The formula for the mode and the median of grouped data

Any student at GCSE level in the UK needs to know about the three averages- the mean, median and mode. Of these, the mode is the simplest to find- it is the most occurring data point. The median is not much more difficult- it's the middle data point. Even at the higher end of the GCSE spectrum, when you need to find averages from tables in which the data has been grouped, finding the mode is easy: just look for the group with the largest associated frequency. Finding the median is, again, not much more involved: work out in which group the middle value falls. Or is it as simple as that?

Don't get me wrong- the above paragraph describes what you, as a higher level GCSE student, need to understand in order to find the mode and median for a grouped frequency table, and if you're feeling confused you probably shouldn't read any further. But strictly speaking, you haven't found the mode or the median. You've found the modal class and median class of the data. That is, the class (or group) in which the median and mode lie.


So how do I find the mode and median?
Let me just stress again that for the purposes of your GCSE, what you've been told to do to find the modal and median classes is exactly what you need to do. This post is purely for those who want to stretch themselves a little further than the bounds of their GCSE course.

It must be noted that, such is the nature of grouped frequency tables, it is not possible to calculate a definite average for the data. A value calculated for the mean, median or mode of grouped data must be referred to as an estimate.

The mode for grouped data
You can calculate the mode for a grouped frequency table by using the following formula:






Where:

• L is the lower class boundary of the modal class.
• f_m is the frequency associated with the modal class.
• f₁ is the frequency of the class before to the modal class.
• f2 is the frequency of the class after the modal class.
• h is the difference between the upper and lower bounds of the modal class.

The median for grouped data
You can calculate the median for a grouped frequency table using the following formula:

Where:

• L is the lower class boundary of the median class.
• f is the frequency associated with the median class.
• n is the total number of observations (i.e. the total of the Frequency column).
• c is the cumulative frequency up to the class before the median class.
• h is the difference between the upper and lower bounds of the median class.

Where do these formulae come from?

For the mode, imagine plotting the frequency of each group with its midpoint, and then joining the points up with a  smooth curve. With most distributions you would see a single peak to this curve. The formula calculates an estimate for the point on the x-axis that is directly below this peak.

For the median, imagine plotting a cumulative frequency graph of the data in your table. To find an estimate of the median, you would find half of your total frequency on the cumulative frequency axis, draw a line horizontally to your hand-drawn c.f. curve, and then drop vertically to the x-axis. Whatever this value is would be your estimate for the median. The formula just does this for you.

Finding the nth term of a sequence (the easy kind)

Everyone dreads the nth term questions, but once you get the hang of them, they're not actually all that hard.

A typical question might be: A sequence of numbers starts 2, 5, 8, 11... . What is the next number in the sequence? Find an expression for the nth term of the sequence.

The method

The next number is 14, obviously - you just add three each time. The difference between the terms is 3 - a term is simply a number in the sequence.

To find the nth term of the sequence, we look for a different sequence that goes up in threes - a good one would be the three times table.

A sidetrack that explains things

The nth term of the three times table is 3n. What does that even mean? It means that if you pick a number - say, 74 - I can tell you that the 74th number in the three times table is three times 74, or 222.

In that case, you picked 74 to be n and I replaced the n in the equation with 74. If I wanted the 100th term, n would be 100.

But the three times table isn't the same as this sequence!

Quite right. The sequence goes 2, 5, 8, 11... and the three times table goes 3, 6, 9, 12.... To get from the three times table to the sequence, we have to take away one. So the expression for the nth term of that series is 3n - 1.

Can you give me another example?

Why, of course! Let's look at -4, 3, 10, 17.... Here, the difference between terms is 7, so we're going to start from the seven times table - which has the expression 7n.

To get from 7, 14, 21... to -4, 3, 10..., we need to take away 11 - so the expression for the nth term of the sequence -4, 3, 10... is 7n - 11.

Derren Brown and Lottery Magic

So, Derren Brown. Guessing lottery numbers. Correctly*. What's the probability of him getting everything right by blind luck?

Let's say Derren's less-talented brother, Gordon, also played the lotto on Wednesday and - by a sheer fluke - also guessed that 2, 11, 23, 28, 35 and 39 would be drawn.

What's probability of the first ball drawn being on Gordon's ticket?

There are six balls on the ticket, and 49 in the machine - Sapphire - they're using for the draw. The probability of any particular ball coming up is one out of 49, and we have six that would be good news - so the probability of the first ball being on the ticket is 6/49.

And the second ball?

Now we only have five balls left in the machine that we want to see. There is also one fewer ball in the machine, so the probability of Gordon liking the second ball is 5/48.

How about the rest of them?

The rest work the same way. Each time there is one fewer ball that helps us to win, so the top goes down by one; each time there is also one fewer ball in the machine, so the bottom goes down by one each time as well. The third ball would be 4/47, the fourth 3/46, the fifth 2/45 and the final ball 1/44.

Now what do we do with those?

We just multiply them all together. We get 720 / 10,068,347,520, which works out to be 1 / 13,983,816 - just a little better than the one in fourteen million you might have heard mentioned.

How do you know it was a trick?

Easy. Derren Brown is a magician. Also, if he knew the numbers before the draw, he'd have shown them before the draw. And probably bought a ticket.

* Of course it's a trick. The deep maths he's talking about? Nuh-uh.

Factorising quadratics

What is a quadratic?

You need to be able to recognise these. One of the tell-tale signs is that your teacher will ask 'what kind of expression/equation is this?'. The other is that it has an x² in it* - the word 'quadratic' comes from the Latin for square.

What do you do with them?

Most often, you have to factorise them (put into brackets) or to solve them (figure out what x has to be). In this post, we're not going to talk about solving them using the quadratic formula, which you use when they ask for answers to a number of significant figures or decimal places. Instead, we'll look at putting them into brackets.

What's the method?

1. Take the number that doesn't have any xs attached to it, including any minus sign


2. List the factor pairs of this number (remember to think about positive and negative options), and what the numbers add up to


3. Find a pair that adds up to the middle number


4. Rewrite the x term of the expression with the x split up into the numbers you've just found (see below for an example)


5. You should now have an expression with four parts to it. Find a common factor for the first two and a common factor for the second two - bracket them off


6. Tidy up the brackets


7. If it's an equation, figure out what x has to be to make the brackets equal 0

Let's start with an example.

Say we have to solve x² -5x + 4 = 0**.

Step 1: We want the number on its own, which is +4
Step 2: What are the factor pairs of +4? We could have:

Factors	Times to make	Add to make
+1 and +4	+4	+5
+2 and +2	+4	+4
-1 and -4	+4	-5
-2 and -2	+4	+4

Step 3: We want them to add to the middle number - in this equation, it's -5. So the pair we want is -1 and -4.
Step 4: We can rewrite -5x as (-1x + -4x): that makes our equation x² -1x -4x + 4
Step 5: The first two terms have an x in common, so we can write them as x ( x-1 ). The second two terms have a -4*** we can take out, so we can write them as -4 (x-1). Hey, look! The bracket is the same for both of them!
Step 6: That means we can write the whole equation as (x-4)(x-1) = 0.
Step 7: It's an equation (there's an equals sign), so we think about what could make each bracket 0. If x=4, the first bracket is 0 and 0 multiplied by anything is 0 so the equation works. If x=1, the second bracket is 0, so the equation also works. The two answers are x=4 and x=1.

What if there's a number in front of the x²?

Excellent question, I was hoping you'd ask that. The method is almost identical, except that instead of just taking the number on its own, you multiply it by the number of x²s there are. For instance, if you had 4x² + 5x -6=0, instead of taking -6 as your number, you'd take 4 times -6 = -24.

Let's do that.

-24 has a lot of factors.

Factors	Times to make	Add to make
+1 and -24	-24	-23
+2 and -12	-24	-10
+3 and -8	-24	-5
+4 and -6	-24	-2
-1 and +24	-24	+23
-2 and +12	-24	+10
-3 and +8	-24	+5
-4 and +6	-24	+2

We want -3 and +8 because they add up to +5: the middle number. We can split up the 5x into -3x + 8x to get:

4x² - 3x + 8x - 6 = 0

If we take an x out of the first pair and a 2 out of the second, we get:

x(4x - 3) + 2(4x-3) = (x+2)(4x-3) = 0

And we're done! x = -2 or x = 3/4.

* Strictly, the x² has to be the highest power - if there's an x³ in the equation, it's a cubic, x⁴ is a quartic - you don't need to know those, you just need to know that they're not quadratics.

** CHORUS: "We have to solve x² - 5x + 4 = 0!"

*** We could take out a 4, sure. However, we really want the bracket afterwards to be the same - generally, we want to take out the minus sign if there is one).

What is a google / googol?

The famous web search engine Google took its name from the maths word 'googol'. But what does it mean? Well, just like words such as 'four,' 'pi,' 'six thousand' and 'one hundred and thirty billion,' 'one googol' is a number. Specifically, the word 'googol' belongs in the same family as hundred, thousand, million, billion and so on: it is a power of ten. What's a power of 10? That's a bigger question for another post (please ask us to get going on it if you really want to know!), but a quick recap: Powers tell you how many times to multiply a number by itself (for example, 5³ means 5 x 5 x 5). A power of 10 is just the number ten 'raised to a power' (for example, 10⁴ is read as "ten to the power of 4," and means 10 x 10 x 10 x 10, which is ten thousand (10,000)). So what's a googol? A googol is

• "ten to the power of one hundred," or
• 10¹⁰⁰, which means:

10 x 10 x 10 x 10 x 10 x 10 x 10 x 10 x 10 x 10 x 10 x 10 x 10 x 10 x 10 x 10 x 10 x 10 x 10 x 10 x 10 x 10 x 10 x 10 x 10 x 10 x 10 x 10 x 10 x 10 x 10 x 10 x 10 x 10 x 10 x 10 x 10 x 10 x 10 x 10 x 10 x 10 x 10 x 10 x 10 x 10 x 10 x 10 x 10 x 10 x 10 x 10 x 10 x 10 x 10 x 10 x 10 x 10 x 10 x 10 x 10 x 10 x 10 x 10 x 10 x 10 x 10 x 10 x 10 x 10 x 10 x 10 x 10 x 10 x 10 x 10 x 10 x 10 x 10 x 10 x 10 x 10 x 10 x 10 x 10 x 10 x 10 x 10 x 10 x 10 x 10 x 10 x 10 x 10 x 10 x 10 x 10 x 10 x 10 x 10

• Which equals:

10,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000 That's quite big.

Perpendicular Bisectors

What is a perpendicular bisector?
Look at the words by themselves, first of all:

• Perpendicular: This means 'at right-angles to.' For example, walls are usually perpendicular to the floor, and table legs are usually perpendicular to the table top.
• Bisector: This means 'something that cuts exactly in half.' For example, if you snap a KitKat finger and the two pieces you have left over are exactly the same size, you have bisected it.

When you're asked to construct a perpendicular bisector, you'll be expected to draw a line that cuts another line exactly in half. Sometimes you'll be given two points and asked to find where something can go if it needs to be the same distance between them. Just pretend that the two points are opposite ends of a line* and use the same method as outlined below.


Tools
Wait! Before you start, you'll need the following tools:

• Pencil (diagrams must be drawn in pencil if you're working towards an exam, and your pencil needs to be well sharpened);
• A ruler or other straight edge;
• A pair of compasses**;
• A rubber (eraser) in case you make mistakes, or to rub out construction lines.
  

How to construct a perpendicular bisector
I'll be assuming that you've already got a line to bisect. If not, now's the time to draw one. Draw any line you like, but if you need a suggestion, go for one that's about 10cm long and draw it in the middle of your page to give yourself plenty of space. Now follow these steps:

Step 1.
Open up your compasses. It doesn't matter how wide, as long as it's at least half the distance between the two ends of your line.

Step 2.
Place the point of your compasses on one end of the line that you're trying to bisect; it doesn't matter which end. Now draw a circle with your compasses in the usual way.

Step 3.
DON'T CHANGE THE SIZE OF YOUR COMPASSES! <--- This bit is really, really, really important. It won't work if you close your compasses or open it further, and you'll have to start again.
Place the point of your compasses on the other end of your line, and draw a circle in the same way.

• If you've done it correctly, you should have two circles that overlap and are exactly the same size. If they're not the same size, or they don't overlap, you'll need to start again and see what you've missed out from these instructions!

Step 4.
Use your ruler to help you join up the two places where the circles cross over with a straight line. Extend your line as far as you can on either side of the diagram

• You've drawn your perpendicular bisector! The next bit is just about tidying up.

Step 5.
You might want to use your rubber to get rid of the circles and make it easier to see the result. Be warned though, that many exam questions will ask you to leave your construction lines visible - in this case, your circles are your construction lines and you should not rub them out!

Now take a look at the perpendicular bisectors example question to see a 'real life' example.


More information
When you become more confident at drawing this type of construction, it'll become clearer that it's not necessary to draw the whole of each circle in order to find the perpendicular bisector. With a bit of practise you'll get used to how much of the circle (or what size arc) you need to draw to get things done.

Other uses
The method for drawing a perpendicular bisector can also be used to:

• Find the midpoint of a line
• Find the midpoint between two points (draw a line between the two points, then find the midpoint of that)
• Find the locus of points equidistant from two given points
  

* Strictly speaking, a line segment.
** That's one set of the pointy type of compass, not two of the magnetic type. If you'd like help setting up your compasses, please comment and we'll sort out a post for it!