I found an I.Q. score question that has been bothering me: If the population mean is 100, the standard deviation is 10, and given a sample of six scores, what is the probability that three of the scores are less than 90, two are between 90 and 120, and one is greater than 120. Can this problem be solved without using Z-scores, using instead the pdf formula for the normal distribution.

What I immediately thought was: $P(X<90)^3 * P(90<X<120)^2 * P(X>120)$, again using the pdf function for the normal distribution.

Is this logical?


That is pretty close, but that is the probability for obtaining three from the first kind, two from the second, and then one fom the third kind in that particular arrangement.

You do also have to consider the other possible arrangements so multiply by $6!/(3!~2!~1!)$, that is, $60$.

EG: The probability of tossing two heads and a tail in no particular order (in a toss of three coins), is $3(\tfrac 12)^2\tfrac 12$, ie $3/8$, because we measure the probabilities for the event $\sf \{HHT, HTH, THH\}$.

  • $\begingroup$ Of course I forgot about combinations, thanks! $\endgroup$ Jun 19 '17 at 23:58

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