"An urn contains 80 marbles, of which 60% are red and 40% are white. Out of 50 samples of 20 marbles, each selected with replacement from the urn, how many samples can be expected to consist of (a) equal numbers of red and white marbles?"

I did it and got my answer = 9, but the correct answer is 6.

My insight: Since every sample has a fixed size of 20, if I calculate the probability of getting 10 red marbles in a sample, then necessarily the other 10 marbles must be white.

So $p = 0.6$, $q = 0.4$. The standard deviation is $ (pq/N)^1/2 = 0.1095 $

Standardizing the distribution, you get $(0.5 - 0.6)/0.1095 = -0.91$. Then $Z(-0.91) = 0.1814$. Now, multiplying this probability by the amount of samples $(50)$, the result is $9.07$ samples, or $9$ when you round up.

Would anyone point out what I'm doing wrong? I'm using the method given by the chapter of the textbook (Schaum's Outlines of Statistics, 6th edition), and it worked on previous exercises.

  • 1
    $\begingroup$ Why not just use the binomial probabilities directly, instead of approximating the binomial distribution with a normal? $\endgroup$ – Matthew Drury Jun 3 '17 at 23:39
  • $\begingroup$ Just work on the binomial distribution, do not think about normal approximation, you will get the correct answer. Your result (9 samples) seems 9 samples with less than 10 red marbles (0-9 red marbles). $\endgroup$ – user158565 Jun 3 '17 at 23:40
  • $\begingroup$ Because in the case I'm particularly interested in using the normal approximation to solve the problem, not just solving the problem. $\endgroup$ – Victor Jun 3 '17 at 23:48
  • 1
    $\begingroup$ Then convert 10 red marbles into an interval on the real number. Based on this interval to use normal approximation, instead of on the one point. $\endgroup$ – user158565 Jun 4 '17 at 0:35

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