There were lots of us on the Mayor’s London Cycle Sky ride today. I was pulling a Halfords Cycle trailer with the granddaughter proudly ensconced – who was waving regally to the attendant crowds. Grandson was on his yellow machine – along with full suspension and fearsome brakes. At one time we were told that there were over eighty thousand riding the paved streets of London.
Dick Whittington would have loved to have been with us. He became the Lord Mayor of London. I am sorry to say that we did not see the present Mayor of London as we swept past the Houses of Parliament and Buckingham Palace. I would have prized an opportunity to say thank you for the privilege of traffic free riding. I can, however, say thank you to the hundred of volunteers lining the route who made the ride such an enjoyable experience.
On the journey back my mind wandered, as it is wont to do on occasions, on how an eager eleven plus child would work out the average age of the cyclists. There was a wide spectrum of ages – with male and female seemingly equally distributed. Our eleven plus mathematician could not have approached all eighty thousand and asked their ages. He or she could, however, have selected a sample drawn over a period of time as the cyclists crossed the starting line. The prospective actuary could have closed his or her eyes and on a signal opened them to select a group of nine cyclists. Answers about the ages of a random sample of nine participants could have sought and recorded.
The ages could then be placed into rank order with the oldest at one end and the youngest at the other. The middle number of the nine would be the median age – because of the ranked order. Suppose that ten ages were obtained – because an enthusiastic member of the cycling fraternity also wanted to be entered into the calculation. The middle number would then be a number between the middle two ages.
(We saw a man on a unicycle but we were not counting wheels but ages.)
So if the question on the eleven plus paper, by chance, does not have an odd number we would need this formula.
The median of a set of N numbers which have been ranked in order is equal to the odd number in the middle. If the range of numbers is not odd, the median is half the sum of the middle two numbers.
Using this method we could have found the age of the cyclist in the very middle of the range of all eighty thousand and one cyclist. If, however, there were exactly eighty thousand some poor soul may have needed to be cut in half.
Who said the eleven plus is easy?