Mean vs Median Lab

Incomes follow a log-normal distribution with σ = 1.5. Work out what fraction of people earn less than the mean. Then show that you never needed σ — a published mean and median are enough.

1. Log-normal means the logarithm is normal: ln X is N(μ, σ²). Everything below is a statement about ln X translated back, and that is the only trick in the problem.

   $$\ln X\sim N(\mu ,{\sigma }^2)$$

2. The median is easy. Half of ln X lies below μ, and the logarithm preserves order, so half of X lies below e^μ. The median is e^μ.

   $$\text{median}=e^{\mu },\text{mean}=e^{\mu +{\sigma }^2/2}$$

3. The mean is not e^μ, and that gap is the whole subject. For a log-normal it is e^(μ + σ²/2) — the mean sits above the median by a factor of e^(σ²/2), which at σ = 1.5 is e^1.125 = 3.0802.

   $$\frac{\text{mean}}{\text{median}}=e^{{\sigma }^2/2}=e^{1.125}=3.0802$$

4. Now the share. Asking what fraction earn below the mean is asking P(X < e^(μ + σ²/2)), and taking logs of both sides turns it into P(ln X < μ + σ²/2), which is a normal question with a normal answer.

   $$P\left(X<e^{\mu +{\sigma }^2/2}\right)=P\left(\ln X<\mu +\frac{{\sigma }^2}{2}\right)$$

5. Standardise it: subtract μ and divide by σ, and the μ cancels. The answer is Φ(σ/2), and at σ = 1.5 that is Φ(0.75) = 0.7734. 77.34% earn below the mean.

   $$=\Phi \left(\frac{\mu +{\sigma }^2/2-\mu }{\sigma }\right)=\Phi \left(\frac{\sigma }{2}\right)=\Phi (0.75)=0.7734$$

6. μ vanished, so scale is irrelevant: a population with a median of 1, or 148, or 22,026 gives the same 77.3% the panel prints. And run the argument backwards — σ = √(2 ln(mean/median)) — so the ratio alone gives σ, and σ gives the share.

   $$\sigma =\sqrt{2\ln \left(\frac{\text{mean}}{\text{median}}\right)}\Rightarrow \text{share}=\Phi \left(\frac{\sigma }{2}\right)$$

A village of five reports incomes of 10, 20, 30, 40 and 5,000. Work out both averages, then suppose one of the five figures is a typing error and decide what you can still say.

1. The mean is the total shared out: 5,100 across five households is 1,020. Notice what that does before going further. The average is larger than four of the five numbers it was built from.

   $$\bar{x}=\frac{10+20+30+40+5000}{5}=\frac{5100}{5}=1020$$

2. The median is a position rather than a total. Sort the five and take the third, which is 30. Four households sit below the mean and two sit below the median, on the same data.

   $$x_{(1)},\dots ,x_{(5)}=10,20,30,40,5000⟹\mathrm{med}=x_{(3)}=30$$

3. The Mean / Median ratio card reads 34.00. On symmetric data it sits at 1, so a reading of 34 is the page saying the two summaries have stopped describing the same village.

   $$\frac{\bar{x}}{\mathrm{med}}=\frac{1020}{30}=34.00$$

4. Drag the fifth value and watch the mean. It moves by a fifth of every unit added, with nothing to stop it: from one end of the slider to the other the mean runs 22 to 1,020.

   $$\frac{\partial \bar{x}}{\partial x_5}=\frac{1}{5}=0.2⟹x_5:10\to 5000\Rightarrow \bar{x}:22\to 1020$$

5. The median over that same drag takes two values, 20 and 30, and the sorting is the reason. Anything below 20 leaves 20 in third place, anything above 30 leaves 30 there, and in between the fifth value is the median itself. Whether the household earns 5,000 or five million, the median is trapped in a window ten wide.

   $$x_5\le 20\Rightarrow \mathrm{med}=20;20\le x_5\le 30\Rightarrow \mathrm{med}=x_5;x_5\ge 30\Rightarrow \mathrm{med}=30$$

6. Now the harder version of the question. You know one of the five figures is wrong but not which one. Work through all five cases and the corrected median has to land between 20 and 40, whatever the true value was. The corrected mean could be anything at all.

   $$\mathrm{med}\in [20,40],\bar{x}\in (-\infty ,\infty )$$