Normal Distribution

• SBS261P

  16 Sep 10, 09:34

  I came across this question and still cannot resolve the 2 arguments, hope that someone can enlighten me here!

  A duck shop sells duck that have weight normally distributed with mean weight 2.7kg and variance 0.424.

   

  Given that 1 kg of duck costs $4, find the probability that out of 4 randomly chosen ducks, the most expensive duck will be below $11.

   

  Let X ~ N(2.7, 0.424)

  Then 4X ~ N(10.8, 6.784)

   

  (Probably the correct solution)

   

  P(most expensive duck below $11)

  = P(all ducks cost less than $11)

  = [P(4X <$11)]^4

  = 0.0792648307

   

  But as I thought about it more...

   

  P(most expensive duck below $11)

  = 1 - P (1 duck cost >= $11)

  = 1 - P (4X>=$11)

  =0.530603529

   

  Reasoning is, if there is one duck that cost >=$11, the other ducks can be any condition, most expensive will still be >=$11 guaranteed (i.e. use pigeonhole principle - if there is 1 duck >=$11, no matter where the other ducks go (<$11 or >$11), there will be at least 1 duck >=$11)

   

  But the thing is, the second method is saying that it doesn't matter how many ducks you choose, the probability is still the same! (e.g. most expensive out of 100 ducks is still 0.5306)

  Why is there a discrepancy? ...
• wee_ws

  16 Sep 10, 12:51

  Hi,

    In your first method, you considered the multiplication principle.

    In your second method, you considered the mutual exclusion principle (where you make the claim that one duck may cost $11 or more).

    If the most expensive duck is below $11, certainly all others must cost less. Therefore, it is logical to consider method 1 in that all 4 ducks cost below $11.

    In addition, it is utypical to encounter a probability that is constant regardless of the event (e.g. number of ducks).

    Thanks.

  Cheers,
  Wen Shih

• SBS261P

  16 Sep 10, 15:22

  my question is, isn't the two cases the same thing? means I'm overcounting the second case? but where is the overcounting?
• Mad Hat

  16 Sep 10, 16:19

  Hi SBS261P

  In the second case, when you say P(1 duck cost ≥ 11), the real meaning is P(At least 1 duck cost ≥ 11). This is not the same as P(4X ≥11).

  P(most expensive duck below $11)

  = 1 - P (1 duck cost >= $11)

  = 1 - P (4X>=$11)

  =0.530603529

  
  

  To find P(At least 1 duck cost ≥ 11), you must consider four cases:

  
  1 duck cost ≥ 11, 3 ducks cost < 11
  2 ducks cost ≥ 11, 2 ducks cost < 11
  3 ducks cost ≥ 11, 1 duck cost < 11
  4 ducks cost ≥ 11

  
  

  and then sum their probabilities.
  
• SBS261P

  16 Sep 10, 16:26

  4X is defined as the price of a duck. the argument put forward is since there is 1 duck already, it doesn't matter how much the other ducks cost (by pigeonhole principle). sounds intuitively correct bt inherently wrong, but I can't find counterexample to disprove this working. I thought of the following to refute this working: -are we considering order of ducks when we do this? -is there conditional probability? the 2nd working was put forward by one of my students and I can't seem to find a way to explain it is wrong... =(
• wee_ws

  17 Sep 10, 11:30

  Hi,

    Mutual exclusion does not apply for this problem, period. That is all the student should be advised on.

    Probabilistic reasoning is hard and abstract at times. Use logic to appeal to the student instead.

    Btw, the Pigeonhole Principle is not assessed in H2 Mathematics, in case the problem originated from that.

    Thanks.

  Cheers,
  Wen Shih

• TenSaru

  19 Sep 10, 00:27

  1 - P(1 duck >11) is the same as P(1 duck < 11) but that doesn't mean that it is the most expensive duck...

  So if 1 duck is >11 that means the rest of the duck do not need to be considered

  but if 1 duck < 11 there is still a chance that the other 3 ducks can have 1 that is >11

• Konnichiwa

  22 Sep 10, 05:04

  A duck shop sells duck that have weight normally distributed with mean weight 2.7kg and variance 0.424.Given that 1 kg of duck costs $4, find the probability that out of 4 randomly chosen ducks, the most expensive duck will be below $11.

  This is two step prob... 1st step -->normal dist

  2nd step -->Binomal distri

  First find the probability that

  P(most exp duck>$11)= success

  Q= 1-P, Q is failure.

  Y~B(n,p)

  To find prob of P(X>$11) where X is the price of duck.

  Using normal distribution,

  mean= 4 * 2.7 = 10.8

  Variance = 4* 0.424 = 1.696

  X~N(10.8, 1.696)

  P(X>$11)=P[z>(11-10.8)/(sgrt1.696)]

  from z table, we can find the Prob... then 1-P is the area that is >$11.

  So.. we have the Prob of price of duck >$11.

  2nd Step: Use Binomal formula to find the prob... it is success/failure event...

  Second step is a binomal prob.... so... if X is the price... 4 times of the price... is not the same as randomly out of 4....

  His/her 1-P here is the area of one duck price under the normal distribution curve...less than $11....

  P(most expensive duck below $11)

  = 1 - P (1 duck cost >= $11)

  = 1 - P (4X>=$11)

  =0.530603529