00:01
Once again, welcome for new problem.
00:04
This time we're dealing with probability.
00:08
And under probability, we have quantification of chance processes.
00:18
So we're quantifying chance processes.
00:21
For example, probability of heads becomes one half in probability of tails also becomes one half.
00:28
These are equally likely events.
00:34
So these events are equally likely.
00:38
And we could also have what we call probability distribution.
00:44
So we can have a probability distribution.
00:51
And when you think about a probability distribution, this involves a discrete, a discrete.
01:00
Create random variable x where each possible value value of x is accompanied with probability.
01:28
So it's accompanied with its probability.
01:31
And so the value the x takes one takes one value during a trial experiment and the other thing that happens with probability distribution is that each probability is between 0 and 1.
02:03
And one so that's a requirement a fast requirement for the probability and the second requirement is the sum of all the probabilities has to be equivalent to one or equal to one if you if you want to say that we can talk about the mean of the probability distribution which is what you call the expected value and so the expected value of the probability distribution is new.
02:36
Remember, new stands for population.
02:40
Population means.
02:42
So if it's new, then we have sum of x, px.
02:46
Remember, x is the random variable, x is the random variable, and px is the probability.
02:57
So we have both the probability and the random variable that's the expected value for for the progbilt distribution and then besides the expected value we could also compute the variance of the probability distribution and the variance is the sum of x minus nm squared p of x or we could also have another formula where we take a sum of x squared p of x minus mu squared sum of x squared p of x minus mu squared so those are different types of formulas and we could also have a formula for the standard deviation for this type of distribution so we have a standard deviation and the standard deviation happens to be sum of x so we're taking it's pretty much the square of the variance, the standard deviation is the square root of the variance.
04:11
So this is sum of x squared px minus n squared you can see that.
04:22
So now we have a new problem and in this particular problem we have a couple of questions that we're looking for.
04:30
We have a series of questions that we're looking for.
04:34
And remember x stands for the number of children in randomly selected south african households and then p of x is the probability of this selection of probability of hunting the x the value which is a specific number of children that's the probability that we're looking for so in a table we have x and we have p of x so we have one two three four and five children and the probability is point two five point with me we have 0 .25, point 33.
05:58
We also have 0 .5, and then we have 0 .17.
06:12
No, we have point right here, we have 0 .17.
06:19
So, 0 .17.
06:30
Of course, we do have 0 .15, and we also have 0 .10.
06:38
So if you sum up these probability, you see that each one of the probabilities falls between 0 and 1.
06:50
So we've looked at that.
06:53
And then this is 5 and 2 is 8, 8 and 7 is 15, 15 and 5 is 20.
07:00
2 and 2 is 4, 4 and 3 is 7, and then 10.
07:05
So it fits the deal.
07:07
And we have a bunch of questions that we want to respond to the first one determine the probability that they have two or more children who have more than two children not two or more, we'll have more than two children, determine the probability that so this is the probability that that a household has two, has more than two children.
07:48
You want to say more than two children, more than two children.
07:58
And then in number two, we want to find the probability that the household will have between two and four children, inclusive...