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The mean and the median selling prices of single-family homes were reported to be \$163,900 and \$210,900. The report was not clear about which was which. \$163,900 is the mean and \$210,900 is the median. Is this true

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  • $\begingroup$ Your question is unclear. Is this for some class? Did you leave anything out? $\endgroup$ – Glen_b Jan 16 '15 at 1:27
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Not necessarily; it depends on the distribution—you can have all three cases. Here are some examples:

  1. Symmetric distributions have their mean equal to their median. An example of this is the standard normal distribution with mean = median (= mode) = 0.
  2. Some right skewed distributions have their mean greater than their median. An example of this is the exponential distribution as parameterized $f(x) = \frac{1}{\theta}e^{-\frac{x}{\theta}}$. This has mean $\theta$ and median $\theta\ln(2)$. As $\ln2 < 1$ the mean is thus greater than the median
  3. Some distributions will have their median greater than their mean. I don't know a rule off-hand, but there are Poisson distributions which will have medians greater than their mean, as the median of a Possion must fall in the interval $\lambda - \ln 2 < \nu < \lambda + \frac{1}{3}$ and the mean is always $\lambda$.

In your specific case, it is likely that we are dealing with a heavily right-skewed distribution (one or two mansions will skew the statistics greatly), so it is more likely that the larger number is the mean and the smaller is the median.

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