Mixed Problems: Set 1 (Statistics)

Transcript

0:00Introduction

Hello everyone! Let us start with a small puzzle. Twenty five observations have mean thirty and standard deviation four. But one value, forty two, was recorded as twenty four; the two digits got swapped. Just one small slip in twenty five values. Will the variance barely move, or will it change a lot? Keep your guess ready, because Example two settles it. This is Set one of our mixed statistics problems, in the JEE Main pattern, and questions like these come up often. So, let's begin.

0:35Set 1: what it holds

So, here is what Set one holds. Fifteen solved problems, on the mean, the median, the mode, the mean deviation and the variance. They are all mixed up, so you choose the tool yourself, just like in the exam. Okay? And the course target is one to two minutes per problem. So first, let us sharpen two tools that this set uses again and again, and then we solve two problems together. With me?

1:01Tool 1: shift and scale

Tool number one, from Chapter one: shift and scale. Suppose every observation, x i, is multiplied by the constant a, and then the constant b is added, to give y i, the new observation. The small i written low is a subscript; it just numbers the observations. And the mean does exactly the same thing: y bar equals a times x bar, plus b, where the bar on top means the mean. Okay? Now the spread. If every single value goes up by the same amount, b, does the spread change at all? It does not change at all, because all the values move together. Every gap between them stays exactly the same. So adding a constant changes nothing in the spread. Right? Only the multiplier acts on the spread. Sigma, the Greek letter, is the standard deviation, and the small y and x written low tell us which data. Sigma y equals mod a times sigma x; the vertical bars are the modulus, because a standard deviation is never negative. And the variance, sigma y squared, equals a squared times sigma x squared, where the small two means squared.

2:11Tool 2: rebuild the two totals

Clear? Tool number two, also from Chapter one: the working formula for the variance. The big capital sigma means the sum of. So sigma squared equals the sum of x i squared, over n, minus x bar squared. In words, the mean of the squares minus the square of the mean. And we divide by n, the number of values, as our course does. Okay? Now, many questions give you only n, the mean and the standard deviation, and no data at all. Can you still find the sum of the values, and the sum of their squares? Yes, you can, and it is really simple. The mean is the sum over n, so the sum of x i equals n times x bar. Right? And for the squares, just turn the first line around. The sum of x i squared equals n times, in brackets, sigma squared plus x bar squared. Here is the topper's habit for this set: always rebuild these two totals first. Got it? And when one value was recorded wrongly, fix both totals: take out the wrong value and its square, and put in the right value and its square. Then recompute, and use the new mean, not the old one. Remember this, because it is the trap in Example two. With me?

3:28Example 1: add 2, multiply by 4, find new mean and SD

Now, Example one, from Set one, marked easy. Twenty observations have mean ten and S D three, where S D is short for standard deviation. Each is increased by two and then multiplied by four. Find the new mean and the new standard deviation. Pause the video and try it first. Done? Okay, let's check. Watch the order: first add two, then multiply by four. So y i equals four times, in brackets, x i plus two, which is four times x i, plus eight. So the multiplier is four, and the constant added is eight. Okay? The mean follows the very same rule. Four times, ten plus two, gives forty eight. Right? Now the standard deviation. Many students do the same thing again: four times, three plus two, which is twenty. Is that right? Oh no, careful, that is the trap! Adding two moves every value together, so it does nothing to the spread; only the four acts. Four times three, and the dot means times, gives twelve. So the new mean is forty eight and the new standard deviation is twelve, the course answer. Clear?

4:46Example 2: 42 recorded as 24, correct mean and variance

Now, remember our puzzle? Here it is, and this one is marked hard. Twenty five observations have mean thirty and S D four. One value, forty two, was recorded as twenty four. Find the correct mean and variance. Pause the video and try it first. Done? Okay, let's check. First, rebuild the totals. The sum of x is twenty five times thirty, which is seven hundred and fifty. Careful here: the question gives the standard deviation, four, so the variance is four squared, which is sixteen. Thirty squared is nine hundred, so the sum of x squared is twenty five times, sixteen plus nine hundred, which gives twenty two thousand nine hundred. Okay? Now read the words carefully: forty two was recorded as twenty four. So which value is sitting in our totals right now, and must come out? Twenty four, yes, exactly! It was recorded, so it went into the totals. Out goes twenty four, and in comes forty two. Seven hundred and fifty, minus twenty four, plus forty two, gives seven hundred and sixty eight. Right? Now the squares, and please do not skip them. Twenty four squared is five hundred and seventy six, and forty two squared is one thousand seven hundred and sixty four. Twenty two thousand nine hundred, minus five hundred and seventy six, plus one thousand seven hundred and sixty four, equals twenty four thousand and eighty eight. Got it? So the correct mean is seven hundred and sixty eight over twenty five, which gives thirty point seven two. It went up a little, since a bigger value replaced a smaller one. Clear? Now the variance. We take the mean of the squares, and subtract the square of the mean. But which mean, the old thirty, or the new thirty point seven two? The new one, of course, because the old mean belongs to the wrong data. Twenty four thousand and eighty eight over twenty five is nine hundred and sixty three point five two. Thirty point seven two squared is nine hundred and forty three point seven one eight four. Subtract, and the correct variance is nineteen point eight zero one six. So, our puzzle is solved: one swapped pair of digits took the variance from sixteen to nineteen point eight. That is a big change from one small slip!

7:09Quick revision: shift-scale rule, totals, wrong value fix

So, quick revision. First, if y equals a times x plus b, the mean becomes a times x bar plus b, but the standard deviation becomes mod a times sigma, because adding b does nothing to the spread. Second, rebuild the two totals first: the sum of x is n times x bar, and the sum of x squared is n times, sigma squared plus x bar squared. Third, for a wrong value, fix both totals, then use the new mean. Now try the practice problem on this page, and then attempt the rest of Set one, at one to two minutes each. All the best!