WEBVTT
Kind: captions
Language: en
00:00:14.380 --> 00:00:18.950
To compute that double integral, I make use
of the polar coordinates, that is, I make
00:00:18.950 --> 00:00:22.320
the
transformation x is equal to r cos theta y
00:00:22.320 --> 00:00:25.880
is equal to r sine theta. Then, you know,
that dx
00:00:25.880 --> 00:00:32.150
dy, the element of area in x-y plane gets
transformed to r dr d theta, that is the element
00:00:32.150 --> 00:00:34.970
of
this term. Your r varies from 0 to infinity
00:00:34.970 --> 00:00:40.910
because r is a non-negative variable and theta
varies from 0 to 2 pi. This is your polar
00:00:40.910 --> 00:00:47.790
coordinate transformation.
So, therefore I square, then becomes 1 upon
00:00:47.790 --> 00:00:54.660
2 pi 0 to infinity 0 to 2 pi and e raise to
minus half r square because this was x square
00:00:54.660 --> 00:00:59.570
plus y square will become r square cos
square theta plus r square sine square theta
00:00:59.570 --> 00:01:01.940
since cos square theta plus sine square theta
is
00:01:01.940 --> 00:01:08.320
1. So, it reduce to r square. And this r dr
d theta, right. And again, this is in a nice
00:01:08.320 --> 00:01:11.829
form
because now you see, you have r here and you
00:01:11.829 --> 00:01:16.740
have e raise to minus half r square.
So, again we will make this transformation,
00:01:16.740 --> 00:01:19.909
right, that r square is equal to t that is
what I
00:01:19.909 --> 00:01:30.350
am doing here. And so, your 2 dr, dr will
be dt. So, r dr will be half dt and so, that
00:01:30.350 --> 00:01:31.679
is
what I have written here, half dt and this
00:01:31.679 --> 00:01:35.421
will go to e raise to minus half t. So, this
is and
00:01:35.421 --> 00:01:44.770
you write, and see the limits are 0 to 2 pi
0 to this thing.
00:01:44.770 --> 00:01:49.270
.Now, here what I have done is, already theta
does not appear here anywhere. So,
00:01:49.270 --> 00:01:54.969
therefore, this is simply 1. So, with respect
to theta this just integrates. So, 2 pi, which
00:01:54.969 --> 00:01:58.600
I
have written here and then is only integration
00:01:58.600 --> 00:02:04.499
respect to r. And to integrate this I make
the transformation r square is equal to t
00:02:04.499 --> 00:02:08.410
and so, this helps me to reduce it further.
And
00:02:08.410 --> 00:02:17.010
this is, you know, minus 2 times e raise minus
half t 0 to infinity. So, this reduces to
00:02:17.010 --> 00:02:20.610
this,
which is equal to 1. So, we have verified,
00:02:20.610 --> 00:02:25.110
that the normal p d f, that we have defined
is
00:02:25.110 --> 00:02:33.530
valid p d f.
And then, now the second step is to compute
00:02:33.530 --> 00:02:41.060
the expectation and instead of computing
for e x we will compute expectation of x minus
00:02:41.060 --> 00:02:46.040
mu by sigma. That will be easier because
we know, that the expectation of x is actually
00:02:46.040 --> 00:02:51.510
mu. So, we will show, that this expectation
is 0 and therefore, get the answer immediately.
00:02:51.510 --> 00:03:01.951
So, here the, only in this integral you get
this term x minus mu by sigma and so, again
00:03:01.951 --> 00:03:09.970
I make the substitution x minus mu by
sigma is equal to y and that reduces the integral
00:03:09.970 --> 00:03:15.190
to this. Again it is of the same form. You
see, that is why the expressions may look
00:03:15.190 --> 00:03:21.420
cumbersome, but the working is not very
difficult and it is just a question of little
00:03:21.420 --> 00:03:27.150
patience and you start seeing where the
calculations are going. So, now we have this.
00:03:27.150 --> 00:03:31.730
So, again you make the transformation y square
equal to t, the same steps, and you get
00:03:31.730 --> 00:03:38.150
this. And so, so now, what I am saying is,
that before I make this substitution I will
00:03:38.150 --> 00:03:41.771
break
up this integral to minus infinity. See, one
00:03:41.771 --> 00:03:47.970
way I can immediately from here conclude,
that this integral is 0 because this is an
00:03:47.970 --> 00:03:52.040
odd integral, right. y is here, this is here,
the
00:03:52.040 --> 00:03:56.850
change of sign will not matter. But here the
change of sign will matter since this is from
00:03:56.850 --> 00:04:02.360
minus infinity to infinity, this integral
will be 0, you know one of the… So, I just
00:04:02.360 --> 00:04:06.040
thought, that I will show you the steps.
So, what I am doing is, I am breaking it,
00:04:06.040 --> 00:04:08.790
breaking up this integral from minus infinity
to
00:04:08.790 --> 00:04:16.630
0 plus 0 to infinity and then, when you make
that transformation, that y square is t, then
00:04:16.630 --> 00:04:22.940
you see, this becomes plus infinity because
when y is minus infinity square will be plus
00:04:22.940 --> 00:04:26.970
infinity. So, this is from, so this would
be infinity to infinity to 0 and the integral
00:04:26.970 --> 00:04:30.320
will be
e minus half t, just as we did here and then,
00:04:30.320 --> 00:04:35.630
this will be 0 to infinity. So, now, it is
a same
00:04:35.630 --> 00:04:41.850
integral expect that here the limits are upside
down. So, when you change the limits it
00:04:41.850 --> 00:04:45.580
will become minus sign. So, there will be
a minus sign here and the same integral with
00:04:45.580 --> 00:04:50.480
the plus sign. So, when you add, the result
is 0, ok.
00:04:50.480 --> 00:04:55.640
.So, I just showed you the steps, in case
you are not very sure, but otherwise we could
00:04:55.640 --> 00:05:01.440
have concluded our computation at this step
only and said that this is equal to 0. So,
00:05:01.440 --> 00:05:06.940
since this 0 expectation of x minus mu by
sigma, sigma is a constant, goes out.
00:05:06.940 --> 00:05:13.650
Therefore, this implies, that expectation
of x is equal to mu. So, I had shown you that
00:05:13.650 --> 00:05:18.360
the
normal p d f is symmetric about mu and this
00:05:18.360 --> 00:05:21.280
is also the mean. And in fact, mu has all
the
00:05:21.280 --> 00:05:25.160
properties, that too now I will show some
more properties of mu.
00:05:25.160 --> 00:05:26.160
.
00:05:26.160 --> 00:05:33.030
Similarly, to compute the variance of x, which
is actually, since mu is the mean, it is
00:05:33.030 --> 00:05:40.919
actually expectation of x minus mu whole square,
right. And this I will write in this form.
00:05:40.919 --> 00:05:45.320
So, now here what you have to do is, again
if you have first make the transformation,
00:05:45.320 --> 00:05:54.930
that x minus mu by sigma is t, so dx by sigma
can be replaced by dt, right. And the limits
00:05:54.930 --> 00:05:59.870
remain the same. So, this is the simple integral
and this way comes sigma square because
00:05:59.870 --> 00:06:04.260
you have x minus mu whole square. So, this
is sigma square t square. This is it.
00:06:04.260 --> 00:06:10.560
Now, what I do here is, the t square I break
up into t and t because this integral we have
00:06:10.560 --> 00:06:20.330
handled already, right. Well, computing this
thing and so, therefore, we will do
00:06:20.330 --> 00:06:24.220
integration by parts. This will be my first
function. That means, I take the integral
00:06:24.220 --> 00:06:27.030
of this,
multiply by this and then, the derivative
00:06:27.030 --> 00:06:29.260
of this into the integral of this. This is
your
00:06:29.260 --> 00:06:37.669
formula, right. So, this integral I have shown
you is e raise to minus half t square, this
00:06:37.669 --> 00:06:44.490
t
into this thing, and ok, no, no, why am I,
00:06:44.490 --> 00:06:47.800
ok, I am actually computing this for you.
So, I
00:06:47.800 --> 00:06:53.110
.am saying t e raise to minus half t square.
dt is half, when I make the transformation
00:06:53.110 --> 00:06:57.490
t
square is s, then 2 dt is ds. So, this reduces
00:06:57.490 --> 00:07:03.210
to this and therefore, is this right.
Now, applying integration by parts. The integral
00:07:03.210 --> 00:07:08.680
of this I have already computed for you
is, this minus e raise to minus 1 by 2 s and
00:07:08.680 --> 00:07:12.180
so, this will be t into, when I transform
back
00:07:12.180 --> 00:07:17.229
again substitute for s from here, this is
t square. So, then t into e raise to minus
00:07:17.229 --> 00:07:20.100
half t
square minus infinity to infinity, you can
00:07:20.100 --> 00:07:26.169
see, that this is 0, right, because this is
t square.
00:07:26.169 --> 00:07:32.110
And so, in the denominator t square e raise
t e raise to infinity is much, much larger
00:07:32.110 --> 00:07:35.770
than
t in the numerator. So, therefore, this will
00:07:35.770 --> 00:07:39.360
be 0. You will be left with this
And then, here again this is what, this is
00:07:39.360 --> 00:07:43.600
your in a normal p d f where your mu is 0
and
00:07:43.600 --> 00:07:48.810
sigma is 1. So, the 1 upon root 2 pi is missing.
So, therefore, this integral will be equal
00:07:48.810 --> 00:07:52.080
to
root 2 pi because with 1 upon root 2 pi this
00:07:52.080 --> 00:07:54.900
will become 1, right. So, therefore, this
whole
00:07:54.900 --> 00:07:59.530
thing is 1. So, this is, this integral is
equal to root 2 pi, so I write root 2 pi,
00:07:59.530 --> 00:08:02.870
which cancels
with this and this is sigma square. So, therefore,
00:08:02.870 --> 00:08:08.440
the variance of the random variable. So,
therefore, the parameters are, now it is very
00:08:08.440 --> 00:08:14.950
clear what the parameters denote, mu is the
mean and we say normal mu sigma square mu
00:08:14.950 --> 00:08:18.500
is the mean and sigma square is the
variance.
00:08:18.500 --> 00:08:25.440
So, we just saw, that normal distribution,
the parameter mu is the mean and sigma square
00:08:25.440 --> 00:08:32.709
is the variance. Now, suppose x is n mu sigma
square and we consider the random
00:08:32.709 --> 00:08:38.060
variable y equal to alpha x plus beta where
alpha is a positive number and alpha and beta
00:08:38.060 --> 00:08:44.070
are some real numbers, right. So, I mean,
I am continuing with the properties of the
00:08:44.070 --> 00:08:48.600
normal distribution.
So, if you want to find out the p d f of y,
00:08:48.600 --> 00:08:53.820
then we start with the cumulative distribution
function. So, probability y less than or equal
00:08:53.820 --> 00:08:57.080
to t is probability alpha x plus beta less
than
00:08:57.080 --> 00:09:03.330
or equal to t, which reduces to this and since
alpha is positive, the inequality remains
00:09:03.330 --> 00:09:06.980
in
that. So, this is t minus beta upon alpha,
00:09:06.980 --> 00:09:13.360
ok. And so, this is your cumulative distribution
function y, which is equal to the cumulative
00:09:13.360 --> 00:09:17.140
density function of x, but the parameter the
t
00:09:17.140 --> 00:09:23.800
replaces is, gets replaced by t minus beta
by alpha, right. So, now, if you differentiate
00:09:23.800 --> 00:09:32.160
both sides, that means, differentiate with
respect to t, then this will become the p
00:09:32.160 --> 00:09:33.170
d f,
right.
00:09:33.170 --> 00:09:40.680
And this is d dt of f x t minus beta by alpha,
which will be 1 by alpha into f x of t minus
00:09:40.680 --> 00:09:47.400
beta by alpha, right. The derivative of capital
F x will be the p d f of the random variable
00:09:47.400 --> 00:09:53.610
.x. And so, when you substitute, you write
down the expression for this, it will be 1
00:09:53.610 --> 00:09:56.890
upon
alpha 1 upon under root 2 pi into sigma e
00:09:56.890 --> 00:10:00.160
raise to minus 1 by 2 sigma square and your
t
00:10:00.160 --> 00:10:05.450
gets replaced by t minus beta by alpha minus
mu whole square. This simplify the
00:10:05.450 --> 00:10:10.810
expression, this gives you t minus beta minus
alpha mu whole square. Here, you have
00:10:10.810 --> 00:10:17.450
alpha sigma and this is 2, sorry, the alpha
in the denominator comes here. So, this will
00:10:17.450 --> 00:10:21.959
be, there will be a into, into alpha square
also.
00:10:21.959 --> 00:10:27.750
I hope you can read it, anyway I am speaking
it out, may be let me just rewrite it whole
00:10:27.750 --> 00:10:42.710
thing here. Yeah, this is 2 alpha square sigma
square. And so, by our definition of the
00:10:42.710 --> 00:10:50.360
normal p d f this will be n, the mean now
becomes beta plus alpha mu at the various
00:10:50.360 --> 00:10:54.190
becomes alpha square sigma square instead
of sigma, right.
00:10:54.190 --> 00:11:01.120
So, therefore, you see, that if you make this
transformation where x is normal mu sigma
00:11:01.120 --> 00:11:07.970
square, then for y, the expectation will become
alpha mu plus beta, which anyway, you
00:11:07.970 --> 00:11:14.720
can show from here also. This is alpha expectation
of x plus beta because beta is a
00:11:14.720 --> 00:11:22.430
constant and so, this is alpha mu plus beta.
And the variance will be, just think because
00:11:22.430 --> 00:11:26.550
the constant will not matter, since you will
see, when you write down the variance, you
00:11:26.550 --> 00:11:35.110
will write down this minus, I mean, this minus
this. So, beta will cancel, alpha will come
00:11:35.110 --> 00:11:40.460
outside and so, it will become square.
And so, either way you can verify, that the,
00:11:40.460 --> 00:11:45.240
for the, for this random variable then mean
will be beta plus alpha mu and the variance
00:11:45.240 --> 00:11:54.500
will be alpha square sigma square. So, you
can see, that you can carry on the properties
00:11:54.500 --> 00:11:56.180
of normal variant.
00:11:56.180 --> 00:11:57.180
..
00:11:57.180 --> 00:12:03.030
Quite easily, now immediate consequence of
this result is, that if x is n mu sigma square,
00:12:03.030 --> 00:12:11.190
then z, which you write as x minus mu by sigma
will be normal 0, 1, right, because what
00:12:11.190 --> 00:12:18.269
is happening? Your, here alpha is actually
1 by sigma. If you compare it with the
00:12:18.269 --> 00:12:29.380
expression alpha x plus beta and your beta
is minus mu by sigma. So, now if you
00:12:29.380 --> 00:12:36.480
substitute, because we said, that the mean
will become alpha mu plus beta for that one.
00:12:36.480 --> 00:12:45.600
So, here it will become mu by sigma minus
mu by sigma, so it is 0, right. And similarly,
00:12:45.600 --> 00:12:52.649
you can show, that the variance will be 1.
So, the transformation x minus mu by sigma
00:12:52.649 --> 00:12:58.550
results in a standard normal variant and which
we refer to as n 0 1.
00:12:58.550 --> 00:13:05.490
Now, we have tables for computing the various
probabilities for normal 0 1 and you see,
00:13:05.490 --> 00:13:12.480
that you can then compute the probabilities
for any random, normal random variant
00:13:12.480 --> 00:13:17.690
through this and that is what, because, because
of that transformation, right. And I will
00:13:17.690 --> 00:13:23.120
work out few examples to show you how it goes.
So, anyway this is the standard notation
00:13:23.120 --> 00:13:29.540
for, for a standard normal variant. This is
your probability minus infinity to x. So,
00:13:29.540 --> 00:13:31.220
that
means, this is a cumulative density function.
00:13:31.220 --> 00:13:37.850
So, this will be 1 upon root 2 pi minus
infinity to x e raise to minus half y square
00:13:37.850 --> 00:13:45.959
d y. So, this probability is given the notation.
Now, the tables are given for x non-negative,
00:13:45.959 --> 00:13:55.070
you know, for values of x going up to…
We will, we will also later on see, that we
00:13:55.070 --> 00:14:01.490
do not need the values to be tabled for very
large values of x. Then, for x less than 0
00:14:01.490 --> 00:14:05.089
we use symmetry of the p d f around because
00:14:05.089 --> 00:14:13.620
.this standard normal. So, this is, this is
symmetric about the origin, right. And this
00:14:13.620 --> 00:14:18.880
is
your, this is this, right.
00:14:18.880 --> 00:14:24.390
Now, the symmetry means, that if you have
x here, then the area to the right of this
00:14:24.390 --> 00:14:30.390
number is the same as the area to the left
of minus x. This is what symmetry means.
00:14:30.390 --> 00:14:36.170
Because this area, this area are equal, therefore,
and since this is half area and this is 0.5,
00:14:36.170 --> 00:14:42.350
so therefore, this shaded portion here same
as the shaded portion here, and so the formula
00:14:42.350 --> 00:14:49.709
is this. So, if you tabled your values for
x positive, then for x negative you can get
00:14:49.709 --> 00:14:55.860
by this
and we can verify this formula right away.
00:14:55.860 --> 00:15:02.350
If you want to compute phi of minus x, then
that is minus infinity to minus x of fx dx.
00:15:02.350 --> 00:15:11.610
Now, if you write y as minus x, then x becomes
minus y. So, the limits go from infinity
00:15:11.610 --> 00:15:19.850
to y, right, infinity to y f of minus y dy,
right, and there is a minus sign here. So,
00:15:19.850 --> 00:15:22.160
you
interchange the limits. This becomes y to
00:15:22.160 --> 00:15:29.769
infinity f of minus y dy, which is 1 of minus,
no, phi minus y, right, because this is now
00:15:29.769 --> 00:15:33.589
y to infinity. So, therefore, by the formula
this
00:15:33.589 --> 00:15:40.640
is this, but phi of minus y is phi of x. So,
therefore, I have shown you, that this is
00:15:40.640 --> 00:15:43.540
1 minus
phi x. So, this formula has been verified.
00:15:43.540 --> 00:15:50.990
So, now you can get values of the cumulative
density function for negative, positive,
00:15:50.990 --> 00:16:01.600
both of x, right. So, let us look at few examples.
Suppose x is normal (2, 4), that means,
00:16:01.600 --> 00:16:07.660
the mean is 2 and the variance is 4. So, then
you want to find the probability, that x is
00:16:07.660 --> 00:16:16.209
between 2 and 4. So, I will use that transformation.
So, here, sorry, this should be z, that
00:16:16.209 --> 00:16:19.769
means, what I am doing is, I show, I will
write in detail.
00:16:19.769 --> 00:16:29.579
I am subtracting 2 and dividing by 2, so less
than x minus 2 by 2 less than 4 minus 2 by
00:16:29.579 --> 00:16:37.600
2. So, this probability goes here, right.
And now, so this is 2 minus 2, 0. And this
00:16:37.600 --> 00:16:40.660
is your
standard normal variant, right. x minus mu
00:16:40.660 --> 00:16:46.620
by sigma is normal 0, 1 for which we always
have this notation of z, right. And this is
00:16:46.620 --> 00:16:53.780
4 minus 2 by 2, which is 1. So, this probability
you can compute in terms of the standard normal
00:16:53.780 --> 00:17:01.750
variant in this way and by our notation
this is phi of 1 minus phi of 0.
00:17:01.750 --> 00:17:08.639
And from tables I get, that phi of 1 is 0.8413
and phi 0 will be 0.5 because we have said,
00:17:08.639 --> 00:17:15.919
that the standard normal is symmetric, the
p d f is symmetric about the origin. So, this
00:17:15.919 --> 00:17:21.240
portion of the, so this area under the curve
will be equal to 0.5 and this will be also
00:17:21.240 --> 00:17:26.419
0.5.
So, phi 0 will always be 0.5 because this
00:17:26.419 --> 00:17:29.360
is the area for phi 0, right, less than or
equal to
00:17:29.360 --> 00:17:34.760
.x less than or equal to 0. So, that would
be half of the area, which is 0.5, right.
00:17:34.760 --> 00:17:37.090
So,
therefore, this is the probability.
00:17:37.090 --> 00:17:41.830
So, therefore, very conveniently, once you
have the tables available for this standard
00:17:41.830 --> 00:17:48.860
normal, you can compute it for any normal
variant, right. And again, find probability
00:17:48.860 --> 00:17:53.030
x
less than 0 for this variable. So, here again
00:17:53.030 --> 00:17:59.800
I make the transformation x minus 2 by 2 less
than, so this will be minus 2 by 2. So, this
00:17:59.800 --> 00:18:03.830
is probability z less than minus 1, so which
is
00:18:03.830 --> 00:18:10.900
phi of minus 1. And then, I use this formula,
this formula. So, phi of minus 1 is 1 of
00:18:10.900 --> 00:18:19.580
minus phi 1. phi 1, I already know, is 0.8413.
So, this is 1 minus 0.8413, which is
00:18:19.580 --> 00:18:21.880
0.1587, right.
.
00:18:21.880 --> 00:18:30.330
Now, when you want to compute absolute x minus
1 greater than 3, so what are you
00:18:30.330 --> 00:18:36.730
saying here? You are saying, that your x minus
1 should be greater than 3 and x minus 1
00:18:36.730 --> 00:18:44.080
should be less than, that means, x minus 1
should be less than minus 3 and x minus 1
00:18:44.080 --> 00:18:50.100
should be greater than 3, yeah. So, from here
what do you get, that x should be greater
00:18:50.100 --> 00:18:55.799
than 4. So, from 4 to infinity. And from here
you get, that x should be less than minus
00:18:55.799 --> 00:18:59.020
2.
So, that means, it should be from minus infinity
00:18:59.020 --> 00:19:04.970
to minus 2. So, that is what we have
done. I have written, that this event is equal
00:19:04.970 --> 00:19:12.120
to the, that means, x must be either here,
minus infinity to minus 2 or it must be in
00:19:12.120 --> 00:19:18.160
the set 4 to infinity. Well, I have used a
different notation here than here, does not
00:19:18.160 --> 00:19:19.160
matter.
00:19:19.160 --> 00:19:23.940
.Now, since these two sets are disjoint, you
can see, right, minus infinity to minus 2
00:19:23.940 --> 00:19:27.650
and
this is 4 to infinity. So, I can write the
00:19:27.650 --> 00:19:33.380
probability as the sum of the individual
probabilities and so, I get this. And again,
00:19:33.380 --> 00:19:37.960
I transform my variable x minus 2 upon 2.
So,
00:19:37.960 --> 00:19:43.080
this remains minus infinity, this is minus
2, so minus 2 by 2.
00:19:43.080 --> 00:19:48.720
And here, again I use the same transformation
to reduce it to standard normal variant and
00:19:48.720 --> 00:19:59.180
therefore, this becomes phi of minus 2, yeah,
because phi of minus infinity is 0 plus 1
00:19:59.180 --> 00:20:06.070
minus, this will be what, this is 1 and this
is infinity. So, here phi infinity is 1 and
00:20:06.070 --> 00:20:10.780
this is
minus phi of 1, right. Because yes, you all
00:20:10.780 --> 00:20:14.669
agreed, that this is phi of infinity, I mean,
this
00:20:14.669 --> 00:20:21.410
phi of infinity will be 1. So, therefore,
and phi of minus 2 you can write is 1 minus
00:20:21.410 --> 00:20:24.980
phi 2,
again by our formula for computing negative
00:20:24.980 --> 00:20:30.220
probabilities in terms of positive this thing
and therefore, this is it.
00:20:30.220 --> 00:20:37.740
So, I just substitute the values. phi 2 from
the tables is 0.9772, this is 0.51 and so,
00:20:37.740 --> 00:20:42.100
this is
the answer. So, one can go on and therefore,
00:20:42.100 --> 00:20:47.820
the whole idea would be, that you sit down,
calculate a few of such probabilities by yourself
00:20:47.820 --> 00:20:55.491
to become familiar.
Now, let us just take an example here. The
00:20:55.491 --> 00:20:59.370
annual rainfall in inches in a certain region
is
00:20:59.370 --> 00:21:06.500
normally distributed with mean 40 inches and
sigma, that is the standard, we call by the
00:21:06.500 --> 00:21:16.220
word, I did not name it, but sigma, this is
under root sigma square is referred to as
00:21:16.220 --> 00:21:26.429
standard deviation, this is also known as
standard deviation. So, standard deviation
00:21:26.429 --> 00:21:31.390
is 4
that means, variance is 16. What is the probability
00:21:31.390 --> 00:21:38.660
that starting with this year it will take
over 10 years before a year occurs having
00:21:38.660 --> 00:21:43.180
rainfall of over 50 inches?
So, let us understand the problem first. They
00:21:43.180 --> 00:21:47.820
are saying, that the annual rainfall is
normally distributed. That means, if every
00:21:47.820 --> 00:21:56.110
year you add up the total rainfall in that
particular region, then those numbers will
00:21:56.110 --> 00:22:03.230
be fitting a normal distribution, which has
mean 40 inches and standard deviation 4, right.
00:22:03.230 --> 00:22:10.070
So, the numbers that you get as the total
rainfall in a year for different years, so
00:22:10.070 --> 00:22:12.370
that has a normal distribution, right. Now,
they
00:22:12.370 --> 00:22:18.890
are asking for a probability, that starting
with this year you go on for 10 years and
00:22:18.890 --> 00:22:23.900
within
those 10 years, no year will have rainfall
00:22:23.900 --> 00:22:29.660
more than 50 inches. So, it is, you know,
actually compounded event, it is not a straight
00:22:29.660 --> 00:22:33.960
forward. So, let us see how we go about
computing this probability.
00:22:33.960 --> 00:22:39.230
So, to compute the probability x greater than
or equal to 50, I will do the same trick,
00:22:39.230 --> 00:22:44.860
reduce this probability in terms of a standard
normal variant. So, since mean rainfall was
00:22:44.860 --> 00:22:50.350
.40 inches, so x minus 40, standard deviation
was 4, so this now reduces to a standard
00:22:50.350 --> 00:22:56.179
normal variant. So, therefore, the required
probability is the same as probability z greater
00:22:56.179 --> 00:23:03.920
than or equal to 5 by 2, which is oh, oh,
oh this is 1 minus, sorry, this has to be
00:23:03.920 --> 00:23:16.350
z greater
than 5 by 2 would be 1 minus 5 phi by 2, right,
00:23:16.350 --> 00:23:26.940
and so, this will be 1 minus of this 1
minus 0.9938 and so, this will not be the
00:23:26.940 --> 00:23:28.650
required probability. We will have to write
the
00:23:28.650 --> 00:23:31.890
number, anyway.
So, what we will do is, yeah, I will continue
00:23:31.890 --> 00:23:37.830
to say, that p is 1 minus 0.9938 or maybe,
I
00:23:37.830 --> 00:23:53.010
will just do this. So, this is 1 minus 0.9938.
So, this is the probability of, is the
00:23:53.010 --> 00:23:58.250
probability of rainfall. So, the probability
would be standard normal variant greater than
00:23:58.250 --> 00:24:06.610
or equal 5 by 2, which will be 1 minus phi
of, phi by 2, 5 by 2 and that is 1 minus 0.9938
00:24:06.610 --> 00:24:11.680
because phi 5 by 2 is 0.9938. So, this is
probability is.
00:24:11.680 --> 00:24:17.230
Now, this we can look upon as the probability
of rainfall being more than 50 inches in a
00:24:17.230 --> 00:24:23.190
year and this we will treat as a success.
That means, now you can say, that in a year
00:24:23.190 --> 00:24:26.409
the
experiment is to measure the rain and if the
00:24:26.409 --> 00:24:28.820
rainfall is more than 50 inches, we treat
that
00:24:28.820 --> 00:24:35.770
is a success, right. So, that is what I said,
that the problem requires because the event
00:24:35.770 --> 00:24:36.770
is
complex.
00:24:36.770 --> 00:24:42.100
So, first we computed the probability of the
rainfall being more than 50 inches in a year,
00:24:42.100 --> 00:24:48.309
which came out to be this. Now, I want to
find out, that in 10 years, rainfall not more
00:24:48.309 --> 00:24:54.380
than 50 inches in any year. So, that means,
if I treat those 10 every year as a trial,
00:24:54.380 --> 00:24:57.130
that
means, the trial is, you add up the total
00:24:57.130 --> 00:24:59.690
number of rainfall in that year and then you
say,
00:24:59.690 --> 00:25:06.409
you are saying, that in 10 years there should
be none of the years, that you count from
00:25:06.409 --> 00:25:11.210
beginning from this year, the rainfall is
more than 50 inches. That means, in other
00:25:11.210 --> 00:25:15.120
words,
there is no success in these 10 trials and
00:25:15.120 --> 00:25:20.990
so, the probability of no success in 10 trials
would be 1 minus p raise to 10, which must
00:25:20.990 --> 00:25:27.440
be 0.9938 raise to 10. So, again please
compute this number because I had made a mistake.
00:25:27.440 --> 00:25:32.559
So, here this is, this will, whatever this
number, you can now use your calculators to
00:25:32.559 --> 00:25:39.490
compute this probability to say that that
will be the probability, that in 10 years
00:25:39.490 --> 00:25:43.520
the
rainfall will be less than 50. So, this is.
00:25:43.520 --> 00:25:46.419
So, I am just trying to show you, that how
you
00:25:46.419 --> 00:25:52.130
first use the normal of approximation, and
then I mean the standardization and then,
00:25:52.130 --> 00:25:57.570
you,
you use the binomial random variable to compute
00:25:57.570 --> 00:25:58.790
the actual probability.
00:25:58.790 --> 00:25:59.790
..
00:25:59.790 --> 00:26:04.760
So, median and mode of a normal distribution.
This, again see, that is why, I mean, when
00:26:04.760 --> 00:26:10.120
you look at all these properties of distribution,
you, you realize, that it, it is a very
00:26:10.120 --> 00:26:17.820
interesting one. And this is, suppose, what
we mean by the median of distribution is the
00:26:17.820 --> 00:26:28.250
number for which the, for which your cumulative
density function has the value 0.5. That
00:26:28.250 --> 00:26:35.650
means, the area of the, if you have this thing
here when this is the, this is the point where
00:26:35.650 --> 00:26:40.250
this is x naught. So, this area is 0.5 and
this area is 0.5.
00:26:40.250 --> 00:26:47.640
And since we have already said, that the normal
distribution is symmetric about x equal
00:26:47.640 --> 00:26:55.930
to mu, so that immediately clear, that the
area to the left of mu is 0.5 and area to
00:26:55.930 --> 00:27:00.211
the right
of mu is 0.5. So, immediately we know, that
00:27:00.211 --> 00:27:07.230
a median, the median is also at x equal to
mu, right. Now, to define the, to obtain the
00:27:07.230 --> 00:27:11.500
mode point, this is the point at which F x,
the
00:27:11.500 --> 00:27:16.920
probability density function attains its maximum
value. We did this for the binomial also,
00:27:16.920 --> 00:27:24.730
remember, and what was the number at which
the random variable attains its maximum
00:27:24.730 --> 00:27:28.720
probability.
So, now, in this case, we will have to find
00:27:28.720 --> 00:27:32.480
out the maximum, the value, the value of x
at
00:27:32.480 --> 00:27:38.470
which this function attains its maximum value.
So, this will be done by finding out the,
00:27:38.470 --> 00:27:45.890
differentiating it, finding out the critical
point and putting this equal to 0. The derivative,
00:27:45.890 --> 00:27:52.220
now you see, that here this is the derivative.
So, this portion is not 0. So, therefore,
00:27:52.220 --> 00:27:54.169
this is
the only portion, which will be 0, and therefore
00:27:54.169 --> 00:27:59.409
that gives you x is equal to mu. So, x, x
equal to mu is the point of maxima or minima
00:27:59.409 --> 00:28:01.620
essentially, or it could be a point of
00:28:01.620 --> 00:28:08.260
.inflection. But, so now, we have to look
at further the second order derivative to
00:28:08.260 --> 00:28:13.270
determine the nature of this critical point.
And I have written down the expression for
00:28:13.270 --> 00:28:18.789
second derivative, that means, you differentiate
this expression again and compute it at,
00:28:18.789 --> 00:28:24.880
evaluate it at x is equal to mu.
So, then you see this, this portion goes 0
00:28:24.880 --> 00:28:27.450
because x equal to mu and here also, when
you
00:28:27.450 --> 00:28:34.080
put x equal to mu, you essentially get this.
So, this is again e raise to 0, which is 1.
00:28:34.080 --> 00:28:38.320
So,
you simply get minus 1 upon 2 pi sigma square,
00:28:38.320 --> 00:28:44.659
which is less than 0. So, if, if the second
derivative has negative sign at a critical
00:28:44.659 --> 00:28:47.799
point, that point must be a point of maxima.
So,
00:28:47.799 --> 00:28:54.919
that much from calculus we can obtain. I am
sure, therefore, it is really interesting,
00:28:54.919 --> 00:28:59.230
that x
equal to mu is the mean, median and mode.
00:28:59.230 --> 00:29:05.800
So, it is all the things combined into 1.
Now, another important, as I told you, I have
00:29:05.800 --> 00:29:12.750
been, I have mentioned, that de Moivre was
used to, you initially defined this normal
00:29:12.750 --> 00:29:18.470
distribution to approximate binomial
probabilities. So, let us, formula is the
00:29:18.470 --> 00:29:27.330
procedure here. So, this is de Moivre Laplace
limit theorem, which is, that if S n is the
00:29:27.330 --> 00:29:36.590
number of successes in n trials of a binomial
random variable (n, p), then for any a less
00:29:36.590 --> 00:29:39.380
then b. So, here of course, for the binomial
the
00:29:39.380 --> 00:29:45.590
mean is n p and the variance is n p q, right.
So, for any a less then b if you want to look
00:29:45.590 --> 00:29:51.550
at the probability of a less than S n minus
n p upon root n p q less than or equal to
00:29:51.550 --> 00:29:54.049
b.
So, you see, what we have done is, we have
00:29:54.049 --> 00:30:02.500
standardized this binomial random variable
of successes in n trials, number of successes
00:30:02.500 --> 00:30:05.520
in n trials. So, it will be S n minus n p
upon
00:30:05.520 --> 00:30:12.870
and root n p q right. So, then it is the de
Moivres Laplace theorem says, that this
00:30:12.870 --> 00:30:19.610
probability can be approximated by phi b minus
phi a where phi is your cumulative
00:30:19.610 --> 00:30:27.669
distribution function for the standard normal
distribution or for the standard normal
00:30:27.669 --> 00:30:31.960
variant. So, this will be phi b minus phi
a as n goes to infinity.
00:30:31.960 --> 00:30:38.400
So, that means, for large n you can approximate
this probability by phi b minus phi a and
00:30:38.400 --> 00:30:44.840
later on, we will show, that this actually
is a very general, I mean, you can talk about
00:30:44.840 --> 00:30:47.940
a
more general result when you do not really
00:30:47.940 --> 00:30:55.880
need to have the distribution as binomial.
And for any distribution, this kind of thing,
00:30:55.880 --> 00:30:57.820
which we call as a central limit theorem,
we
00:30:57.820 --> 00:31:00.620
will talk about it later.
But right now, de Moivres Laplace limit theorem
00:31:00.620 --> 00:31:05.940
simply say, that if you have a binomial
random variable, then if you want to compute
00:31:05.940 --> 00:31:12.010
the probability, that S n minus n p upon
under root n p q. So, this lies between a
00:31:12.010 --> 00:31:15.030
and b, a strictly less than this, then this
can be
00:31:15.030 --> 00:31:20.950
.approximated by phi b minus phi a. This is
the Laplace theorem. So, now, we will use
00:31:20.950 --> 00:31:21.950
this.
.
00:31:21.950 --> 00:31:26.659
And so we say, that now suppose, you want
to compute the probability, that c is strictly
00:31:26.659 --> 00:31:32.140
less than S n is less than or equal to d and
here, of course, it is understood c is also
00:31:32.140 --> 00:31:36.440
less
strictly less than d, then we will standardize
00:31:36.440 --> 00:31:40.350
the whole inequality in the sense, that we
will transform.
00:31:40.350 --> 00:31:46.940
This to c minus n p upon under root n p q
and this will be less than S n minus n p upon
00:31:46.940 --> 00:31:52.100
under root n p q, which will be less than
or equal to d minus n p under root n p q.
00:31:52.100 --> 00:31:56.030
And
now, you see, this becomes standard normal
00:31:56.030 --> 00:32:00.320
variant and so, you’re a gets replaced by
this
00:32:00.320 --> 00:32:06.660
in the theorem and your b gets replaced by
this, right. And so, I can say, that by the
00:32:06.660 --> 00:32:09.710
de
Moivre’s Laplace theorem, that this probability,
00:32:09.710 --> 00:32:16.600
which is the same as this can be
approximated by phi of d minus n p upon under
00:32:16.600 --> 00:32:23.580
root n p q and phi of c minus n p upon
under root n p q. So, this is the whole idea.
00:32:23.580 --> 00:32:28.890
So, therefore, and these are, these values
are tabulated. So, given c d have n n p, you
00:32:28.890 --> 00:32:32.179
can
look up the tables and compute this probability.
00:32:32.179 --> 00:32:39.260
And of course, if you want to compute
the actual probability, then we will see through
00:32:39.260 --> 00:32:44.340
an example, that it can be very, very
cumbersome. So, in fact, this is a very useful
00:32:44.340 --> 00:32:51.039
theorem and it in a very simple way allows
you to compute, approximate these, compute
00:32:51.039 --> 00:32:53.140
this probability, right.
00:32:53.140 --> 00:33:01.270
.And of course, then we will continue with
the computations for approximating these
00:33:01.270 --> 00:33:09.090
probabilities and what they said is, that
this is good enough as long as your n p into
00:33:09.090 --> 00:33:11.590
1
minus p is greater than or equal to 10. That
00:33:11.590 --> 00:33:17.280
means, you do not require too many very
large values of n, but as long as, of course,
00:33:17.280 --> 00:33:19.100
if this number is larger than 10, you get
a
00:33:19.100 --> 00:33:24.830
better approximation and that you can also
for yourself experiment with problems where
00:33:24.830 --> 00:33:32.529
you try to increase the value of n and then
see, that your, the estimate will improve.
00:33:32.529 --> 00:33:34.929
So,
anyway, but this gives good approximation
00:33:34.929 --> 00:33:40.129
as long as n p into 1 minus p is greater than
or equal to 10.
00:33:40.129 --> 00:33:41.129
.
00:33:41.129 --> 00:33:47.549
Now, there is important aspect to this approximation
and that is the continuity correction
00:33:47.549 --> 00:33:54.779
factor, which I will discuss here. So, you
see, look at this curve. So, I have the binomial
00:33:54.779 --> 00:34:01.980
curve here, the bar chart. And then, the red
line curve is the normal approximation and
00:34:01.980 --> 00:34:07.970
you see, what we are saying here is, that
if you are asking for the probability x greater
00:34:07.970 --> 00:34:13.139
than c. This is the shaded portion that you
are looking for.
00:34:13.139 --> 00:34:20.470
So, here you want to say, that probability
x greater than 7, if you are asking for this
00:34:20.470 --> 00:34:26.529
probability x greater than 7, yeah. So, then
if x greater than 7 means, that x is greater
00:34:26.529 --> 00:34:32.079
than or equal to 8. So, to get that event,
to get that probability, if you want a good
00:34:32.079 --> 00:34:38.980
approximation, see we are the rectangle, the
bar, that represents the probability at 0.8.
00:34:38.980 --> 00:34:45.349
So, that is starting from 7.5. You want to
cover that whole area because you are wanting
00:34:45.349 --> 00:34:51.889
probability x greater than or equal to 8.
So, therefore, you will need to say, that
00:34:51.889 --> 00:34:53.230
your
00:34:53.230 --> 00:34:59.099
.approximate, approximating standard normal
variant should b greater than or equal to
00:34:59.099 --> 00:35:01.680
7.5
and not 8.
00:35:01.680 --> 00:35:06.660
So, because the, so you will want to say,
for the continuity factor you will say, that
00:35:06.660 --> 00:35:10.640
you
will do it from 7.5 because you want to use
00:35:10.640 --> 00:35:16.940
a, approx, include the area, which is at the
point a. The probability, which is represented
00:35:16.940 --> 00:35:24.839
by, at the point a by this bar, should get
added to the, to your estimate, right. And
00:35:24.839 --> 00:35:28.250
similarly, if you are wanting, let us say
I am.
00:35:28.250 --> 00:35:33.020
So, I have that table here, just look at the
example. So, if you have x equal to 6, that
00:35:33.020 --> 00:35:39.950
means, a single value, if you have x equal
to 6, then you would want it between 5.5 and
00:35:39.950 --> 00:35:46.170
6.5 because that is the rectangle, the bar
that you constructed at 6 that stands at,
00:35:46.170 --> 00:35:51.219
from 5.5
to 6.5 and the length, the height is the probability
00:35:51.219 --> 00:35:55.980
of that x equal to 6, right.
So, therefore, the continuity factor, that
00:35:55.980 --> 00:35:58.920
you require would be for probability x equal
to 6,
00:35:58.920 --> 00:36:05.869
it will be 5.5 and 6.5. Your x must vary,
then because see, the discrete situation,
00:36:05.869 --> 00:36:09.579
you are
now approximating by a continuous situation.
00:36:09.579 --> 00:36:15.329
Then, as I said, x greater than 6 would be
x greater than 6.5 and if x is greater than
00:36:15.329 --> 00:36:17.969
or equal to 6, then you are including 6, then
you
00:36:17.969 --> 00:36:22.579
will have to go little further down, right.
That means, you will have to start from 5.5.
00:36:22.579 --> 00:36:25.380
If it
is x greater than or equal to 6, then you
00:36:25.380 --> 00:36:28.519
will begin from 5.5. So, x would be greater
than
00:36:28.519 --> 00:36:33.979
5.5 if x is less than 6. That means, x is
less than or equal to 5, then again you will
00:36:33.979 --> 00:36:37.369
say,
that x is less than 5.5. So, that is very
00:36:37.369 --> 00:36:41.619
clear from this graph. And x less than or
equal to 6
00:36:41.619 --> 00:36:45.510
would be x again.
See, the moment you have equal, then you have
00:36:45.510 --> 00:36:49.999
to go a little ahead of it, 6.5 and if it,
it is
00:36:49.999 --> 00:36:54.180
strict in equality, then. So, the rule is
very clear. And by looking at this figure
00:36:54.180 --> 00:36:56.950
you can
always… So, if you keep this in mind, then
00:36:56.950 --> 00:37:01.299
you will not go wrong because you will
realize, that you have to, you have to include
00:37:01.299 --> 00:37:05.720
the area under the bar or the rectangle,
which is for that particular value, the limiting
00:37:05.720 --> 00:37:10.959
end value. And so, you will have to
accordingly to give the correction factor,
00:37:10.959 --> 00:37:16.019
right. So, this is called the continuity correction
factor because you are estimating a discrete
00:37:16.019 --> 00:37:22.789
situation by this thing.
So, if you use the continuity correction factor,
00:37:22.789 --> 00:37:30.479
now let, let us give this thing, let us give,
let me give you, show you the calculations
00:37:30.479 --> 00:37:37.849
through an example. And so, let me show you
an example through an example how we, what
00:37:37.849 --> 00:37:40.079
I mean by the continuity correction factor.
00:37:40.079 --> 00:37:46.239
.So, let x be the number of times a fair coin,
so fair coin means, that the probability of
00:37:46.239 --> 00:37:52.400
showing a head and a tale are the same and
that equals half, right. So, lands heads when
00:37:52.400 --> 00:37:59.170
it is flipped 49 times. x is the number of
times head shows up when I have flipped fair
00:37:59.170 --> 00:38:06.519
coin 49 times. Now, find the probability,
that x is equal to 25. Use the normal
00:38:06.519 --> 00:38:14.309
approximation and then compare it with the
exact probability. So, I want to compute
00:38:14.309 --> 00:38:19.170
probability x equal to 25.
And as I told, I just discussed with you,
00:38:19.170 --> 00:38:25.829
that since this is the binomial random variable,
you are trying to approximate it by a normal
00:38:25.829 --> 00:38:29.910
distribution. So, the bar is actually, when
x
00:38:29.910 --> 00:38:40.469
equal to 25, the bar starts from, 20, 24.5
and ends at 25.5. So, this is the bar, right,
00:38:40.469 --> 00:38:42.859
and
this, the height is the probability that you
00:38:42.859 --> 00:38:47.509
associate with x equal to 25. So, 24.5 and
25.5.
00:38:47.509 --> 00:38:53.619
So, therefore I have to change the two approximate
again. I should say this, that the
00:38:53.619 --> 00:38:58.910
approximate this thing will be 24.5 less than
or equal to x less than or equal to 25.5.
00:38:58.910 --> 00:39:03.190
So, I
am now approximating this event by this event,
00:39:03.190 --> 00:39:11.130
right, because of the continuity factor
and then, because the mean, yeah, I am sorry,
00:39:11.130 --> 00:39:14.250
this should not be 49. I have to write it
correctly here.
00:39:14.250 --> 00:39:20.259
So, now for this binomial random variable,
your number of times, you flipping the coin
00:39:20.259 --> 00:39:29.839
is 49 probability, that your p is half. So,
therefore, n p is 49 into 1 by 2, which is
00:39:29.839 --> 00:39:35.970
24.5,
right. So, it should be 24.5, right and your
00:39:35.970 --> 00:39:48.400
variance is n p q. So, that is 49 into 4 divided
by, divided by 4. And so, under root of that
00:39:48.400 --> 00:39:51.469
would be and that is why, I choose this 49
to
00:39:51.469 --> 00:39:58.440
make it perfect square. So, that is 7 by 2.
So, this is now standardized. So, x minus
00:39:58.440 --> 00:40:03.799
24.5
divided by 7 by 2 would make it standard normal.
00:40:03.799 --> 00:40:09.650
And so, this is the event that you want
to, you want to now find out the probability.
00:40:09.650 --> 00:40:13.190
So, this is this, right. And therefore, this
is
00:40:13.190 --> 00:40:20.559
equal to 0 and this is 1 upon 2 by 7. So,
therefore, this is equal to phi of 2 by 7
00:40:20.559 --> 00:40:21.890
minus phi
0.
00:40:21.890 --> 00:40:30.440
Now, this is phi of 0.28 minus phi 0, phi,
phi of 0.28 from the normal tables is equal
00:40:30.440 --> 00:40:36.529
to
0.6103 minus 0.5, right. And so, this comes
00:40:36.529 --> 00:40:45.829
out to be 0.1103. So, this is our probability
that we have obtained for x equal to 25 through
00:40:45.829 --> 00:40:52.500
normal approximation. And remember,
the condition was, that your n p q must be
00:40:52.500 --> 00:40:59.650
greater than or equal to 10.
So, in our case our n p q is how much? n p
00:40:59.650 --> 00:41:09.490
q number is 49 by 4. So, which is more than
10, right, that is, our n p q is 49 by 4,
00:41:09.490 --> 00:41:14.729
which is equal to 42, sorry, 12 point something,
12
00:41:14.729 --> 00:41:23.680
.4’s are 48. So, 0.25. In fact, so this
is more than 10, right. And so, we applied
00:41:23.680 --> 00:41:26.170
the
approximation and this is the result that
00:41:26.170 --> 00:41:29.739
we got, right.
Now, exact probability if you want to compute
00:41:29.739 --> 00:41:36.930
using your binomial computations, then
you see, it is actually 49 choose 25 and 1
00:41:36.930 --> 00:41:41.259
by 2 raise to 49, 49 trials and your p and
q are
00:41:41.259 --> 00:41:48.640
the same. If you write out this number, 49
into 48 up to 25 because that will be 49 minus
00:41:48.640 --> 00:41:56.479
25 plus 1 and then, 20 factorial 1 upon 2
raise to 49. It took me almost half an hour
00:41:56.479 --> 00:42:00.749
to
compute from, sitting at the computers and
00:42:00.749 --> 00:42:06.599
this number comes out to be 0.11275.
So, if you just look at this here and that
00:42:06.599 --> 00:42:09.650
is what I am saying, now you compare it with,
so
00:42:09.650 --> 00:42:15.509
third decimal place, the number differs the
value differs and therefore, I would say,
00:42:15.509 --> 00:42:18.640
that
this is a good approximation. And here, you
00:42:18.640 --> 00:42:28.369
see our n p q was only, 10, 12.25. And if
you take larger n, that means, if you had
00:42:28.369 --> 00:42:35.390
flipped the coin more than 49, then this number
would have improved.
00:42:35.390 --> 00:42:41.479
And just see the phenomenal calculations you
had to do even for n equal to 49, I mean,
00:42:41.479 --> 00:42:48.019
multiplying these numbers, 24 numbers getting
25 factorial multiplying 2 49 times and
00:42:48.019 --> 00:42:52.599
then dividing. So, this this the amount of
calculation, that you would have to do for
00:42:52.599 --> 00:42:56.450
the
exact probability certainly is not required.
00:42:56.450 --> 00:42:59.949
If you can approximate this probability by
this
00:42:59.949 --> 00:43:05.259
simple method because this table, these values
are already available to you. And so, this
00:43:05.259 --> 00:43:13.019
is what I am trying to say, that you know,
as we go on, you will see the numerous
00:43:13.019 --> 00:43:18.459
applications, uses of this concept of normal
distribution and its computations.
00:43:18.459 --> 00:43:19.459
.
00:43:19.459 --> 00:43:26.569
.So, I will continue with the examples of
you know, binomial, the normal approximation
00:43:26.569 --> 00:43:32.290
of binomial probabilities and this is, these
are two good examples from the book by Ross
00:43:32.290 --> 00:43:36.410
Sheldon. As I told you, that this is the book
on probability theory and at the end of the
00:43:36.410 --> 00:43:42.080
course I will give you all these references.
So, I just thought I will, you know, give
00:43:42.080 --> 00:43:46.440
you
feeling of, you know, some more examples to,
00:43:46.440 --> 00:43:53.890
to reinforce the idea that the
approximations of discrete probabilities can
00:43:53.890 --> 00:44:00.069
be done by the continuous random variables
and in the very effective way. So, let us
00:44:00.069 --> 00:44:05.640
look at this example.
The ideal size of a 1st year class at a particular
00:44:05.640 --> 00:44:10.759
college is 150 students. From past
experience the college knows, that on the
00:44:10.759 --> 00:44:16.430
average, 30 percent of those accepted for
admission will actually attend the college.
00:44:16.430 --> 00:44:21.519
So, therefore, you know, people tend to apply
to more than one college and then they, of
00:44:21.519 --> 00:44:27.239
course, decide which is the best one for them?
So, the college has the experience, that only
00:44:27.239 --> 00:44:31.809
30 percent of the people who have been
accepted for admission will actually attend
00:44:31.809 --> 00:44:40.339
the college. So, therefore, the college adopts
the policy of giving admission to 450, 450
00:44:40.339 --> 00:44:46.660
students, so that finally, when they, you
know, drop out, after the dropout rate, they
00:44:46.660 --> 00:44:49.460
will still be left with the, with the class
of
00:44:49.460 --> 00:44:56.859
size 150. This is the idea. So, they decide
to offer admission to 450 students. Now,
00:44:56.859 --> 00:45:02.859
compute the probability, that more than 150
1st year students attend the class, right.
00:45:02.859 --> 00:45:10.239
So, given that 30 percent is the, so this
is the probability of a person attending who
00:45:10.239 --> 00:45:12.329
has
been given admission, will attend college
00:45:12.329 --> 00:45:16.289
is 0.3. So, therefore we want to find out.
So,
00:45:16.289 --> 00:45:22.979
here, of course, we will say x is the number
of students who attend college and so, the
00:45:22.979 --> 00:45:31.520
value of x will be equal to the, we can treat
the person who has been given admission and
00:45:31.520 --> 00:45:37.640
attends college as a success. So, out of 450
people x will be the number of students who
00:45:37.640 --> 00:45:44.390
attend college. So, this would be a binomial
random variable with n as your 450 and your
00:45:44.390 --> 00:45:52.549
p as 0.3 right. So, this is binomial (450,
0.3), because we are treating, that the experiment
00:45:52.549 --> 00:45:57.130
has been performed. That means, admissions
have been offered to 450 students.
00:45:57.130 --> 00:46:02.079
And out of this those who attend, who all
actually come attend the college is a success.
00:46:02.079 --> 00:46:07.859
So, the number of successes we are saying
is a binomial random variable, right. So,
00:46:07.859 --> 00:46:12.749
therefore, you have to compute the probability.
So, actually you want that the class size
00:46:12.749 --> 00:46:17.250
should be, so you have to compute the probability
of more than 150. That means, it
00:46:17.250 --> 00:46:22.579
should be 151, 152 and so on. So, therefore,
when you write probability x greater than
00:46:22.579 --> 00:46:25.589
or
equal to 151, then your continuity factor
00:46:25.589 --> 00:46:29.529
when you apply, then this will become 150.5,
00:46:29.529 --> 00:46:33.770
.right because you are actually asking for
the probability, that x is greater than or
00:46:33.770 --> 00:46:37.489
equal to
151 because more than 150.
00:46:37.489 --> 00:46:44.449
So, therefore, we just standardize this, you
know, this variant and that would subtract
00:46:44.449 --> 00:46:49.480
x
minus n p. n p is 450 into 0.3 and then divide
00:46:49.480 --> 00:46:54.589
it by n p q. So, 450 into 0.3 into 0.7 and
do
00:46:54.589 --> 00:46:59.290
this to the right hand side also. So, you
get this and then, of course, the advantage
00:46:59.290 --> 00:47:01.849
of
taking in solved example is, that you have
00:47:01.849 --> 00:47:04.660
the calculations done for you. So, here this
is
00:47:04.660 --> 00:47:09.900
actually the phi of 1.59. So, this number
reduces to 1.5.
00:47:09.900 --> 00:47:15.849
So, therefore, this is your standard normal
probability from minus infinity to, sorry,
00:47:15.849 --> 00:47:18.319
this
is, yeah, to 1.59. So, we are writing this
00:47:18.319 --> 00:47:22.910
as 1 minus phi of, so this number is 1.59.
So, this
00:47:22.910 --> 00:47:32.949
required probability is 1 minus phi of 1.59,
which comes out to be this, 0.559. So, hence
00:47:32.949 --> 00:47:40.229
this reduces to 6 percent of the time more
than 150 of the 450 accepted students will
00:47:40.229 --> 00:47:45.450
attend college. So, the college is in good
situation, because it is only 6 percent chance,
00:47:45.450 --> 00:47:51.260
that more than 150 people will actually come
and attend the college those who have been
00:47:51.260 --> 00:47:53.660
offered admission. So, this was one situation.
.
00:47:53.660 --> 00:48:00.439
Now, another interesting problem. This is,
see to determine the effectiveness of a certain
00:48:00.439 --> 00:48:06.559
diet in reducing the amount of cholesterol
in the blood stream, 100 people are put on
00:48:06.559 --> 00:48:10.439
a
diet. After they have been on the diet for
00:48:10.439 --> 00:48:17.710
a sufficient length of time, their cholesterol
count will be taken. The nutritionist running
00:48:17.710 --> 00:48:25.430
the experiment has decided to endorse the
diet if at least 65 percent of the people
00:48:25.430 --> 00:48:27.849
have a lower cholesterol count, right.
00:48:27.849 --> 00:48:33.439
.So, after the trial period, at the end of
the trial period, you will again take a, test
00:48:33.439 --> 00:48:36.329
their
cholesterol in the blood stream. And if the
00:48:36.329 --> 00:48:43.910
count is lower for 65 percent of the people,
then after going on the diet, so the nutritionist
00:48:43.910 --> 00:48:49.700
will, have I made the sentence complete,
running the experiment has decided to endorse,
00:48:49.700 --> 00:48:56.779
so the nutritionist will endorse the diet
and say, yes, it is thus proven to lower the
00:48:56.779 --> 00:49:02.859
cholesterol in the blood.
So, now what we want to find out is, what
00:49:02.859 --> 00:49:05.420
is the probability, that the nutritionist
will
00:49:05.420 --> 00:49:12.779
endorse the diet. That the nutritionist endorses
the diet, will endorse, there I write, will,
00:49:12.779 --> 00:49:17.730
will endorse the diet if, in fact, it has
no effect on the cholesterol level.
00:49:17.730 --> 00:49:22.390
So, what we are saying is that suppose the
diet has had no effect on the cholesterol,
00:49:22.390 --> 00:49:25.339
but
what is the probability, that the nutritionist
00:49:25.339 --> 00:49:29.640
will still endorse it. So, therefore, now
the
00:49:29.640 --> 00:49:34.950
way we are arguing out and so, we have to
now decide how to model the situation. And
00:49:34.950 --> 00:49:41.670
the idea here is, that you know, it is either
way. See, people may on their own have their
00:49:41.670 --> 00:49:46.789
cholesterol count come down and so, the chance
of that happening is, we know, equally
00:49:46.789 --> 00:49:52.459
likely, either your cholesterol count goes
up or it comes down. So, that is the equally
00:49:52.459 --> 00:49:59.049
likely situation and therefore, it is being
said that ok, you can just take p equal to
00:49:59.049 --> 00:50:02.880
half.
So, therefore we are assuming, that the diet
00:50:02.880 --> 00:50:08.740
has had no affect on the cholesterol level.
So, but the since we, the count is going to
00:50:08.740 --> 00:50:11.660
be taken after the trial period and then,
if it
00:50:11.660 --> 00:50:16.740
turns out that 65 percent of people have lowered
their count and they will be, then the
00:50:16.740 --> 00:50:21.549
diet will be endorsed . So, therefore, we
will work with p equal to half. So, I hope
00:50:21.549 --> 00:50:24.460
it is
clear. So, what we are assuming is, that chance
00:50:24.460 --> 00:50:29.079
of the count being going up and down is
equally likely. So, therefore, we will take
00:50:29.079 --> 00:50:37.150
p to be half. And x is the number of people
whose, whose cholesterol level is lower, right.
00:50:37.150 --> 00:50:41.410
Then, see what, what we are looking for
is, so the binomial random variable.
00:50:41.410 --> 00:50:49.420
And therefore, this is, this is the probability,
sigma 65 to 100 100 i half raise to 100,
00:50:49.420 --> 00:51:00.619
right, because r and n minus r, both are will
up to end. This is p is half. So, this is
00:51:00.619 --> 00:51:03.119
nothing
and of course, you can see, that this is ((Refer
00:51:03.119 --> 00:51:05.670
Time: 51:03)) stupendous task, you know,
trying to compute it this way. So, therefore
00:51:05.670 --> 00:51:10.119
we will say, that essentially, we are looking
for probability x greater than or equal to
00:51:10.119 --> 00:51:18.329
65 and again add the correction, continuity
correction factor, so it will be 64.5 and
00:51:18.329 --> 00:51:22.959
then n p would be 100 into half and, and p
q will
00:51:22.959 --> 00:51:29.249
be 100 into half into half, which becomes
25. So, under root 5 and this is 50. So, this
00:51:29.249 --> 00:51:32.369
is
what you are looking for and this number comes
00:51:32.369 --> 00:51:33.509
out to be 2.9.
00:51:33.509 --> 00:51:39.759
.So, therefore, it is the required probability
is 1 minus phi of the normal standard normal
00:51:39.759 --> 00:51:49.089
probability 2.9 from minus infinity to 2.9
and this comes out to be 0.0019. So, which
00:51:49.089 --> 00:51:52.589
is
very small and therefore, the chance that
00:51:52.589 --> 00:51:57.619
you know, with p as half the chance, that
the 65
00:51:57.619 --> 00:52:03.719
people out of those 100 will have their cholesterol
lowered is very low and therefore, the
00:52:03.719 --> 00:52:08.559
diet will not get endorsed.
So, anyway because the diet was not having
00:52:08.559 --> 00:52:11.009
as, as we said, that probably the diet has
no
00:52:11.009 --> 00:52:16.269
affect on the cholesterol level, so therefore,
it does not get endorsed. So, no loss, nobody
00:52:16.269 --> 00:52:22.209
is lost. So, this is how, you know, one can
go on and look at different situations and
00:52:22.209 --> 00:52:25.319
then
try to see. So, I just thought, that these
00:52:25.319 --> 00:52:31.499
two will also add to your, you know, experience
of handling, you know, these problems and
00:52:31.499 --> 00:52:41.180
also reinforced the idea of, you know,
computing, you know computing these, what
00:52:41.180 --> 00:52:48.739
shall I say, messy probabilities by you
know, approximating through continuous random
00:52:48.739 --> 00:52:57.869
variables and making your task easy.
00:52:57.869 --> 00:52:58.869
.