PL9fwy3NUQKwaIc7KWc8buW344941GStQL/QKT3u32x4wE_raw.srt

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Today, inshallah, we'll start chapter six. Chapter

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six talks about the normal distribution. In this

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chapter, there are mainly two objectives. The

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first objective is to compute probabilities from

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normal distribution. And mainly we'll focus on

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objective number one. So we are going to use

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normal distribution in this chapter. And we'll

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know how can we compute probabilities if the data

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set is normally distributed. You know many times

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you talked about extreme points or outliers. So

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that means if the data has outliers, that is the

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distribution is not normally distributed. Now in

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this case, If the distribution is normal, how can

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we compute probabilities underneath the normal

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curve? The second objective is to use the normal

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probability plot to determine whether a set of

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data is approximately normally distributed. I mean

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beside box plots we discussed before. Beside this

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score, how can we tell if the data point or

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actually the entire distribution is approximately

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normally distributed or not. Before we learn if

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the point is outlier by using backsplot and this

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score. In this chapter we'll know how can we

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determine if the entire distribution is

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approximately normal distributed. So there are two

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objectives. One is to compute probabilities

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underneath the normal curve. The other, how can we

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tell if the data set is out or not? If you

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remember, first class, we mentioned something

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about data types. And we said data has mainly two

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types. Numerical data, I mean quantitative data.

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and categorical data, qualitative. For numerical

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data also it has two types, continuous and

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discrete. And discrete takes only integers such as

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number of students who take this class or number

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of accidents and so on. But if you are talking

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about Age, weight, scores, temperature, and so on.

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It's continuous distribution. For this type of

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variable, I mean for continuous distribution, how

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can we compute the probabilities underneath the

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normal? So normal distribution maybe is the most

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common distribution in statistics, and it's type

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of continuous distribution. So first, let's define

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continuous random variable. maybe because for

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multiple choice problem you should know the

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definition of continuous random variable is a

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variable that can assume any value on a continuous

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it can assume any uncountable number of values. So

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it could be any number in an interval. For

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example, suppose your ages range between 18 years

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and 20 years. So maybe someone of you, their age

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is about 18 years, three months. Or maybe your

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weight is 70 kilogram point five, and so on. So

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it's continuous on the variable. Other examples

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for continuous, thickness of an item. For example,

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the thickness.

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This one is called thickness. Now, the thickness

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may be 2 centimeters or 3 centimeters and so on,

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but it might be 2.5 centimeters. For example, for

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this remote, the thickness is 2.5 centimeters or 2

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.6, not exactly 2 or 3. So it could be any value.

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Range is, for example, between 2 centimeters and 3

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centimeters. So from 2 to 3 is a big range because

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it can take anywhere from 2.1 to 2.15 and so on.

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So thickness is an example of continuous random

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variable. Another example, time required to

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complete a task. Now suppose you want to do an

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exercise. Now the time required to finish or to

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complete this task may be any value between 2

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minutes up to 3 minutes. So maybe 2 minutes 30

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seconds, 2 minutes 40 seconds and so on. So it's

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continuous random variable. Temperature of a

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solution. height, weight, ages, and so on. These

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are examples of continuous random variable. So

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these variables can potentially take on any value

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depending only on the ability to precisely and

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accurately measure. So that's the definition of

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continuous random variable. Now, if you look at

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the normal distribution, It looks like bell

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-shaped, as we discussed before. So it's bell

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-shaped, symmetrical. Symmetrical means the area

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to the right of the mean equals the area to the

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left of the mean. I mean 50% of the area above and

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50% below. So that's the meaning of symmetrical.

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The other feature of normal distribution, the

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measures of center tendency are equal or

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approximately equal. Mean, median, and mode are

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roughly equal. In reality, they are not equal,

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exactly equal, but you can say they are

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approximately equal. Now, there are two parameters

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describing the normal distribution. One is called

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the location parameter. location, or central

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tendency, as we discussed before, location is

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determined by the mean mu. So the first parameter

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for the normal distribution is the mean mu. The

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other parameter measures the spread of the data,

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or the variability of the data, and the spread is

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sigma, or the variation. So we have two

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parameters, mu and sigma. The random variable in

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this case can take any value from minus infinity

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up to infinity. So random variable in this case

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continuous ranges from minus infinity all the way

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up to infinity. I mean from this point here up to

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infinity. So the values range from minus infinity

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up to infinity. And if you look here, the mean is

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located nearly in the middle. And mean and median

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are all approximately equal. That's the features

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or the characteristics of the normal distribution.

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Now, how can we compute the probabilities under

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the normal killer? The formula that is used to

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compute the probabilities is given by this one. It

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looks complicated formula because we have to use

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calculus in order to determine the area underneath

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the cube. So we are looking for something else. So

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this formula is it seems to be complicated. It's

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not hard but it's complicated one, but we can use

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it. If we know calculus very well, we can use

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integration to create the probabilities underneath

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the curve. But for our course, we are going to

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skip this formula because this

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formula depends actually on mu and sigma. A mu can

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take any value. Sigma also can take any value.

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That means we have different normal distributions.

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Because the distribution actually depends on these

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two parameters. So by varying the parameters mu

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and sigma, we obtain different normal

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distributions. Since we have different mu and

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sigma, it means we should have different normal

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distributions. For this reason, it's very

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complicated to have tables or probability tables

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in order to determine these probabilities because

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there are infinite values of mu and sigma maybe

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your edges the mean is 19. Sigma is, for example,

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5. For weights, maybe the mean is 70 kilograms,

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the average is 10. For scores, maybe the average

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is 65, the mean is 20, sigma is 20, and so on. So

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we have different values of mu and sigma. For this

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reason, we have different normal distributions.

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Because changing mu shifts the distribution either

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left or to the right. So maybe the mean is shifted

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to the right side, or the mean maybe shifted to

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the left side. Also, changing sigma, sigma is the

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distance between the mu and the curve. The curve

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is the points, or the data values. Now this sigma

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can be increases or decreases. So if sigma

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increases, it means the spread also increases. Or

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if sigma decreases, also the spread will decrease.

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So the distribution or the normal distribution

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depends actually on these two values. For this

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reason, since we have too many values or infinite

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values of mu and sigma, then in this case we have

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different normal distributions. There is another

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distribution. It's called standardized normal.

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Now, we have normal distribution X, and how can we

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transform from normal distribution to standardized

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normal distribution? The reason is that the mean

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of Z, I mean, Z is used for standardized normal.

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The mean of Z is always zero, and sigma is one.

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Now it's a big difference. The first one has

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infinite values of Mu and Sigma. Now, for the

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standardized normal distribution, the mean is

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fixed value. The mean is zero, Sigma is one. So,

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the question is, how can we actually transform

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from X, which has normal distribution, to Z, which

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has standardized normal with mean zero and Sigma

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of one. Let's see. How can we translate x which

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has normal distribution to z that has standardized

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normal distribution? The idea is you have just to

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subtract mu of x, x minus mu, then divide this

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result by sigma. So we just subtract the mean of

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x. and dividing by its standard deviation now so

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if we have x which has normal distribution with

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mean mu and standard deviation sigma to transform

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or to convert to z score use this formula x minus

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the mean then divide by its standard deviation now

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all of the time we are going to use z for

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standardized normal distribution and always z has

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mean zero and all and sigma or standard deviation.

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So the z distribution always has mean of zero and

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sigma of one. So that's the story of standardizing

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the normal value. Now the Formula for this score

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becomes better than the first one, but still we

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have to use calculus in order to determine the

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probabilities under the standardized normal k. But

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this distribution has mean of zero and sigma of

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one. So we have a table on page 570. Look at page

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570. We have table or actually there are two

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tables. One for negative value of Z and the other

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for positive value of Z. So we have two tables for

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positive and negative values of Z on page 570 and

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571.

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Now the table on page 570 looks like this one. The

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table you have starts from minus 6, then minus 5,

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minus 4.5, and so on. Here we start from minus 3.4

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all the way down up to 0. Look here, all the way

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up to 0. So these scores here. Also we have 0.00,

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0.01, up to 0.09. Also, the other page, page 571,

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gives the area for positive z values. Here we have

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0.0, 0.1, 0.2, all the way down up to 3.4 and you

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have up to 6. Now let's see how can we use this

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table to compute the probabilities underneath the

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normal curve.

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First of all, you have to know that Z has mean

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zero, standard deviation of one. And the values

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could be positive or negative. Values above the

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mean, zero, have positive Z values. The other one,

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values below the mean, have negative Z values. So

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Z score can be negative or positive. Now this is

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the formula we have, z equals x minus mu divided

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by six.

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Now this value could be positive if x is above the

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mean, as we mentioned before. It could be a

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negative if x is smaller than the mean or zero.

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Now the table we have gives the area to the right,

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to the left, I'm sorry, to the left, for positive

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and negative values of z. Okay, so we have two

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tables actually, one for negative on page 570, and

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the other one for positive values of z. I think we

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discussed that before when we talked about these

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scores. We have the same formula.

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Now let's look at this, the next slide. Suppose x

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is distributed normally with mean of 100. So the

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mean of x is 100. and the standard deviation of

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50. So sigma is 50. Now let's see how can we

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compute the z-score for x equals 200. Again the

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formula is just x minus mu divided by sigma x 200

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minus 100 divided by 50 that will give 2. Now the

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sign of this value is positive That means x is

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greater than the mean, because x is 200. Now,

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what's the meaning of 2? What does this value tell

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you?

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Yeah, exactly. x equals 200 is two standard

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deviations above the mean. Because if you look at

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200, the x value, The mean is 100, sigma is 50.

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Now the difference between the score, which is

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200, and the mu, which is 100, is equal to

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standard deviations, because the difference is

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100. 2 times 50 is 100. So this says that x equals

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200 is 2 standard deviations above the mean. If z

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is negative, you can say that x is two standard

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deviations below them. Make sense? So that's how

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can we compute the z square. Now, when we

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transform from normal distribution to

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standardized, still we will have the same shape. I

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mean the distribution is still normally

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distributed. So note, the shape of the

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distribution is the same, only the scale has

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changed. So we can express the problem in original

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units, X, or in a standardized unit, Z. So when we

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have X, just use this equation to transform to

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this form.

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Now, for example, suppose we have normal

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distribution and we are interested in the area

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00:18:26,040 --> 00:18:32,660
between A and B. Now, the area between A and B, it

255
00:18:32,660 --> 00:18:34,700
means the probability between them. So

256
00:18:34,700 --> 00:18:39,140
statistically speaking, area means probability. So

257
00:18:39,140 --> 00:18:42,700
probability between A and B, I mean probability of

258
00:18:42,700 --> 00:18:45,380
X greater than or equal A and less than or equal B

259
00:18:45,380 --> 00:18:49,420
is the same as X greater than A or less than B.

260
00:18:50,450 --> 00:18:57,210
that means the probability of X equals A this

261
00:18:57,210 --> 00:19:02,510
probability is zero or probability of X equals B

262
00:19:02,510 --> 00:19:06,930
is also zero so in continuous distribution the

263
00:19:06,930 --> 00:19:10,630
equal sign does not matter I mean if we have equal

264
00:19:10,630 --> 00:19:15,130
sign or we don't have these probabilities are the

265
00:19:15,130 --> 00:19:19,390
same so I mean for example if we are interested

266
00:19:20,310 --> 00:19:23,450
for probability of X smaller than or equal to E.

267
00:19:24,850 --> 00:19:30,370
This probability is the same as X smaller than E.

268
00:19:31,330 --> 00:19:33,730
Or on the other hand, if you are interested in the

269
00:19:33,730 --> 00:19:39,010
area above B greater than or equal to B, it's the

270
00:19:39,010 --> 00:19:44,770
same as X smaller than E. So don't worry about the

271
00:19:44,770 --> 00:19:48,660
equal sign. Or continuous distribution, exactly.

272
00:19:49,120 --> 00:19:53,820
But for discrete, it does matter. Now, since we

273
00:19:53,820 --> 00:19:58,200
are talking about normal distribution, and as we

274
00:19:58,200 --> 00:20:01,320
mentioned, normal distribution is symmetric around

275
00:20:01,320 --> 00:20:05,900
the mean, that means the area to the right equals

276
00:20:05,900 --> 00:20:09,340
the area to the left. Now the entire area

277
00:20:09,340 --> 00:20:12,940
underneath the normal curve equals one. I mean

278
00:20:12,940 --> 00:20:16,500
probability of X ranges from minus infinity up to

279
00:20:16,500 --> 00:20:21,500
infinity equals one. So probability of X greater

280
00:20:21,500 --> 00:20:26,920
than minus infinity up to infinity is one. The

281
00:20:26,920 --> 00:20:31,480
total area is one. So the area from minus infinity

282
00:20:31,480 --> 00:20:38,080
up to the mean mu is one-half. The same as the

283
00:20:38,080 --> 00:20:42,600
area from mu up to infinity is also one-half. That

284
00:20:42,600 --> 00:20:44,760
means the probability of X greater than minus

285
00:20:44,760 --> 00:20:48,300
infinity up to mu equals the probability from mu

286
00:20:48,300 --> 00:20:52,120
up to infinity because of symmetry. I mean you

287
00:20:52,120 --> 00:20:56,160
cannot say that for any distribution. Just for

288
00:20:56,160 --> 00:20:59,000
symmetric distribution, the area below the mean

289
00:20:59,000 --> 00:21:03,780
equals one-half, which is the same as the area to

290
00:21:03,780 --> 00:21:07,110
the right of the mean. So the entire Probability

291
00:21:07,110 --> 00:21:11,330
is one. And also you have to keep in mind that the

292
00:21:11,330 --> 00:21:17,570
probability always ranges between zero and one. So

293
00:21:17,570 --> 00:21:20,030
that means the probability couldn't be negative.

294
00:21:22,870 --> 00:21:27,730
It should be positive. It shouldn't be greater

295
00:21:27,730 --> 00:21:31,710
than one. So it's between zero and one. So always

296
00:21:31,710 --> 00:21:39,020
the probability lies between zero and one. The

297
00:21:39,020 --> 00:21:44,500
tables we have on page 570 and 571 give the area

298
00:21:44,500 --> 00:21:46,040
to the left side.

299
00:21:49,420 --> 00:21:54,660
For negative or positive z's. Now for example,

300
00:21:54,940 --> 00:22:03,060
suppose we are looking for probability of z less

301
00:22:03,060 --> 00:22:08,750
than 2. How can we find this probability by using

302
00:22:08,750 --> 00:22:12,210
the normal curve? Let's go back to this normal

303
00:22:12,210 --> 00:22:16,410
distribution. In the second page, we have positive

304
00:22:16,410 --> 00:22:17,070
z-scores.

305
00:22:23,850 --> 00:22:33,390
So we ask about the probability of z less than. So

306
00:22:33,390 --> 00:22:40,690
the second page, gives positive values of z. And

307
00:22:40,690 --> 00:22:44,590
the table gives the area below. And he asked about

308
00:22:44,590 --> 00:22:49,550
here, B of z is smaller than 2. Now 2, if you

309
00:22:49,550 --> 00:22:54,910
hear, up all the way down here, 2, 0, 0. So the

310
00:22:54,910 --> 00:23:00,530
answer is 9772. So this value, so the probability

311
00:23:00,530 --> 00:23:02,130
is 9772.

312
00:23:03,990 --> 00:23:05,390
Because it's 2.

313
00:23:09,510 --> 00:23:14,650
It's 2, 0, 0. But if you ask about what's the

314
00:23:14,650 --> 00:23:20,590
probability of Z less than 2.05? So this is 2.

315
00:23:23,810 --> 00:23:30,370
Now under 5, 9, 7, 9, 8. So the answer is 9, 7.

316
00:23:34,360 --> 00:23:38,900
Because this is two, and we need five decimal

317
00:23:38,900 --> 00:23:44,820
places. So all the way up to 9798. So this value

318
00:23:44,820 --> 00:23:54,380
is 2.05. Now it's about, it's more than 1.5,

319
00:23:55,600 --> 00:23:56,880
exactly 1.5.

320
00:24:02,140 --> 00:24:04,880
1.5. This is 1.5.

321
00:24:08,800 --> 00:24:09,720
9332.

322
00:24:12,440 --> 00:24:16,300
1.5. Exactly 1.5. So 9332.

323
00:24:18,780 --> 00:24:27,990
What's about probability less than 1.35? 1.3 all

324
00:24:27,990 --> 00:24:35,250
the way to 9.115. 9.115. 9.115. 9.115. 9.115. 9

325
00:24:35,250 --> 00:24:35,650
.115. 9.115.

326
00:24:41,170 --> 00:24:42,430
9.115. 9.115. 9.115. 9.115. 9.115. 9.115. 9.115. 9

327
00:24:42,430 --> 00:24:42,450
.115. 9.115. 9.115. 9.115. 9.115. 9.115. 9.115. 9

328
00:24:42,450 --> 00:24:42,450
.115. 9.115. 9.115. 9.115. 9.115. 9.115. 9.115. 9

329
00:24:42,450 --> 00:24:44,050
.115. 9.115. 9.115. 9.115. 9.115. 9.115. 9.115. 9

330
00:24:44,050 --> 00:24:50,530
.115. 9.115. 9.115. 9.115. 9.115. 9.115. 9.115. 9

331
00:24:50,530 --> 00:24:54,980
.115. 9. But here we are looking for the area to

332
00:24:54,980 --> 00:25:01,280
the right. One minus one. Now this area equals

333
00:25:01,280 --> 00:25:05,660
one minus because

334
00:25:05,660 --> 00:25:11,420
since suppose

335
00:25:11,420 --> 00:25:18,760
this is the 1.35 and we are interested in the area

336
00:25:18,760 --> 00:25:24,030
to the right or above 1.35. The table gives the

337
00:25:24,030 --> 00:25:28,230
area below. So the area above equals the total

338
00:25:28,230 --> 00:25:31,970
area underneath the curve is 1. So 1 minus this

339
00:25:31,970 --> 00:25:39,050
value, so equals 0.0885,

340
00:25:39,350 --> 00:25:42,250
and so on. So this is the way how can we compute

341
00:25:42,250 --> 00:25:47,850
the probabilities underneath the normal curve. if

342
00:25:47,850 --> 00:25:51,090
it's probability of z is smaller than then just

343
00:25:51,090 --> 00:25:55,910
use the table directly otherwise if we are talking

344
00:25:55,910 --> 00:26:00,390
about z greater than subtract from one to get the

345
00:26:00,390 --> 00:26:04,870
result that's how can we compute the probability

346
00:26:04,870 --> 00:26:13,750
of z less than or equal now

347
00:26:13,750 --> 00:26:18,890
let's see if we have x and x that has normal

348
00:26:18,890 --> 00:26:22,070
distribution with mean mu and standard deviation

349
00:26:22,070 --> 00:26:26,250
of sigma and let's see how can we compute the

350
00:26:26,250 --> 00:26:33,790
value of the probability mainly

351
00:26:33,790 --> 00:26:38,190
there are three steps to find the probability of x

352
00:26:38,190 --> 00:26:42,490
greater than a and less than b when x is

353
00:26:42,490 --> 00:26:47,000
distributed normally first step Draw normal curve

354
00:26:47,000 --> 00:26:54,880
for the problem in terms of x. So draw the normal

355
00:26:54,880 --> 00:26:58,140
curve first. Second, translate x values to z

356
00:26:58,140 --> 00:27:03,040
values by using the formula we have. z x minus mu

357
00:27:03,040 --> 00:27:06,440
divided by sigma. Then use the standardized normal

358
00:27:06,440 --> 00:27:15,140
table on page 570 and 571. For example, Let's see

359
00:27:15,140 --> 00:27:18,420
how can we find normal probabilities. Let's assume

360
00:27:18,420 --> 00:27:23,760
that X represents the time it takes to download an

361
00:27:23,760 --> 00:27:28,580
image from the internet. So suppose X, time

362
00:27:28,580 --> 00:27:33,760
required to download an image file from the

363
00:27:33,760 --> 00:27:38,460
internet. And suppose we know that the time is

364
00:27:38,460 --> 00:27:42,060
normally distributed for with mean of eight

365
00:27:42,060 --> 00:27:46,130
minutes. And standard deviation of five minutes.

366
00:27:46,490 --> 00:27:47,510
So we know the mean.

367
00:27:50,610 --> 00:27:59,670
Eight. Eight. And sigma of five minutes. And they

368
00:27:59,670 --> 00:28:03,410
ask about what's the probability of X smaller than

369
00:28:03,410 --> 00:28:07,990
eight one six. So first thing we have to compute,

370
00:28:08,170 --> 00:28:12,190
to draw the normal curve. The mean lies in the

371
00:28:12,190 --> 00:28:18,060
center. which is 8. He asked about probability of

372
00:28:18,060 --> 00:28:22,580
X smaller than 8.6. So we are interested in the

373
00:28:22,580 --> 00:28:27,920
area below 8.6. So it matched the table we have.

374
00:28:29,980 --> 00:28:34,900
Second step, we have to transform from normal

375
00:28:34,900 --> 00:28:37,280
distribution to standardized normal distribution

376
00:28:37,280 --> 00:28:42,120
by using this form, which is X minus mu divided by

377
00:28:42,120 --> 00:28:51,430
sigma. So x is 8.6 minus the mean, 8, divided by

378
00:28:51,430 --> 00:28:57,130
sigma, gives 0.12. So just straightforward

379
00:28:57,130 --> 00:29:02,890
calculation, 8.6 is your value of x. The mean is

380
00:29:02,890 --> 00:29:12,810
8, sigma is 5, so that gives 0.12. So now, the

381
00:29:12,810 --> 00:29:17,210
problem becomes, instead of asking x smaller than

382
00:29:17,210 --> 00:29:25,110
8.6, it's similar to z less than 0.12. Still, we

383
00:29:25,110 --> 00:29:26,310
have the same normal curve.

384
00:29:29,450 --> 00:29:32,990
8, the mean. Now, the mean of z is 0, as we

385
00:29:32,990 --> 00:29:39,230
mentioned. Instead of x, 8.6, the corresponding z

386
00:29:39,230 --> 00:29:43,000
value is 0.12. So instead of finding probability

387
00:29:43,000 --> 00:29:48,580
of X smaller than 8.6, smaller than 1.12, so they

388
00:29:48,580 --> 00:29:53,760
are equivalent. So we transform here from normal

389
00:29:53,760 --> 00:29:56,980
distribution to standardized normal distribution

390
00:29:56,980 --> 00:29:59,980
in order to compute the probability we are looking

391
00:29:59,980 --> 00:30:05,820
for. Now, this is just a portion of the table we

392
00:30:05,820 --> 00:30:06,100
have.

393
00:30:10,530 --> 00:30:18,530
So for positive z values. Now 0.1 is 0.1. Because

394
00:30:18,530 --> 00:30:25,670
here we are looking for z less than 0.1. So 0.1.

395
00:30:27,210 --> 00:30:32,950
Also, we have two. So move up to two decimal

396
00:30:32,950 --> 00:30:38,190
places, we get this value. So the answer is point.

397
00:30:42,120 --> 00:30:45,860
I think it's straightforward to compute the

398
00:30:45,860 --> 00:30:49,460
probability underneath the normal curve if X has

399
00:30:49,460 --> 00:30:53,160
normal distribution. So B of X is smaller than 8.6

400
00:30:53,160 --> 00:30:56,740
is the same as B of Z less than 0.12, which is

401
00:30:56,740 --> 00:31:02,680
around 55%. Makes sense because the area to the

402
00:31:02,680 --> 00:31:07,080
left of 0 equals 1 half. But we are looking for

403
00:31:07,080 --> 00:31:12,440
the area below 0.12. So greater than zero. So this

404
00:31:12,440 --> 00:31:16,600
area actually is greater than 0.5. So it makes

405
00:31:16,600 --> 00:31:20,440
sense that your result is greater than 0.5.

406
00:31:22,320 --> 00:31:22,960
Questions?

407
00:31:25,480 --> 00:31:30,780
Next, suppose we are interested of probability of

408
00:31:30,780 --> 00:31:35,380
X greater than. So that's how can we find normal

409
00:31:35,380 --> 00:31:41,980
upper tail probabilities. Again, the table we have

410
00:31:41,980 --> 00:31:46,580
gives the area to the left. In order to compute

411
00:31:46,580 --> 00:31:50,880
the area in the upper tail probabilities, I mean

412
00:31:50,880 --> 00:31:55,620
this area, since the normal distribution is

413
00:31:55,620 --> 00:32:00,160
symmetric and The total area underneath the curve

414
00:32:00,160 --> 00:32:04,680
is 1. So the probability of X greater than 8.6 is

415
00:32:04,680 --> 00:32:11,640
the same as 1 minus B of X less than 8.6. So first

416
00:32:11,640 --> 00:32:17,020
step, just find the probability we just have and

417
00:32:17,020 --> 00:32:21,680
subtract from 1. So B of X greater than 8.6, the

418
00:32:21,680 --> 00:32:25,930
same as B of Z greater than 0.12. which is the

419
00:32:25,930 --> 00:32:30,370
same as 1 minus B of Z less than 0.5. It's 1 minus

420
00:32:30,370 --> 00:32:36,230
the result we got from previous one. So this value

421
00:32:36,230 --> 00:32:39,410
1 minus this value gives 0.452.

422
00:32:41,610 --> 00:32:45,090
So for the other tail probability, just subtract 1

423
00:32:45,090 --> 00:32:47,690
from the lower tail probabilities.

424
00:32:51,930 --> 00:32:55,750
Now let's see how can we find Normal probability

425
00:32:55,750 --> 00:33:01,750
between two values. I mean if X, for example, for

426
00:33:01,750 --> 00:33:06,610
the same data we have, suppose X between 8 and 8

427
00:33:06,610 --> 00:33:13,360
.6. Now what's the area between these two? Here we

428
00:33:13,360 --> 00:33:17,220
have two values of x, x is 8 and x is 8.6.

429
00:33:24,280 --> 00:33:33,780
Exactly, so below 8.6 minus below 8 and below 8 is

430
00:33:33,780 --> 00:33:40,840
1 half. So the probability of x between 8

431
00:33:40,840 --> 00:33:47,340
and And 8.2 and 8.6. You can find z-score for the

432
00:33:47,340 --> 00:33:52,480
first value, which is zero. Also compute the z

433
00:33:52,480 --> 00:33:55,540
-score for the other value, which as we computed

434
00:33:55,540 --> 00:34:01,580
before, 0.12. Now this problem becomes z between

435
00:34:01,580 --> 00:34:04,540
zero and 0.5.

436
00:34:07,480 --> 00:34:15,120
So B of x. Greater than 8 and smaller than 8.6 is

437
00:34:15,120 --> 00:34:20,800
the same as z between 0 and 0.12. Now this area

438
00:34:20,800 --> 00:34:25,320
equals b of z smaller than 0.12 minus the area

439
00:34:25,320 --> 00:34:26,520
below z which is 1.5.

440
00:34:31,100 --> 00:34:37,380
So again, b of z between 0 and 1.5 equal b of z

441
00:34:37,380 --> 00:34:42,840
small. larger than 0.12 minus b of z less than

442
00:34:42,840 --> 00:34:46,520
zero. Now, b of z less than 0.12 gives this

443
00:34:46,520 --> 00:34:53,060
result, 0.5478. The probability below zero is one

444
00:34:53,060 --> 00:34:56,160
-half because we know that the area to the left is

445
00:34:56,160 --> 00:34:59,320
zero, same as to the right is one-half. So the

446
00:34:59,320 --> 00:35:04,240
answer is going to be 0.478. So that's how can we

447
00:35:04,240 --> 00:35:07,540
compute the probabilities for lower 10 directly

448
00:35:07,540 --> 00:35:12,230
from the table. upper tail is just one minus lower

449
00:35:12,230 --> 00:35:18,990
tail and between two values just subtracts the

450
00:35:18,990 --> 00:35:21,970
larger one minus smaller one because he was

451
00:35:21,970 --> 00:35:26,310
subtracted bz less than point one minus bz less

452
00:35:26,310 --> 00:35:29,430
than or equal to zero that will give the normal

453
00:35:29,430 --> 00:35:36,850
probability another example suppose we are looking

454
00:35:36,850 --> 00:35:49,350
for X between 7.4 and 8. Now, 7.4 lies below the

455
00:35:49,350 --> 00:35:55,270
mean. So here, this value, we have to compute the

456
00:35:55,270 --> 00:36:00,130
z-score for 7.4 and also the z-score for 8, which

457
00:36:00,130 --> 00:36:04,090
is zero. And that will give, again,

458
00:36:07,050 --> 00:36:13,710
7.4, if you just use this equation, minus

459
00:36:13,710 --> 00:36:17,690
the mean, divided by sigma, negative 0.6 divided

460
00:36:17,690 --> 00:36:21,150
by 5, which is negative 0.12.

461
00:36:22,730 --> 00:36:31,410
So it gives B of z between minus 0.12 and 0. And

462
00:36:31,410 --> 00:36:35,700
that again is B of z less than 0. minus P of Z

463
00:36:35,700 --> 00:36:40,140
less than negative 0.12. Is it clear? Now here we

464
00:36:40,140 --> 00:36:42,260
converted or we transformed from normal

465
00:36:42,260 --> 00:36:45,960
distribution to standardized. So instead of X

466
00:36:45,960 --> 00:36:52,100
between 7.4 and 8, we have now Z between minus 0

467
00:36:52,100 --> 00:36:57,480
.12 and 0. So this area actually is the red one,

468
00:36:57,620 --> 00:37:03,740
the red area is one-half. Total area below z is

469
00:37:03,740 --> 00:37:10,700
one-half, below zero, and minus z below minus 0

470
00:37:10,700 --> 00:37:17,820
.12. So B of z less than zero minus negative 0.12.

471
00:37:18,340 --> 00:37:21,940
That will give the area between minus 0.12 and

472
00:37:21,940 --> 00:37:28,860
zero. This is one-half. Now, B of z less than

473
00:37:28,860 --> 00:37:33,270
negative 0.12. look you go back to the normal

474
00:37:33,270 --> 00:37:37,650
curve to the normal table but for the negative

475
00:37:37,650 --> 00:37:42,310
values of z negative point one two negative point

476
00:37:42,310 --> 00:37:53,290
one two four five two two it's four five point

477
00:37:53,290 --> 00:37:56,630
five minus point four five two two will give the

478
00:37:56,630 --> 00:37:58,370
result we are looking for

479
00:38:01,570 --> 00:38:06,370
So B of Z less than 0 is 0.5. B of Z less than

480
00:38:06,370 --> 00:38:12,650
negative 0.12 equals minus 0.4522. That will give

481
00:38:12,650 --> 00:38:14,290
0 forcibility.

482
00:38:16,790 --> 00:38:23,590
Now, by symmetric, you can see that this

483
00:38:23,590 --> 00:38:28,470
probability between

484
00:38:28,470 --> 00:38:38,300
Z between minus 0.12 and 0 is the same as the

485
00:38:38,300 --> 00:38:43,340
other side from 0.12 I mean this area the red one

486
00:38:43,340 --> 00:38:46,200
is the same up to 8.6

487
00:38:55,600 --> 00:38:58,840
So the area between minus 0.12 up to 0 is the same

488
00:38:58,840 --> 00:39:04,920
as from 0 up to 0.12. Because of symmetric, since

489
00:39:04,920 --> 00:39:09,680
this area equals the same for the other part. So

490
00:39:09,680 --> 00:39:15,660
from 0 up to 0.12 is the same as minus 0.12 up to

491
00:39:15,660 --> 00:39:19,100
0. So equal, so the normal distribution is

492
00:39:19,100 --> 00:39:23,200
symmetric. So this probability is the same as B of

493
00:39:23,200 --> 00:39:27,980
Z between 0 and 0.12. Any question?

494
00:39:34,520 --> 00:39:36,620
Again, the equal sign does not matter.

495
00:39:42,120 --> 00:39:45,000
Because here we have the complement. The

496
00:39:45,000 --> 00:39:49,250
complement. If this one, I mean, complement of z

497
00:39:49,250 --> 00:39:53,350
less than, greater than 0.12, the complement is B

498
00:39:53,350 --> 00:39:56,350
of z less than or equal to minus 0.12. So we

499
00:39:56,350 --> 00:40:00,070
should have just permutation, the equality. But it

500
00:40:00,070 --> 00:40:04,830
doesn't matter. If in the problem we don't have

501
00:40:04,830 --> 00:40:07,470
equal sign in the complement, we should have equal

502
00:40:07,470 --> 00:40:11,430
sign. But it doesn't matter actually if we have

503
00:40:11,430 --> 00:40:14,510
equal sign or not. For example, if we are looking

504
00:40:14,510 --> 00:40:19,430
for B of X greater than A. Now what's the

505
00:40:19,430 --> 00:40:25,950
complement of that? 1 minus less

506
00:40:25,950 --> 00:40:32,450
than or equal to A. But if X is greater than or

507
00:40:32,450 --> 00:40:37,870
equal to A, the complement is without equal sign.

508
00:40:38,310 --> 00:40:40,970
But in continuous distribution, the equal sign

509
00:40:40,970 --> 00:40:44,990
does not matter. Any question?

510
00:40:52,190 --> 00:40:58,130
comments. Let's move to the next topic which talks

511
00:40:58,130 --> 00:41:05,510
about the empirical rule. If you remember before

512
00:41:05,510 --> 00:41:16,750
we said there is an empirical rule for 68, 95, 95,

513
00:41:17,420 --> 00:41:23,060
99.71. Now let's see the exact meaning of this

514
00:41:23,060 --> 00:41:23,320
rule.

515
00:41:37,580 --> 00:41:40,460
Now we have to apply the empirical rule not to

516
00:41:40,460 --> 00:41:43,020
Chebyshev's inequality because the distribution is

517
00:41:43,020 --> 00:41:48,670
normal. Chebyshev's is applied for skewed

518
00:41:48,670 --> 00:41:52,630
distributions. For symmetric, we have to apply the

519
00:41:52,630 --> 00:41:55,630
empirical rule. Here, we assume the distribution

520
00:41:55,630 --> 00:41:58,390
is normal. And today, we are talking about normal

521
00:41:58,390 --> 00:42:01,330
distribution. So we have to use the empirical

522
00:42:01,330 --> 00:42:02,410
rules.

523
00:42:07,910 --> 00:42:13,530
Now, the mean is the value in the middle. Suppose

524
00:42:13,530 --> 00:42:16,900
we are far away. from the mean by one standard

525
00:42:16,900 --> 00:42:22,720
deviation either below or above and we are

526
00:42:22,720 --> 00:42:27,040
interested in the area between this value which is

527
00:42:27,040 --> 00:42:33,040
mu minus sigma so we are looking for mu minus

528
00:42:33,040 --> 00:42:36,360
sigma and mu plus sigma

529
00:42:53,270 --> 00:42:59,890
Last time we said there's a rule 68% of the data

530
00:42:59,890 --> 00:43:06,790
lies one standard deviation within the mean. Now

531
00:43:06,790 --> 00:43:10,550
let's see how can we compute the exact area, area

532
00:43:10,550 --> 00:43:15,250
not just say 68%. Now X has normal distribution

533
00:43:15,250 --> 00:43:18,390
with mean mu and standard deviation sigma. So

534
00:43:18,390 --> 00:43:25,280
let's compare it from normal distribution to

535
00:43:25,280 --> 00:43:29,700
standardized. So this is the first value here. Now

536
00:43:29,700 --> 00:43:34,940
the z-score, the general formula is x minus the

537
00:43:34,940 --> 00:43:40,120
mean divided by sigma. Now the first quantity is

538
00:43:40,120 --> 00:43:45,660
mu minus sigma. So instead of x here, so first z

539
00:43:45,660 --> 00:43:49,820
is, now this x should be replaced by mu minus

540
00:43:49,820 --> 00:43:55,040
sigma. So mu minus sigma. So that's my x value,

541
00:43:55,560 --> 00:44:00,240
minus the mean of that, which is mu, divided by

542
00:44:00,240 --> 00:44:07,900
sigma. Mu minus sigma minus mu mu cancels, so plus

543
00:44:07,900 --> 00:44:13,520
one. And let's see how can we compute that area. I

544
00:44:13,520 --> 00:44:16,980
mean between minus one and plus one. In this case,

545
00:44:17,040 --> 00:44:23,180
we are interested or we are looking for the area

546
00:44:23,180 --> 00:44:28,300
between minus one and plus one this area now the

547
00:44:28,300 --> 00:44:31,360
dashed area i mean the area between minus one and

548
00:44:31,360 --> 00:44:39,460
plus one equals the area below one this area minus

549
00:44:39,460 --> 00:44:44,980
the area below minus one that will give the area

550
00:44:44,980 --> 00:44:48,200
between minus one and plus one now go back to the

551
00:44:48,200 --> 00:44:52,500
normal table you have and look at the value of one

552
00:44:52,500 --> 00:45:02,620
z and one under zero what's your answer one point

553
00:45:02,620 --> 00:45:11,520
one point now without using the table can you tell

554
00:45:11,520 --> 00:45:17,360
the area below minus one one minus this one

555
00:45:17,360 --> 00:45:17,840
because

556
00:45:23,430 --> 00:45:29,870
Now the area below, this is 1. The area below 1 is

557
00:45:29,870 --> 00:45:31,310
0.3413.

558
00:45:34,430 --> 00:45:37,590
Okay, now the area below minus 1.

559
00:45:40,770 --> 00:45:42,050
This is minus 1.

560
00:45:46,810 --> 00:45:49,550
Now, the area below minus 1 is the same as above

561
00:45:49,550 --> 00:45:50,510
1.

562
00:45:54,310 --> 00:45:58,810
These are the two areas here are equal. So the

563
00:45:58,810 --> 00:46:03,110
area below minus 1, I mean b of z less than minus

564
00:46:03,110 --> 00:46:09,130
1 is the same as b of z greater than 1. And b of z

565
00:46:09,130 --> 00:46:12,650
greater than 1 is the same as 1 minus b of z

566
00:46:12,650 --> 00:46:17,310
smaller than 1. So b of z less than 1 here. You

567
00:46:17,310 --> 00:46:19,710
shouldn't need to look again to the table. Just

568
00:46:19,710 --> 00:46:26,770
subtract 1 from this value. Make sense? Here we

569
00:46:26,770 --> 00:46:30,490
compute the value of B of Z less than 1, which is

570
00:46:30,490 --> 00:46:35,430
0.8413. We are looking for B of Z less than minus

571
00:46:35,430 --> 00:46:39,770
1, which is the same as B of Z greater than 1.

572
00:46:40,750 --> 00:46:43,850
Now, greater than means our tail. It's 1 minus the

573
00:46:43,850 --> 00:46:48,700
lower tail probability. So this is 1 minus. So the

574
00:46:48,700 --> 00:46:52,240
answer again is 1 minus 0.8413.

575
00:46:54,280 --> 00:47:00,040
So 8413 minus 0.1587.

576
00:47:11,380 --> 00:47:17,030
So 8413. minus 1.1587.

577
00:47:21,630 --> 00:47:27,570
Okay, so that gives 0.6826.

578
00:47:29,090 --> 00:47:37,550
Multiply this one by 100, we get 68.1826.

579
00:47:38,750 --> 00:47:44,010
So roughly 60-80% of the observations lie between

580
00:47:44,010 --> 00:47:50,470
one standard deviation around the mean. So this is

581
00:47:50,470 --> 00:47:53,850
the way how can we compute the area below one

582
00:47:53,850 --> 00:47:57,250
standard deviation or above one standard deviation

583
00:47:57,250 --> 00:48:03,790
of the mean. Do the same for not mu minus sigma,

584
00:48:05,230 --> 00:48:11,540
mu plus minus two sigma and mu plus two sigma. The

585
00:48:11,540 --> 00:48:14,600
only difference is that this one is going to be

586
00:48:14,600 --> 00:48:17,280
minus 2 and do the same.

587
00:48:20,620 --> 00:48:23,080
That's the empirical rule we discussed in chapter

588
00:48:23,080 --> 00:48:28,980
3. So here we can find any probability, not just

589
00:48:28,980 --> 00:48:33,660
95 or 68 or 99.7. We can use the normal table to

590
00:48:33,660 --> 00:48:36,900
give or to find or to compute any probability.

591
00:48:48,270 --> 00:48:53,090
So again, for the other one, mu plus or minus two

592
00:48:53,090 --> 00:49:00,190
sigma, it covers about 95% of the axis. For mu

593
00:49:00,190 --> 00:49:03,750
plus or minus three sigma, it covers around all

594
00:49:03,750 --> 00:49:08,450
the data, 99.7. So just do it at home, you will

595
00:49:08,450 --> 00:49:14,210
see that the exact area is 95.44 instead of 95.

596
00:49:14,840 --> 00:49:18,520
And the other one is 99.73. So that's the

597
00:49:18,520 --> 00:49:23,520
empirical rule we discussed in chapter three. I'm

598
00:49:23,520 --> 00:49:32,560
going to stop at this point, which is the x value

599
00:49:32,560 --> 00:49:38,400
for the normal probability. Now, what we discussed

600
00:49:38,400 --> 00:49:43,560
so far, we computed the probability. I mean,

601
00:49:43,740 --> 00:49:49,120
what's the probability of X smaller than E? Now,

602
00:49:49,200 --> 00:49:56,240
suppose this probability is known. How can we

603
00:49:56,240 --> 00:50:01,500
compute this value? Later, we'll talk about that.

604
00:50:06,300 --> 00:50:09,820
It's backward calculations. It's inverse or

605
00:50:09,820 --> 00:50:11,420
backward calculation.

606
00:50:13,300 --> 00:50:14,460
for next time inshallah.