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I was attempting the following self-study question. After repeated tries, my answers were at best 0.0944, however, the model answer seemed to be otherwise at 0.1889.

Will appreciate any pointers as I simply can't figure out even after repeated attempts.

Question

According to the 2010 census in a city, university-educated employee earned a mean annual income of USD 61,000 & standard deviation of USD4,000. If the incomes are normally distributed and two university-educated employees are randomly selected, what is the probability that one of the them earns more than USD66,000 per year and the other earns less than USD66,000 per year?

My Attempt

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    $\begingroup$ Think about binomial distribution and notice that your answer is exactly the half of the correct answer. Update with another hint: C(2, 1) = 2!/1!1! = 2. $\endgroup$
    – Epameinondas
    Commented May 4, 2014 at 7:39
  • $\begingroup$ @Epaminondas thanks for the hint. I see that with 2C1 = 2, if we multiply 0.0944 by 2, we will get the answer 0.1889. However, what I am unclear still is how does binomial come into play in this question where we are both looking for failure and success at the same time? $\endgroup$
    – user1275515
    Commented May 4, 2014 at 10:36

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It is a TRUE/FALSE fact to earn more than USD66,000 per year and you correctly calculated that the probability for a university-educated employee to be one of those people (probability of TRUE) is 0.1056. Now, among the two person you select you should have one TRUE statement. To formalize that situation let X define the number of TRUE statement among the n = 2 repetitions of an experiment with probability 0.1056. You should calculate P(X = 1). The rest should be clear.

Update: your answer would be correct if the position of TRUE statement was fixed in the first or the second selected employee. Since no such restriction applies there are more than one combinations and that is the reason that binomial distribution should be applied.

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  • $\begingroup$ thanks you so much once again. will be sharing my answer based on your method. $\endgroup$
    – user1275515
    Commented May 4, 2014 at 16:29
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You should apply the "Law of total probability" to get the correct answer (0.1889).

Your problem is considering two random employees whose salaries (X and Y) are normally distributed with mean 61000 and st. dev. 4000.

The probability you have calculated is: $$P(Y > 66000 \cap X < 66000)$$ You have to sum it to $P(X > 66000 \cap Y < 66000)$ to get: $$P[(X > 66000 \cap Y < 66000) \cup (Y > 66000 \cap X < 66000)]$$ that is the correct answer of your problem (0.1889).

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