Life Stats

Enter your date of birth and read some interesting statistics about the world since you came into it. Plenty of discussion starters here!

Life Stats:

https://neal.fun/life-stats/

Venus Transits and the Size of the Solar System

This entry in 2020's Big Lock-Down Math-Off hosted by the Aperiodical introducess the Venus transits of the 1600s and their importance in developing an early understanding of the true scale of our Solar system (when the page loads, scroll about half way down to find the article).

Venus Transits and the Size of the Solar System: https://aperiodical.com/2020/05/the-big-lock-down-math-off-match-23/

And you can find out more about Venus at Blogstronomy! 

American Football and Maths

Ex- professional American Footballer John Urschel is studying for a PhD in mathematics. In this Numberphile podcast he discusses his interest and career in both, touching on some of the crossovers between two disciplines many people would assume to be entirely separate.

Mathematical Quotes

Around 400 quotes about mathematics and from mathematicians, collated by Colin Beveridge of Flying Colours Maths.

Mathematical Quotes: https://www.flyingcoloursmaths.co.uk/mathematical-quotes/

100 (ish) of these are available for purchase as an ebook for Amazon Kindle: https://amzn.to/3h0DVhv

Music and Maths: The Score

This episode of the Relatively Prime maths podcast explores just a small part of the intersection between mathematics and music. 

ACMEScience · Relatively Prime: The Score

Mathematics and Medicine: AI & Metabolites

Maths fan and Biotech entrepreneur  Leila Pirhaji shares her plan to build an AI network to improve understanding and treatment of disease. See her TED talk in the player below or find further details, a transcript and further reading on the TED website.

Social Distancing Rules: Bad Maths

In June 2020 the world was dealing with the Covid-19 pandemic and "social distancing" was a buzzword for the time: everyone had to stay a given distance away from each other to reduce the chances of catching the virus or passing it on to someone else. There was much debate as to what this distance actually should be and, in the lead up to the potential reopening of shops in the UK, on 5th June The Guardian published this article:

One of the illustrations from the Guardian article

Plans to reopen shops in England 'in chaos' over social distancing rules

Study finds 10 sq metres of space is required per shopper to stick to government’s 2-metre guidelines

Link: https://www.theguardian.com/business/2020/jun/05/shops-reopen-england-social-distancing-rules-coronavirus

Maths teacher Mark Dawes found the original source for the mentioned study and wrote a critical piece about it and the article linked above, discovering and explaining some of the misconceptions and mistakes spanning both:

More than sums (blog) - Mathematically Distant: https://morethansums.blogspot.com/2020/06/mathematically-distant.html

Is March 23rd an Exceptionally Good Day For Athletes To Be Born?

Here's another real-life-maths guest post by tutor, author and friend-of-the-blog Colin, of Flying Colours Maths. This time, it's Olympic.

Short answer: no.

An article in the Guardian suggests that because Mo Farah, Chris Hoy and Steve Redgrave share a birthday with Jason Kenny, Roger Bannister, Joe Calzaghe and Mike Atherton, March 23rd must be a Good Day to have Olympian kids.

It's unusual, right enough, for three gold medallists from these Games - Farah, Kenny and Hoy- to share March 23rd as a birthday. You can work out the odds against it, roughly, using a binomial distribution: The probability of any one athlete having March 23rd as a birthday is - give or take - 1/365. There were 44 British gold medallists at the games. So the probability of at least three gold medallists sharing March 23rd as a birthday works out to be

1 - 44C0 (1/365)⁰(364/365)⁴⁴ - 44C1 (1/365)¹(364/365)⁴³ - 44C2 (1/365)²(364/365)⁴²

(That is, the probability of there not being 0, 1 or 2 athletes with that birthday).

That works out to be 1 - 0.8863 - 0.1071 - 0.0063, which is about 1 in 4000. Tiny odds!

Trouble is, that's a deeply, deeply flawed experiment. We've started with the knowledge that three of the athletes share that particular birthday. The probability of three athletes sharing any birthday is much higher. Roughly speaking, it's 365 times as high, about one in 11, nowhere near enough to be statistically significant. Actually working it out properly at 8am is a little beyond me, but writing a little simulation isn't. Taking 44 random birthdays, repeatedly, the results look like this:

• No two people share a birthday about 7% of the time
• No more than two people share any birthday about 84% of the time
• Three people share a birthday about 8% of the time
• Four people share a birthday about 0.3% of the time
• Five people share a birthday about 0.005% of the time

(That 8% is pretty close to the one-in-11 I suggested earlier, which makes it look like I'm on the right track).

If, as the article does, you extend it to previous medallists and athletes from other sports, you're bound to be able to cherry-pick more people with the same birthday.

Really, you'd expect the Guardian to run these things past a mathematician before they go live.

Guest Post: Is It Worth Insuring My Mobile?

Here's another guest post from Colin!

I'm in the process of upgrading my phone, and was just offered insurance against losing or damaging my new shiny. "It'll cost you £299 to replace your phone," it says, and offers me peace of mind for just £6 per month over my two year contract.

How much?

Seems reasonable, on the face of it. But let's do some sums, and not think about the fact that my home insurance payment is only double that.

Over the 24 months, with the insurance, I'd end up paying a total of £24 x 6 = £144: nearly half of the value of the phone. So, is it worth it?




Well, that turns into a probability question. The value of the insurance to me is how much the phone would cost to replace multiplied by the probability of me claiming on it. For instance, if I had a 1% chance of losing the phone over the next two years, the insurance would be worth about 0.01 x £299 = £2.99.

So, if I call the value V, the probability p and the cost of the phone C, the equation would be: V = pC.

Should I?

That would be worthwhile if the value was greater than the cost of the insurance (I): V > I, and since we know V is the same thing as pC, I can replace that in the inequality:

pC > I

Now, I know two of those values: C = 299 and I = 144, so, the policy is good value if:

299 p > 144

Dividing both sides by 299 (inequalities work, for the most part, just like equals signs):

p > 144 / 299 ~ 0.4816.

That means, the policy is good value for me if I think I have a more than 48% chance of losing my phone over the next two years. And while I'm clumsy and forgetful, I've never lost or broken a mobile phone.

So, this policy is a really bad idea for me - I'd do much better to put £6 a month into a savings account and use it to buy something nice at the end of it: maybe a new phone!

How Can We Calculate the Angular Size of an Object in the Sky?

This post is written as a sister-post to this one over at Blogstronomy, and is intended to show more explicitly the mathematical process that is alluded to there.

Angular size, as I describe on the Blogstronomy post, is a way that astronomers use to talk about how big things appear in the sky. That post talks about how big Jupiter and its moons Io and Ganymede would appear in the sky above Europa. This post explains how to work it out.

In order to work it out we need to be fairly comfortable with trigonometry - that's that GCSE topic with SOH CAH TOA and right-angled triangles. We also need to know two things about the object we want to look at: its actual size, and its distance from us.

Here's the general maths that we need:

Drawn by me using Paper on the iPad

The grey lines are imaginary lines from our eyeball to the extents of the object in the sky. r stands for the radius of the object we're looking at, and the yellow line (d) represents the distance between us and the centre of the object. Luckily for us, it also cuts the (grey) isosceles triangle in half and gives us a right-angled triangle that we can get to work on with trig.

If we're trying to find the angle that's marked (that's half the angular size), we can label the radius as the 'opposite' side, the yellow line as the 'adjacent' side, and the grey line is the hypotenuse. Let's call the angle "x", to make things easy. We know the radius and distance, so from our school-days "SOH CAH TOA", we know that we have to use the Tangent function:

Tan(x) = Opposite / Adjacent

Or, using more sensible notation from the diagram:

Tan(x) = r / d

We want to find x, though, so we can rearrange:

x = Tan^-1(r / d)

But if we're being honest with ourselves, we want to find the entire angle between the two grey lines, so we need to double whatever result we get (calling the angular size "a"):

a = 2x = 2 x Tan^-1(r / d)

Now all we have to do is substitute the values we know for each of the objects. Here's the data you need:

• Io: r = 1830 km ; d = 249,300 km.
• Ganymede: r = 2631.2 km ; d = 399,300 km.
• Jupiter: r = 69,911 km ; d = 671,100 km.

All you have to do is whack those numbers in the right places into your favourite calculator, then we've found the angular size of these three bodies as viewed from Europa! Just make sure your calculator's set to 'degrees' and neither 'radians' nor 'stun'. The result for Io comes out at 0.841 degrees. If you want to find out what that means for our view and/or find out the answers for the Ganymede and Jupiter (but try them out yourself first, or it's no fun), head over to the post at Blogstronomy!