P(Y) + P(Xnot Y)P(not Y)]
=> (0.03)(0.002) / [(0.03)(0.002)+(.97)(.998)] = 0.0000619 or 0.00619%
(a) What percentage of apples will test as having worms?
(/> Give than an apple has tested as having worms, what is the probability that there is a worm inside?
So for (a) I got this for an answer:
P(YX) = P(XY)P(Y) / P(X) => P(XY)P(Y) / [(P(XY)P(Y) + P(Xnot Y)P(not Y)]
=> (0.03)(0.002) / [(0.03)(0.002)+(.97)(.998)] = 0.0000619 or 0.00619%
I think that your problem is based upon the interpretation of the date. This question doesn't have any complicated parts. Just consider.
 0.2% of all apples have worms.
 99.8% of all apples don't have worms.
 3% of the apples without worms will false positive.
So the total number found to have worms is actually a simple calculation away... whether this actually corresponds to Bayes formula I don't know, except to say that it is an overcomplication. And yes, your answer was much too small
For the second question, this is just a division. Consider the total number of apples flagged, and the total correctly flagged.
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