# Distribution of green bean lengths before and after cutting

Say I have a bag of beans, with different lengths, but following a normal distribution with some mean and some standard deviation.

If I were to take out the beans in pairs, and cut the longer one in half, leave the shorter one alone, and put them both in a different pot (so they won't be subject to more than one inspection or cut), and continue doing this for all the beans, how will the distribution change? (An acceptable result is finding the new mean and standard deviation as functions of the old ones - a simulation is interesting but I can do that myself, and wouldn't give that satisfaction of understanding that a real solution would bring).

Context: I thought of this problem a number of years ago while doing a first year undergraduate course in statistics, while chopping beans for dinner. I later asked my tutor, but he didn't know how to solve it (or didn't have time). I thought of it again a few days ago, again while chopping beans. I never did any further studies in statistics and have forgotten 90% of what I learned so I'm not really equipped to begin solving it myself. Hoping I can get closure on this topic :-)

• Edited, thanks @BruceET – Erin Sep 29 '17 at 13:25
• A simulation would probably tell you more. You end up with a mixture distribution: two-thirds of the resulting beans will be distributed as half the maximum of two iid random variables from the original distribution while one-third will be distributed as the minimum of two iid random variables. Assuming negative length beans are very unlikely, I would expect two peaks in the distribution, with the lower peak being higher as you end up with twice as many half beans as whole beans – Henry Sep 29 '17 at 13:47
• @BruceET: in your example of original mean and SD are $10$cm and $2$cm, I get a resulting mean close to $5 \times \frac23 + 10 \times \frac13 \approx 6.66$ much as I would expect, and a standard deviation just under $2$. Reducing the original standard deviation towards $0$ increases the resulting standard deviation towards $\frac53 \sqrt{2} \approx 2.357$ – Henry Sep 29 '17 at 13:59
• @Henry. Agree with your comments. Either my memory fails me (always possible at my age) or the edit made the problem fundamentally different and clearer. So deleting my Comment to avoid confusion. – BruceET Sep 29 '17 at 15:22
• The dist'n of the max of two normals can lead to messy analysis, but simulation is easy. If you want to explore analytic answers, perhaps assume an (admittedly unrealistic) uniform population distribution. Then max and min are not so difficult. – BruceET Sep 30 '17 at 19:13