# Guesstimation

## Solving the World's Problems on the Back of a Cocktail Napkin

## Lawrence Weinstein, John Adam

- 320 pagine
- English
- ePUB (disponibile sull'app)
- Disponibile su iOS e Android

# Guesstimation

## Solving the World's Problems on the Back of a Cocktail Napkin

## Lawrence Weinstein, John Adam

## Informazioni sul libro

Guesstimation is a book that unlocks the power of approximation--it's popular mathematics rounded to the nearest power of ten! The ability to estimate is an important skill in daily life. More and more leading businesses today use estimation questions in interviews to test applicants' abilities to think on their feet. Guesstimation enables anyone with basic math and science skills to estimate virtually anything--quickly--using plausible assumptions and elementary arithmetic.

Lawrence Weinstein and John Adam present an eclectic array of estimation problems that range from devilishly simple to quite sophisticated and from serious real-world concerns to downright silly ones. How long would it take a running faucet to fill the inverted dome of the Capitol? What is the total length of all the pickles consumed in the US in one year? What are the relative merits of internal-combustion and electric cars, of coal and nuclear energy? The problems are marvelously diverse, yet the skills to solve them are the same. The authors show how easy it is to derive useful ballpark estimates by breaking complex problems into simpler, more manageable ones--and how there can be many paths to the right answer. The book is written in a question-and-answer format with lots of hints along the way. It includes a handy appendix summarizing the few formulas and basic science concepts needed, and its small size and French-fold design make it conveniently portable. Illustrated with humorous pen-and-ink sketches, Guesstimation will delight popular-math enthusiasts and is ideal for the classroom.

## Domande frequenti

## Informazioni

# Chapter 1

How to Solve Problems

**STEP 1:**Write down the answer [4]. In other words, come up with a reasonably close solution. This is frequently all the information you need.

- too big
- too small
- just right

**STEP 2:**If you can’t estimate the answer, break the problem into smaller pieces and estimate the answer for each one. You only need to estimate each answer to within a factor of ten. How hard can that be?

*approximate*geometric mean of any two numbers, just average their coefficients and average their exponents.

^{∗}In the clown case, the geometric mean of one (10

^{0})

^{†}and 100 (10

^{2}) is 10 (10

^{1}) because one is the average of the exponents zero and two. Similarly, the geometric mean of 2 × 10

^{15}and 6 × 10

^{3}is about 4 × 10

^{9}(because

^{∗}If the sum of the exponents is odd, it is a little more complicated. Then you should decrease the exponent sum by one so it is even, and multiply the final answer by three. Therefore, the geometric mean of one and 10

^{3}is 3 × 10

^{1}= 30.

## EXAMPLE 1: MongaMillions Lottery Ticket Stack

^{†}If you stacked up all the possible different lottery tickets, how tall would this stack be? Which distance is this closest to: a tall building (100m or 300 ft), a small mountain (1000 m), Mt Everest (10,000 m), the height of the atmosphere (10

^{5}m), the distance from New York to Chicago (10

^{6}m), the diameter of the Earth (10

^{7}m), or the distance to the moon (4?10

^{8}m)? Imagine trying to pick the single winning ticket from a stack this high.

**Solution:**To solve this problem, we need two pieces of information: the number of possible tickets and the thickness of each ticket. Because your chance of winning is one in 100 million, this means that there are 100 million (10

^{8}) possible different tickets.§ We can’t reliably estimate really thin items like a single lottery ticket (is it 1/16 in. or 1/64 in.? is it 1mm or 0.1 mm?) so let’s try to get the thickness of a pack of tickets.

^{8}tickets is

^{4}m is 20 kilometers or 20 km (which is about 15 miles since 1 mi = 1.6 km).

## EXAMPLE 2: Flighty Americans

**Solution 1:**Start with the number of Americans and estimate how many plane flights each of us take per year. There are 3 × 10

^{8}Americans.

^{∗}Most of us probably travel once a year (i.e., two flights) on vacation or business and a small fraction of us (say 10%) travel much more than that. This means that the number of flights per person per year is between two and four (so we’ll use three). Therefore, the total number of flights per year is

**Solution 2:**Start with the number of airports and then estimate the flights per airport and the passengers per flight. There are several reasonable size airports in a medium-sized state (e.g., Virginia has Dulles, Reagan-National, Norfolk, Richmond, and Charlottesville; and Massachusetts has Boston and Springfield). If each of the fifty states has three airports then there are 150 airports in the US. Each airport can handle at most one flight every two minutes, which is 30 flights per hour or 500 flights per 16-hour day. Most airports will have many fewer flights than the maximum. Each airplane can hold between 50 and 250 passengers. This means that we have about

^{8}, which is close enough to both answers.

## EXAMPLE 3: Piano Tuners in Los Angeles

**Solution:**This is a sufficiently complicated problem that we cannot just estimate the answer. To solve this, we need to break down the problem. We need to estimate (1) how many pianos there are in Los Angeles and (2) how many pianos each tuner can care for. To estimate the number of pianos, we need (1) the population of the city, (2) the proportion of people that own a piano, and (3) the number of schools, churches, etc. that also have pianos. To estimate the number of pianos each tuner can care for, we need to estimate (1) how often each piano is tuned, (2) how much time it takes to tune a piano, and (3) how much time a piano tuner spends tuning pianos.

- population of Los Angeles
- proportion of pianos per person
- how often each piano is tuned per year
- how much time it takes to tune each piano
- how much time each piano tuner works per year

- The population of Los Angeles must be much less than 10
^{8}(since the population of the US is 3 × 10^{8}). It must be much more than 10^{6}(since that is the size of an ordinary big city). We’ll estimate it at 10^{7}. - Pianos will be owned by individuals, schools, and houses of worship. About 10% of the population plays a musical instrument (it’s surely more than 1% and less than 1...