Ebooks

The Ebook Market in Numbers

Over the years the ebook market has grown from a relatively obscure niche to a thrilling billion-dollar mass market. The total ebook revenues went from 64 million $ in 2008 to about 3 billion $ in 2012. That’s a increase by a factor of close to 50 in just a few years.

ebook market revenues

The number of units sold also increased by the same factor (from 10 million units in 2008 to 457 million in 2012).

ebook market units sold

(Source)

However, many experts believe that the ebook market has reached a plateau and the numbers for the first half of 2013 seem to confirm that.

From the revenues and units sold we can also extract the development of the average price for sold ebooks. It strongly increased from 6.4 $ in 2008 to about 8 $ in 2009. After that, it quickly went back down to 7 $ in 2010 and 6.7 $ in 2012. So ebooks have gotten cheaper in the last few years, but are still more expensive than in 2008.

average price ebook

As of 2012, ebooks make up 20 % of the general book market.

21 % of American adults have read an ebook / magazine / newspaper on an e-reader in 2012. This is up from 17 % in the previous year.

A survey, again from 2012, shows that most e-book consumers prefer Amazon’s Kindle Fire (17 %, up from no use) , followed by Apple’s iPad (10 %, same as previous year) and Barnes & Noble’s Nook (7 %, up from 2 %).

The Emerging World Of Ebooks

As we all know, the Internet changed everything. The Net and the Web have brought the world closer together. Many older means of communication have either been replaced or changed so as to co-exist with, and complement, electronic communication. Email, for example, has replaced almost all business and a lot of personal letter writing, though our mailboxes remain filled with lots of mail, most unwanted. A lot of printed newspapers and magazines still exist, but their content is now also available on websites, and the websites are timelier and often offer more detailed information. Printed media is not dead by any means, as millions of people still prefer to curl up with a good book or grab a paper on their way to work. There have been many efforts to popularize ebooks, downloadable books in digital form, but their acceptance remains in its infancy.

But that won’t stay that way. Ebooks make sense. Since books are almost all text, an ebook download is very fast and hundreds of ebooks can fit onto a small storage card. Ebooks do not contribute to cutting down forests, they do not need to be trucked across the country, they do not produce waste, and they are usually a lot less expensive than printed books. Ebooks also have many other advantages. Depending on your ebook reader software, an ebook can be annotated, bookmarked and searched. The latter is especially useful; I often want to go back to a certain quote or paragraph in a book, and electronic search is so much easier than leafing through a printed book.

One of the problems ebooks face is that people do not know how to use them. They are confused by the many different ebook formats or think they need a particular piece of hardware to read them. In fact, the formats are not really a problem. Most computers can read popular ebook formats and ebook reader software is freely available. Hardware is a bit more of an issue. Hardcovers and paperbacks are awfully convenient and they don’t need batteries, so a lot of people shy away from reading on a computer screen or spending the money for a dedicated ebook reader.

This is really too bad as ebooks are clearly the way of the future. They just make too much sense. Those who dismiss ebooks are missing out on a great and increasingly attractive alternative to the printed page. Those who are willing to give ebooks a chance are rewarded with lower costs and the ability to carry an entire library on a device of their choice, be that a notebook computer, a Tablet PC, a dedicated ebook reader, a PDA or even a smartphone. And they have access to a potentially much larger variety of books. That’s because ebooks make self-publishing easy and lots of authors who don’t have a chance of getting picked up by traditional print publishing houses can distribute their books electronically. Best of all, there is no waste and there will never be unsold books that end up on a bargain table or in a landfill.

My advice is to give ebooks a chance. Download a free ebook. Look for sites dedicated to ebooks, especially those with a website design that is appealing. See what format you prefer, and what device you like to read on. But be warned: you may get hooked. Once you get into them, downloading and reading ebooks can become a passion.

Intensity: How Much Power Will Burst Your Eardrums?

Under ideal circumstances, sound or light waves emitted from a point source propagate in a spherical fashion from the source. As the distance to the source grows, the energy of the waves is spread over a larger area and thus the perceived intensity decreases. We’ll take a look at the formula that allows us to compute the intensity at any distance from a source.

Great Formulas_html_7230225e

First of all, what do we mean by intensity? The intensity I tells us how much energy we receive from the source per second and per square meter. Accordingly, it is measured in the unit J per s and m² or simply W/m². To calculate it properly we need the power of the source P (in W) and the distance r (in m) to it.

I = P / (4 · π · r²)

This is one of these formulas that can quickly get you hooked on physics. It’s simple and extremely useful. In a later section you will meet the denominator again. It is the expression for the surface area of a sphere with radius r.

Before we go to the examples, let’s take a look at a special intensity scale that is often used in acoustics. Instead of expressing the sound intensity in the common physical unit W/m², we convert it to its decibel value dB using this formula:

dB ≈ 120 + 4.34 · ln(I)

with ln being the natural logarithm. For example, a sound intensity of I = 0.00001 W/m² (busy traffic) translates into 70 dB. This conversion is done to avoid dealing with very small or large numbers. Here are some typical values to keep in mind:

0 dB → Threshold of Hearing
20 dB → Whispering
60 dB → Normal Conversation
80 dB → Vacuum Cleaner
110 dB → Front Row at Rock Concert
130 dB → Threshold of Pain
160 dB → Bursting Eardrums

No onto the examples.

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We just bought a P = 300 W speaker and want to try it out at maximal power. To get the full dose, we sit at a distance of only r = 1 m. Is that a bad idea? To find out, let’s calculate the intensity at this distance and the matching decibel value.

I = 300 W / (4 · π · (1 m)²) ≈ 23.9 W/m²

dB ≈ 120 + 4.34 · ln(23.9) ≈ 134 dB

This is already past the threshold of pain, so yes, it is a bad idea. But on the bright side, there’s no danger of the eardrums bursting. So it shouldn’t be dangerous to your health as long as you’re not exposed to this intensity for a longer period of time.

As a side note: the speaker is of course no point source, so all these values are just estimates founded on the idea that as long as you’re not too close to a source, it can be regarded as a point source in good approximation. The more the source resembles a point source and the farther you’re from it, the better the estimates computed using the formula will be.

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Let’s reverse the situation from the previous example. Again we assume a distance of r = 1 m from the speaker. At what power P would our eardrums burst? Have a guess before reading on.

As we can see from the table, this happens at 160 dB. To be able to use the intensity formula, we need to know the corresponding intensity in the common physical quantity W/m². We can find that out using this equation:

160 ≈ 120 + 4.34 · ln(I)

We’ll subtract 120 from both sides and divide by 4.34:

40 ≈ 4.34 · ln(I)   

9.22 ≈ ln(I)

The inverse of the natural logarithm ln is Euler’s number e. In other words: e to the power of ln(I) is just I. So in order to get rid of the natural logarithm in this equation, we’ll just use Euler’s number as the basis on both sides:

e^9.22 ≈ e^ln(I)

10,100 ≈ I

Thus, 160 dB correspond to I = 10,100 W/m². At this intensity eardrums will burst. Now we can answer the question of which amount of power P will do that, given that we are only r = 1 m from the sound source. We insert the values into the intensity formula and solve for P:

10,100 = P / (4 · π · 1²)

10,100 = 0.08 · P

P ≈ 126,000 W

So don’t worry about ever bursting your eardrums with a speaker or a set of speakers. Not even the powerful sound systems at rock concerts could accomplish this.

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This was an excerpt from the ebook “Great Formulas Explained – Physics, Mathematics, Economics”, released yesterday and available here: http://www.amazon.com/dp/B00G807Y00.

Computing and Tracking the Amazon Sales Rank

The webpage http://www.novelrank.com/ provides a very neat simple way to track the sales rank of any book on Amazon. This service is completely free.

The sales rank is computed from the sales rate. The more a book sells per day, the lower the rank will be.  Here’s an approximate formula, taken from: http://www.edwardwrobertson.com/2013/02/a-quick-way-to-calculate-amazon-sales.html.

100,000 / rank = sales per day

So if a book is on rank 50,000, it sells about twice a day. As far as I know, a borrow counts as a sale and a free download as one third of a sale.

I use novelrank to track my ebooks. This is what the output looks like (launch of “Great Formulas Explained”):

novelrankgreatformulas

Indeed a neat tool to see how a book is performing. Note that the tracking starts on the day you add it, dates before that are not shown.

As you can see, during the period when no sale is made the sales rank increases more or less linearly at about # 50,000 per day. The average rank during this time can be calculated by the formula: final minus initial rank divided by 2. When a sale is made, the rank makes a discontinuous jump to a lower value.

Quantitative Analysis of Top 60 Kindle Romance Novels

I did a quantitative analysis of the current Top 60 Kindle Romance ebooks. Here are the results. First I’ll take a look at all price related data and conclusions.

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  • Price over rank:

pricerank

There seems to be no relation between price and rank. A linear fit confirmed this. The average price was 3.70 $ with a standard deviation of 2.70 $.

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  • Price frequency count:

pricescount

(Note that prices have been rounded up) About one third of all romance novels in the top 60 are offered for 1 $. Roughly another third for 3 $ or 4 $.

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  • Price per 100 pages over rank:

pricerank

Again, no relation here. The average price per 100 pages was 1.24 $ with a standard deviation of 0.86 $.

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  • Price per 100 pages frequency count:

PPP1

About half of all novels in the top 60 have a price per 100 pages lower than 1.20 $. Another third lies between 1.20 $ and 1.60 $.

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  • Price per 100 pages over number of pages:

PPP2

As I expected, the bigger the novel, the less you pay per page. Romance novels of about 200 pages cost 1.50 $ per 100 pages, while at 400 pages the price drops to about 1 $ per 100 pages. The decline is statistically significant, however there’s a lot of variation.

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  • Review count:

reviewscount

A little less than one half of the top novels have less than 50 reviews. About 40 % have between 50 and 150 reviews. Note that some of the remaining 10 % more than 600 reviews (not included in the graph).

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  • Rating over rank:

rankreviews

There’s practically no dependence of rank on rating among the top 60 novels. However, all have a rating of 3.5 stars or higher, most of them (95 %) 4 stars or higher.

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  • Pages over ranking:

pagesrank

There’s no relation between number of pages and rank. A linear fit confirmed this. The average number of pages was 316 with a standard deviation of 107.

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  • Pages count:

pagescount

About 70 % of the analyzed novels have between 200 and 400 pages. 12 % are below and 18 % above this range.

Mathematics of Explosions

When a strong explosion takes place, a shock wave forms that propagates in a spherical manner away from the source of the explosion. The shock front separates the air mass that is heated and compressed due to the explosion from the undisturbed air. In the picture below you can see the shock sphere that resulted from the explosion of Trinity, the first atomic bomb ever detonated.

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Using the concept of similarity solutions, the physicists Taylor and Sedov derived a simple formula that describes how the radius r (in m) of such a shock sphere grows with time t (in s). To apply it, we need to know two additional quantities: the energy of the explosion E (in J) and the density of the surrounding air D (in kg/m3). Here’s the formula:

r = 0.93 · (E / D)0.2 · t0.4

Let’s apply this formula for the Trinity blast.

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In the explosion of the Trinity the amount of energy that was released was about 20 kilotons of TNT or:

E = 84 TJ = 84,000,000,000,000 J

Just to put that into perspective: in 2007 all of the households in Canada combined used about 1.4 TJ in energy. If you were able to convert the energy released in the Trinity explosion one-to-one into useable energy, you could power Canada for 60 years.

But back to the formula. The density of air at sea-level and lower heights is about D = 1.25 kg/m3. So the radius of the sphere approximately followed this law:

r = 542 · t0.4

After one second (t = 1), the shock front traveled 542 m. So the initial velocity was 542 m/s ≈ 1950 km/h ≈ 1210 mph. After ten seconds (t = 10), the shock front already covered a distance of about 1360 m ≈ 0.85 miles.

How long did it take the shock front to reach people two miles from the detonation? Two miles are approximately 3200 m. So we can set up this equation:

3200 = 542 · t0.4

We divide by 542:

5.90 t0.4

Then take both sides to the power of 2.5:

t 85 s ≈ 1 and 1/2 minutes

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Let’s look at how the different parameters in the formula impact the radius of the shock sphere:

  • If you increase the time sixfold, the radius of the sphere doubles. So if it reached 0.85 miles after ten seconds, it will have reached 1.7 miles after 60 seconds. Note that this means that the speed of the shock front continuously decreases.

For the other two parameters, it will be more informative to look at the initial speed v (in m/s) rather the radius of the sphere at a certain time. As you noticed in the example, we get the initial speed by setting t = 1, leading to this formula:

v = 0.93 · (E / D)0.2

  • If you increase the energy of the detonation 35-fold, the initial speed of the shock front doubles. So for an atomic blast of 20 kt · 35 = 700 kt, the initial speed would be approximately 542 m /s · 2 = 1084 m/s.

  • The density behaves in the exact opposite way. If you increase it 35-fold, the initial speed halves. So if the test were conducted at an altitude of about 20 miles (where the density is only one thirty-fifth of its value on the ground), the shock wave would propagate at 1084 m/s

Another field in which the Taylor-Sedov formula is commonly applied is astrophysics, where it is used to model Supernova explosions. Since the energy released in such explosions dwarfs all atomic blasts and the surrounding density in space is very low, the initial expansion rate is extremely high.

This was an excerpt from the ebook “Great Formulas Explained – Physics, Mathematics, Economics”, released yesterday and available here: http://www.amazon.com/dp/B00G807Y00. You can take another quick look at the physics of shock waves here: Mach Cone.

The Time Value of Money and Inflation

To make a point, I’ll start this blog entry in an unusual way, that is, by talking about vectors. A vector is basically an ordered row of numbers. Consider this expression for example:

(12, 3, 5)

This vector could represent a lot of things. For example a point in a three dimensional coordinate system, with the vector components being the x-, y- and z-values respectively. Or for a company offering three products, it could stand for the sales of these products in a certain year.

Why this talk about vectors? You were probably very surprised when you heard grandma say that she paid only 150 $ for her first car. It seems so amazingly cheap. But it is not. Your dear grandma is talking about 1950’s money, while you are thinking of today’s money. These two have a very different value.

If you want to specify the costs of a good precisely, merely giving an amount of money will not be sufficient. The value of money changes over time and thus to be absolutely precise, you should always couple this amount with a certain year. For example, this is what grandma’s car really cost:

(150 $, 1950)

This is far from (150 $, 2012), which is what you were thinking of when grandma shared the story with you. Using an online inflation calculator, we can conclude that this is actually what the car would cost in today’s money:

(1410 $, 2012)

Not an expensive car, but certainly more than 150 $ in today’s money. Now you can see why I started this chapter using vectors. They allow us to easily and clearly couple an amount with a year. A true pedant would even ask for one more component since we are still missing the respective months. But let’s not get too pedantic.

How can we justify saying that 150 $ in 1950’s money is the same as 1410 $ in today’s money? We can look at how much of a certain good these amounts would buy in the given year. With 150 $ in 1950 you could fill your basket with about as many apples as you can with 1410 $ today. The same goes for most other common goods: oranges, potatoes, water, cinema tickets, and so on.

This is inflation, goods get more expensive each year. At a later point we will take a look at what reasons there are for inflation to occur. But before that, let’s define the rate of inflation and see how it is measured …

This was an excerpt from the ebook “Business Math Basics – Practical and Simple”, available for Kindle here: http://www.amazon.com/dp/B00FXB8QSO.

Physics (And The Formula That Got Me Hooked)

A long time ago, in my teen years, this was the formula that got me hooked on physics. Why? I can’t say for sure. I guess I was very surprised that you could calculate something like this so easily. So with some nostalgia, I present another great formula from the field of physics. It will be a continuation of and a last section on energy.

To heat something, you need a certain amount of energy E (in J). How much exactly? To compute this we require three inputs: the mass m (in kg) of the object we want to heat, the temperature difference T (in °C) between initial and final state and the so called specific heat c (in J per kg °C) of the material that is heated. The relationship is quite simple:

E = c · m · T

If you double any of the input quantities, the energy required for heating will double as well. A very helpful addition to problems involving heating is this formula:

E = P · t

with P (in watt = W = J/s) being the power of the device that delivers heat and t (in s) the duration of the heat delivery.

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The specific heat of water is c = 4200 J per kg °C. How much energy do you need to heat m = 1 kg of water from room temperature (20 °C) to its boiling point (100 °C)? Note that the temperature difference between initial and final state is T = 80 °C. So we have all the quantities we need.

E = 4200 · 1 · 80 = 336,000 J

Additional question: How long will it take a water heater with an output of 2000 W to accomplish this? Let’s set up an equation for this using the second formula:

336,000 = 2000 · t

t ≈ 168 s ≈ 3 minutes

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We put m = 1 kg of water (c = 4200 J per kg °C) in one container and m = 1 kg of sand (c = 290 J per kg °C) in another next to it. This will serve as an artificial beach. Using a heater we add 10,000 J of heat to each container. By what temperature will the water and the sand be raised?

Let’s turn to the water. From the given data and the great formula we can set up this equation:

10,000 = 4200 · 1 · T

T ≈ 2.4 °C

So the water temperature will be raised by 2.4 °C. What about the sand? It also receives 10,000 J.

10,000 = 290 · 1 · T

T ≈ 34.5 °C

So sand (or any ground in general) will heat up much stronger than water. In other words: the temperature of ground reacts quite strongly to changes in energy input while water is rather sluggish. This explains why the climate near oceans is milder than inland, that is, why the summers are less hot and the winters less cold. The water efficiently dampens the changes in temperature.

It also explains the land-sea-breeze phenomenon (seen in the image below). During the day, the sun’s energy will cause the ground to be hotter than the water. The air above the ground rises, leading to cooler air flowing from the ocean to the land. At night, due to the lack of the sun’s power, the situation reverses. The ground cools off quickly and now it’s the air above the water that rises.

Image
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I hope this formula got you hooked as well. It’s simple, useful and can explain quite a lot of physics at the same time. It doesn’t get any better than this. Now it’s time to leave the concept of energy and turn to other topics.

This was an excerpt from my Kindle ebook: Great Formulas Explained – Physics, Mathematics, Economics. For another interesting physics quicky, check out: Intensity (or: How Much Power Will Burst Your Eardrums?).

Typical Per-Page-Prices for Ebooks

I did a little analysis of ebook prices per 100 pages for different categories in the Amazon Kindle store. In each category I looked at the top 12 paid books. This data can help readers to judge prices and authors to set them. Here are the results in increasing order:

Erotica: 1.7 $ per 100 pages (ranging from 1.0 – 3.1 $ per 100 pages)
Sci-Fi and Fantasy: 1.8 $ per 100 pages (ranging from 0.8 – 4.4 $ per 100 pages)
Short Stories: 2.0 $ per 100 pages (ranging from 0.5 – 4.2 $ per 100 pages)
Self-Help: 3.6 $ per 100 pages (ranging from 1.3 – 6.7 $ per 100 pages)
Applied Math: 4.0 $ per 100 pages (ranging from 0.9 – 7.9 $ per 100 pages)
Economy / Business: 7.2 $ per 100 pages (ranging from 3.3 – 17.2 $ per 100 pages)

Typical (and in my opinion fair) prices seem to be 2 $ per 100 pages for fiction and 4 $ per 100 pages for non-fiction. In the special case of business books, prices of 7 $ per 100 pages seem common.

My Fair Game – How To Use the Expected Value

You meet a nice man on the street offering you a game of dice. For a wager of just 2 $, you can win 8 $ when the dice shows a six. Sounds good? Let’s say you join in and play 30 rounds. What will be your expected balance after that?

You roll a six with the probability p = 1/6. So of the 30 rounds, you can expect to win 1/6 · 30 = 5, resulting in a pay-out of 40 $. But winning 5 rounds of course also means that you lost the remaining 25 rounds, resulting in a loss of 50 $. Your expected balance after 30 rounds is thus -10 $. Or in other words: for the player this game results in a loss of 1/3 $ per round.

 Let’s make a general formula for just this case. We are offered a game which we win with a probability of p. The pay-out in case of victory is P, the wager is W. We play this game for a number of n rounds.

The expected number of wins is p·n, so the total pay-out will be: p·n·P. The expected number of losses is (1-p)·n, so we will most likely lose this amount of money: (1-p)·n·W.

 Now we can set up the formula for the balance. We simply subtract the losses from the pay-out. But while we’re at it, let’s divide both sides by n to get the balance per round. It already includes all the information we need and requires one less variable.

B = p · P – (1-p) · W

This is what we can expect to win (or lose) per round. Let’s check it by using the above example. We had the winning chance p = 1/6, the pay-out P = 8 $ and the wager W = 2 $. So from the formula we get this balance per round:

B = 1/6 · 8 $ – 5/6 · 2 $ = – 1/3 $ per round

Just as we expected. Let’s try another example. I’ll offer you a dice game. If you roll two six in a row, you get P = 175 $. The wager is W = 5 $. Quite the deal, isn’t it? Let’s see. Rolling two six in a row occurs with a probability of p = 1/36. So the expected balance per round is:

B = 1/36 · 175 $ – 35/36 · 5 $ = 0 $ per round

I offered you a truly fair game. No one can be expected to lose in the long run. Of course if we only play a few rounds, somebody will win and somebody will lose.

It’s helpful to understand this balance as being sound for a large number of rounds but rather fragile in case of playing only a few rounds. Casinos are host to thousands of rounds per day and thus can predict their gains quite accurately from the balance per round. After a lot of rounds, all the random streaks and significant one-time events hardly impact the total balance anymore. The real balance will converge to the theoretical balance more and more as the number of rounds grows. This is mathematically proven by the Law of Large Numbers. Assuming finite variance, the proof can be done elegantly using Chebyshev’s Inequality.

The convergence can be easily demonstrated using a computer simulation. We will let the computer, equipped with random numbers, run our dice game for 2000 rounds. After each round the computer calculates the balance per round so far. The below picture shows the difference between the simulated balance per round and our theoretical result of – 1/3 $ per round.

Image

(Liked the excerpt? Get the book “Statistical Snacks” by Metin Bektas here: http://www.amazon.com/Statistical-Snacks-ebook/dp/B00DWJZ9Z2)

The Probability of Becoming a Homicide Victim

 Each year in the US there are about 5 homicides per 100000 people, so the probability of falling victim to a homicide in a given year is 0.00005 or 1 in 20000. What are the chances of falling victim to a homicide over a lifespan of 70 years?

 Let’s approach this the other way around. The chance of not becoming a homicide victim during one year is p = 0.99995. Using the multiplication rule we can calculate the probability of this event occurring 70 times in a row:

 p = 0.99995 · … · 0.99995 = 0.9999570

 Thus the odds of not becoming a homicide victim over the course of 70 years are 0.9965. This of course also means that there’s a 1 – 0.9965 = 0.0035, or 1 in 285, chance of falling victim to a homicide during a life span. In other words: two victims in every jumbo jet full of people. How does this compare to other countries?

 In Germany, the homicide rate is about 0.8 per 100000 people. Doing the same calculation gives us a 1 in 1800 chance of becoming a murder victim, so statistically speaking there’s one victim per small city. At the other end of the scale is Honduras with 92 homicides per 100000 people, which translates into a saddening 1 in 16 chance of becoming a homicide victim over the course of a life and is basically one victim in every family.

 It can get even worse if you live in a particularly crime ridden part of a country. The homicide rate for the city San Pedro Sula in Honduras is about 160 per 100000 people. If this remained constant over time and you never left the city, you’d have a 1 in 9 chance of having your life cut short in a homicide.

Liked the excerpt? Get the book “Statistical Snacks” by Metin Bektas here: http://www.amazon.com/Statistical-Snacks-ebook/dp/B00DWJZ9Z2. For more excerpts check out Missile Accuracy (CEP), Immigrants and Crime and Monkeys on Typewriters.