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Sampling- linear combinations

  1. Apr 15, 2014 #1
    The scores X1 and X2 in papers 1 and 2 of an examination are normally distributed with means 24.3 and 31.2 respectively and standard deviations 3.5 and 3.1 respectively The final mark for each candidate is found by calculating 2X1+1.5X2. Find the probability that a random sample of 8candiates will have a mean final mark of less than 60.

    what i have done so far is F~N(95.4,70.6225) and Sample mean of F, N (95.4,70.6225/8)
    P(sample mean of F<60)
    ... then unable to do alrdy because the z-score is too large
  2. jcsd
  3. Apr 15, 2014 #2
    It seems to me that independence would be violated rather badly. :-) Students scoring highly on one test are more likely to score highly on the other. You're probably supposed to assume independence, but you should scold your teacher for not being explicit about it.
  4. Apr 15, 2014 #3
    Chogg, can i ask you another question, which my answer is different from answer given.
    The random variable X is distributed Po(4.1) .Find the probability that the mean of a random sample of 50 observations of X is less than 4.5.

    first, i find out p=4.1/50 variance=npq=3.7638
    then approx to normal, N(4.1,3.7638/50)
    P(sample mean<4.5)
    continuity correction with -1/100
    anything wrong?what is the ans u get?
  5. Apr 15, 2014 #4
    I assume "Po" means Poisson, and 4.1 is the value of [itex]\lambda[/itex].

    Why would you switch to the approximating normal? It's going to be a rather lousy approximation with such a small mean. Why not just use the poisson directly?

    Anyway, you might find it useful to know that the sum of (independent!) draws from Poisson distributions is Poisson-distributed, and its mean is the sum of the means. You also might find it useful to translate the statement about the mean to a statement about the sum of your 50 random draws.
  6. Apr 15, 2014 #5
    But how to do Possion with continuous(4.5) ? r u able to get z-score for this ques 1.34?if yes, can you show me the step pls thx alot...if like u said using sum, then Po~(205) still need to use normal is it?
    Last edited: Apr 15, 2014
  7. Apr 15, 2014 #6

    The output of a Poisson is discrete: 0, 1, 2, etc.

    The mean of a Poisson is a continuous quantity. It could be anything.

    For example, imagine a probability distribution which gives 1 with probability [itex]p[/itex], and 0 with probability [itex]1-p[/itex]. It only ever outputs discrete numbers (1 or 0), but its mean is a continuous quantity ([itex]p[/itex]).

    Why do you care about z-scores? This is a Poisson problem, not a Gaussian problem. z-scores are for Gaussian distributions.

    Also: what does "ques 1.34" mean? Is this a homework problem? (It sure sounds like it.) If it is, you're asking in the wrong forum.

    In any case: if you want to know the probability that a random variable is less than some value, the concept that will help you is the cumulative distribution function, or "CDF". What is the CDF for the Poisson distribution? Does it give a reasonable-sounding answer?
  8. Apr 15, 2014 #7
    im using normal because of central limit theorem, which states that if n is large, the distribution can be approx to normal. Because i duno hw to do poisson for P(X<4.5)...
  9. Apr 15, 2014 #8
    Did you find the CDF for the Poisson distribution from the wikipedia page I linked to?

    Did you try plugging in your numbers to that formula?
  10. Apr 15, 2014 #9
    i know that formula for poisson . P(x<4.5)=P(x=0,1,2,3,4)? but this is a question for sampling? or can you show me how you did?
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