Part C: Concept Checks

Transcript

0:00Introduction

Hello everyone! Here is a quick puzzle. Suppose a teacher multiplies every mark in a test by minus two. The mean gets multiplied by minus two, so the standard deviation does too, and becomes negative, right? Hmm, can a standard deviation ever be negative? Hold that thought. Today we do six quick concept checks from Part C of our statistics and probability problem bank. Checks like these come up often in JEE questions, so let's begin.

0:32Add a constant: does the variance change?

Check number one is about the variance. Sigma squared, with the small Greek letter sigma, is the variance: one over n, times the sum of x sub i minus x bar, whole squared. The big Greek sigma means sum; x sub i, with the small i written low, is each observation; x bar, with a bar on top, is the mean; and n is the number of observations. Okay? Now add the same constant c to every observation, so y sub i equals x sub i plus c. Every value has gone up, so does the variance go up too? No, it does not. The mean also goes up by c, so y bar equals x bar plus c. Then each deviation, y sub i minus y bar, is exactly x sub i minus x bar; the c cancels. With me? Every deviation is unchanged, so the variance is unchanged: sigma y squared equals sigma x squared, with the small y and x written low. Clear?

1:36Mutually exclusive and independent?

Check number two is about two kinds of events. Two events are mutually exclusive when they cannot happen together, like these two circles that do not overlap. So A intersection B, with the upside down U sign, is the empty set, the circle with a slash; and P of A intersection B, the probability of both, is zero. Okay? And two events are independent when P of A intersection B equals P of A, times P of B, got it? Now, if P of A and P of B are both positive, can mutually exclusive events also be independent? They have nothing to do with each other, so surely yes, right? Oh no, careful, the answer is no! If both probabilities are greater than zero, so is their product; the pointed sign means greater than, and the double arrow means implies. With me? But here P of A intersection B is zero, and zero is not equal to a positive product; the slashed equals sign means not equal. So they are never independent. Clear?

2:38If A is a subset of B, compare P(A given B) and P(A)

Check number three is about conditional probability. In the figure, the box U is the whole sample space, and event A sits completely inside event B, the big oval. This is A subset of B, the rounded sign with a line under it. So A intersection B is just A itself. Okay? P of A given B, with the vertical line read as given, is P of A intersection B, over P of B, and it needs P of B greater than zero, right? So which one is bigger: P of A given B, or P of A on its own? The top, P of A intersection B, is just P of A, so P of A given B equals P of A over P of B, okay? And P of B is at most one; the less than sign with a line under it means less than or equal to. Dividing by it cannot make the fraction smaller, so P of A given B is greater than or equal to P of A, with the same sign turned round. Once B has happened, the space shrinks to B, so the share of A can only grow or stay the same. Got it?

3:43Binomial: can the variance exceed the mean?

Check number four: the binomial distribution. Capital X follows capital B of n comma p; the wavy sign is called tilde, and we read it as the word follows. Here n is a fixed number of independent trials, p is the probability of success, the same in every trial and between zero and one, and q, one minus p, is the probability of failure. Okay? The mean is n p, and the variance is n p q, right? So tell me, can the variance of a binomial distribution ever be more than its mean? Look, n p q is n p, times q, which is the mean times q. Clear? Since p is more than zero, q is less than one; the same pointed sign, turned round, means less than. So the variance is always less than the mean, never more. Hence, if a question gives a variance bigger than the mean, no binomial distribution fits. Got it?

4:41Example 1: SD when every value is multiplied by minus 2

Now Example one, our puzzle from the start. What happens to the S D, the standard deviation, when every observation is multiplied by minus two? Pause the video and try it first. Done? Okay, let's check. The new values are y sub i equals minus two times x sub i. The mean is also multiplied by minus two, so each deviation becomes minus two times the old one. Okay? Now, in the variance, each deviation gets squared. So what happens to the minus two? It becomes plus four, yes, exactly! So sigma y squared is one over n, times the sum of four times the old squared deviations, which is four times sigma x squared. Right? Now take the square root, the tick-shaped sign. The square root of four times sigma x squared is two times sigma x. So the S D is doubled, not multiplied by minus two, and it is never negative. That answers our puzzle! With me? In general, if y sub i equals a times x sub i, plus b, then sigma y equals mod a, times sigma x; the vertical bars give the modulus. The added constant b changes nothing, and the sign of the multiplier does not matter.

6:01Example 2: sum of squares vs square of the sum over n

Clear? Now Example two, a little harder. Is the sum of x sub i squared always greater than or equal to one over n, times the square of the sum of x sub i? Here is a hint: think of the variance. Pause the video and try it first. Done? Okay, let's check. Use the course's working formula: sigma squared equals the sum of x sub i squared over n, minus x bar squared, the mean of the squares minus the square of the mean. And x bar is the sum of x sub i, over n. Okay? Multiply it by n, and n times sigma squared equals the sum of x sub i squared, minus n times x bar squared. Right? So, what is n times x bar squared, in terms of the sum of x sub i? Well, x bar squared is the square of the sum, over n squared, and one n cancels. So n times x bar squared is one over n, times the square of the sum. With me? The sum of the squares, minus one over n times the square of the sum, equals n times sigma squared. And a variance is never negative, so this is greater than or equal to zero. Clear? So yes, it is always true; the difference is exactly n times the variance. Whenever the sum of squares and the square of the sum appear together, think of the variance. Got it?

7:29Quick revision: all six statistics and probability checks

So, quick revision. Adding a constant leaves the variance unchanged. Multiplying by a constant scales the S D by its modulus, so minus two doubles it. Mutually exclusive events with positive probabilities are never independent. If A is a subset of B, P of A given B is at least P of A; and for a binomial, the variance n p q is less than the mean n p. And the sum of squares, minus one over n times the square of the sum, is n times sigma squared, never negative. Well done on all six checks! Now try the practice problem on this page. All the best!