PL9fwy3NUQKwaIc7KWc8buW344941GStQL/BRRl21n6QWg_postprocess.srt

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So last time, we discussed how to find the

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probabilities underneath the normal curve for

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three cases. If the point lies in the lower tail,

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as this one, or upper tail, or between. So we can

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do the computations for this kind of

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probabilities. And I think we gave two examples.

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For example, if we are looking for probability of

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Z is smaller than 0.1, and Z, as we mentioned last

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time, is the standardized normal distribution that

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has mean of 0 and sigma is 1. In this case, just

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go to the table you have. Now we are looking for 0

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.12, for example. So here we have 0.1. Under 2, we

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get this result. This value is the probability of

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Z smaller than 0.5. But you have to keep in mind

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that we have to transform first from normal

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distribution to standardized normal distribution.

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For this specific example, if we are looking for

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mean of 8 and standard deviation of 5, that's for

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the normal distribution. In this case, the z-score

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is given by this equation, which is x minus 3

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divided by sigma. So 8.6 minus 8 divided by 5

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gives 0.12. In this case, we can use the

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standardized normal table. Now, for the other

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case, we are looking for the probability of x

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greater than 8.6. So we are looking for the upper

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tail probability. The table we have gives the area

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to the left side. And we know that the total area

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underneath the normal curve is 1. So the

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probability of x greater than 8.6 is the same as 1

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minus the probability of x smaller than x less

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than 8.6. So here, for 8.6, we got z squared to be

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0.12. So the probability of z greater than 0.12 is

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the same as 1 minus z of z is smaller than 0.12.

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So 1 minus the answer we just got from part A will

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get us to something. So this is the probability of

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X is greater than 8.6. So all of the time, if you

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are asking about computing the probability in the

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upper tail, you have to first find the probability

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in the lower tail, then subtract that value from

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1. So that's the probability for the upper tail.

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The last one, we are looking for probability

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between two values. For example, x. What's the

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probability of x greater than 8 and smaller than 8

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.6? So we are looking for this area, the red area.

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So the probability of x between these two values

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actually equals the probability of x smaller than.

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8.6 minus the probability of X smaller than 8A.

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And that, in this case, you have to compute two

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values of this score. One for the first value,

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which is A. This value is zero because the mean is

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zero. And we know that Z can be negative. If X is

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smaller than Mu, Z can positive if X is greater

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than Mu and equals zero only if X equals Mu. In

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this case, X equals Mu, so Z score is zero. The

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other one as we got before is 0.12. So now, we

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transform actually the probability of X between

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8 and 8.6 to z-score between 0 and 0.12. In this

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case, we can use the normal theorem. Now, this

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area is, as we mentioned, just b of x smaller than

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0.12 minus b of x, b of z smaller than 0. This

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probability, we know that, 0.54878. Now, the

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probability of z smaller than 0 is one-half.

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Because the total area underneath the normal curve

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is 1, and 0 divided the curve into two equally

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parts. So the area to the right of 0 is the same

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as the area to the left of 0. So in this case,

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minus 0.5. So this is your answer. So the

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probability of Z between 8 and 8.6 is around 0478.

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I think we stopped last time at this point.

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This is another example to compute the probability

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of X greater than 7.4 and 8. Also, it's the same

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idea here, just find these scores for the two

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values.

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L with B of Z greater than minus 0.12 up to 0. Now

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this red area equals the area below 0. I mean B of

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Z smaller than 0 minus the probability of Z

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smaller than minus 0.12. Now by using symmetric

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probability of the normal distribution, we know

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that The probability of Z smaller than minus 0.12

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equals the probability of Z greater than 0.12.

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Because this area, if we have Z smaller than minus

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0.12, the area to the left, that equals the area

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to the right of the same point, because of

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symmetry. And finally, you will end with this

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result.

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D of z minus 0.12, all the way up to 0, this area,

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is the same as the area from 0 up to 0.5. So this

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area actually is the same as D of z between 0 and

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0.5. So if you have negative sign, and then take

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the opposite one, the answer will be the same

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because the normal distribution is symmetric

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around 0.9. The questions, I think we stopped

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here. And also we talked about empirical rules.

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The one we mentioned in chapter three, in chapter

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three. And we know that, as we mentioned before,

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that is 68.16% of the observations fall within one

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standard deviation around the mean. So this area

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from mu minus one sigma up to mu plus sigma, this

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area covers around 68%. Also

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95% or actually 95.44% of the data falls within

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two standard deviations of the mean. And finally,

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around most of the data, around 99.73% of the data

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falls within three subdivisions of the population

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mean.

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And now the new topic is how can we find the X

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value if the probability is given. It's vice

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versa. In the previous questions, we were asking

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about find the probability, for example, if X is

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smaller than a certain number. Now suppose this

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probability is given, and we are looking to find

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this value. I mean, for example, suppose in the

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previous examples here,

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Suppose we know this probability. So the

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probability is given. The question is, how can we

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find this value? It's the opposite, sometimes

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called backward normal calculations.

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There are actually two steps. to find the x value

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for a certain probability or for a given or for a

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known probability the first step we have to find

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the z score then use this equation to find the x

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value corresponding to the z score you have and x

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is just mu plus sigma times mu so first step you

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have to find the z score

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corresponding to the probability we have. So find

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the z value for the non-probability, then use that

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z score to find the value of x by using this

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equation. So x equals mu plus z sigma. z could be

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negative, could be positive, depends on the

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probability you have. If the probability is above

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0.5, I mean 0.5 and greater than 0.5, this

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corresponds to 10. But if z-score is negative, I'm

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sorry, if z-score is negative, then the

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probability should be smaller than 0.5. So if the

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probability is given less than 0.5, then your z

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-score should be negative, otherwise should be

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positive. So you have to be careful in this case.

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Now look at this example. Let x represent the time

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it takes in seconds. to download an image file

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from the internet. The same example as we did

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before. And here we assume that x is normal

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distribution with mean of 8 and standard deviation

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of 5. Now, let's see how can we find the value of

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x such that 20% of download times are smaller than

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x.

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So, this probability is a fraction. Also, always

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the probability is between 0 and 1. So, the

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probability here is 20%. In this case, your z

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-score should be negative. Because 20% is more

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than 0.5. So, z-score should be in this side, in

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the left side.

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So, again, he asks about finding x-value such that

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20%.

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So here again we are looking for this value, for

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the value of x, which is smaller than the area to

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the left of this x, equals 0.2.

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Now, the first step, we have to find the z-score.

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It's backward, z-score first, then x. Find a z

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-score corresponding to the probability of 0.2.

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The approximate one, the near value, I mean, to

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the 0.2 is 0.2005. Sometimes you have the exact

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value from the table you have, but most of the

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time you don't have it. So you have to look at the

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approximate value, which is very close to the one

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you have. So here, we are looking for 0.2. The

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closest value to 0.2 is 0.2005. Now, the

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corresponding value to this probability is minus 0

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.8 all the way up to 4. So your z-score is

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negative 0.84. So this is the first step. Any

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question? Again.

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Now if we just go back to this equation,

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z equals x minus mu over sigma. A cross

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multiplication, I mean if you multiply both sides

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by sigma, you will get sigma times z equals x

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minus mu.

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Now, in this question,

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he asks about, find the value of x such that 20%

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of download times are less than x.

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Now the probability is less than 0.5, so your z

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-score should be on the left side. So here we need

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to find the value of z first. Go back to the

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normal table you have.

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This is the normal table.

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We are looking for minus 0.2. I'm sorry, we are

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looking for 0.2. So the closest value to 0.2 is

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this one, 0.2005. So this is the closest value.

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So the exact answer is sometimes not given. So the

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approximate one, minus 0.8, all the way up to 4.

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So z-score minus 0.8. Any question?

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So the value of z-score is minus 0.84. So my

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corresponding x-value equals X equal mu. The mu is

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given as A plus Z is minus 0.84 times sigma. Sigma

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is 5. You will end with 3.8. So this means the

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probability of X less than 3.8. Equal point.

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Exactly, equal point.

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So in this case, the probability is given, which

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is 0.20. And we ask about what's the value of x in

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this case. So the first step was find the z-score.

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Then use this value. I mean, plug this value in.

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to find the corresponding X score. That's the

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backward normal calculations.

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Let's do one problem from the practice, which is

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number 18.

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Is it clear?

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The owners of a fish market determined

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that the average weight for a catfish is 3.2 So

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this is the

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value of the mean 40

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So again, the owner of a fish market determined

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that the average weight for a catfish is 3.2

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pounds. So the mean is 3.2 with a standard

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deviation of 0.8. So sigma is 0.8. Now, assuming

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the weights of catfish are normally distributed.

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In this case, you ask about what's the probability

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that a randomly selected catfish will weigh more

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than 4.4. So what's the probability of X More

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than. So greater than 4.

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I just gave the idea to solve this problem. At

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home, you can compute it to find the exact answer.

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So first step, find z score. z is 4.4.

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Divide by z.

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Just compute this value. It's 0.8 divided by 0.8

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equals 1.

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So z-score is 1. So we are looking for the

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probability of z greater than 1.

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1.5. 1.2. 1.5. 1.5. 1.5. So I'm looking for the

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probability of x of z greater than

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1 minus P

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of Z less than or equal to 1.5. Now go back to the

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table.

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Now 1.5 under 0. It's 0.9332.

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So, 1 minus this probability gives 0.668.

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That's the probability of X greater than 4.4.

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So, the answer is 0.0668.

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Now, for the same question.

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What's the probability that a randomly selected

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fish will weigh between 3 and 5 pounds?

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3?

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Up to 5. So first we have to find the score for 3

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out of 5. For it to be just 3 minus 3.2. Divide by

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0.8 is the first z value. Negative 0.2 divided by

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0.8 minus 0.25. The other one, 5 minus 3.2 divided

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by 0.8. 1 minus 0.8 divided by 0.8 equals

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2.25.

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Just double check this result. So here, the

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probability of X between 3 and 5 equals minus 0

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.25, smaller than Z, smaller than 2.5.

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00:22:12,650 --> 00:22:17,210
So it's transformed from normal distribution to

248
00:22:17,210 --> 00:22:20,750
standardized normal distribution. So here instead

249
00:22:20,750 --> 00:22:23,530
of computing the probability of X between three

250
00:22:23,530 --> 00:22:26,070
and five, we are looking for the probability

251
00:22:26,070 --> 00:22:31,350
between Z between actually minus. It's minus

252
00:22:31,350 --> 00:22:34,590
because your value here is smaller than the

253
00:22:34,590 --> 00:22:37,790
average. The average was 3.2, so it should be

254
00:22:37,790 --> 00:22:42,630
negative. So z score between minus 0.25 all the

255
00:22:42,630 --> 00:22:48,150
way up to 2.25. So now, this is the probability we

256
00:22:48,150 --> 00:22:56,590
are looking for. Zero in the middle minus one

257
00:22:56,590 --> 00:23:03,610
-fourth to the left of z of zero, mu of zero. And

258
00:23:03,610 --> 00:23:08,560
this is the value of 2.25. Now we are looking

259
00:23:08,560 --> 00:23:09,940
actually for this probability.

260
00:23:12,960 --> 00:23:18,360
The area between minus 0.25 all the way up to 2.5.

261
00:23:19,980 --> 00:23:25,200
So this area equals the

262
00:23:25,200 --> 00:23:29,000
probability of Z less than 2.25 minus.

263
00:23:34,280 --> 00:23:38,780
And again, use the normal. table to give this

264
00:23:38,780 --> 00:23:39,860
value and another one.

265
00:23:42,980 --> 00:23:52,400
Any questions? So first step here, we compute the

266
00:23:52,400 --> 00:23:56,880
z-score for each value x. So the problem is

267
00:23:56,880 --> 00:24:01,380
transformed from normal distribution to

268
00:24:01,380 --> 00:24:05,060
standardized normal distribution. So it becomes z

269
00:24:05,060 --> 00:24:11,500
between minus 1.25 up to 2.25. Now, this area,

270
00:24:11,960 --> 00:24:19,900
this dashed area equals the area below 2.25 minus

271
00:24:19,900 --> 00:24:25,000
the area below minus 1.25. Now, by using the

272
00:24:25,000 --> 00:24:27,760
similar way we did before, you will compute the

273
00:24:27,760 --> 00:24:30,960
value of z. The probability of z is smaller than 2

274
00:24:30,960 --> 00:24:39,580
.25 by using The normal table. So here, 2.2 up to

275
00:24:39,580 --> 00:24:47,900
5. So 9, 8, 7, 8. 9, 8, 7, 8.

276
00:24:53,900 --> 00:25:00,260
So the area below 2.25, 2.2, this value. All the

277
00:25:00,260 --> 00:25:03,940
way up to 5 gives 987.

278
00:25:05,540 --> 00:25:08,860
Now, what's about the probability of Z smaller

279
00:25:08,860 --> 00:25:15,320
than minus 0.25? If you go back to the Z table,

280
00:25:15,380 --> 00:25:18,960
but for the other one, the negative one.

281
00:25:23,120 --> 00:25:28,540
Minus 2 minus 0.2 up

282
00:25:28,540 --> 00:25:34,780
to 5. 0.4013 minus,

283
00:25:36,620 --> 00:25:43,100
that will give the probability between three and

284
00:25:43,100 --> 00:25:43,280
five.

285
00:25:46,180 --> 00:25:48,900
This is the second part.

286
00:25:51,120 --> 00:25:52,460
So the final answer.

287
00:26:00,630 --> 00:26:05,450
So this is the probability that the selected cash

288
00:26:05,450 --> 00:26:10,650
fish will weigh between three and five pounds.

289
00:26:11,810 --> 00:26:20,770
Now, other question is, for the same problem, you

290
00:26:20,770 --> 00:26:28,020
said a citation Catfish should be one of the top 2

291
00:26:28,020 --> 00:26:33,860
% in the weight. Assuming the weights of catfish

292
00:26:33,860 --> 00:26:38,660
are normally distributed, at what weight in bounds

293
00:26:38,660 --> 00:26:43,680
should the citation, the notation be established?

294
00:26:45,800 --> 00:26:50,600
So in this board, he asked about what's the value

295
00:26:50,600 --> 00:26:52,120
of x, for example.

296
00:26:57,160 --> 00:27:01,680
is greater than what value here. And this

297
00:27:01,680 --> 00:27:07,880
probability equals 2%. Because you said the

298
00:27:07,880 --> 00:27:14,720
citation catfish should be one of the top 2%. So

299
00:27:14,720 --> 00:27:23,560
the area in the right here, this area is 2%.

300
00:27:26,000 --> 00:27:34,080
What's the value of x in this case? So here, the

301
00:27:34,080 --> 00:27:38,420
value of x greater than a equals 0.02, and we are

302
00:27:38,420 --> 00:27:39,460
looking for this value.

303
00:27:45,750 --> 00:27:49,170
gives the area to the left side. So this

304
00:27:49,170 --> 00:27:54,490
probability actually, the area to the right is 2%,

305
00:27:54,490 --> 00:28:00,810
so the area to the left is 98%. So this is the

306
00:28:00,810 --> 00:28:01,410
same as,

307
00:28:04,610 --> 00:28:07,930
as we know, the equal sign does not matter because

308
00:28:07,930 --> 00:28:09,090
we have continuous distribution.

309
00:28:12,050 --> 00:28:14,650
continuous distribution, so equal sign does not

310
00:28:14,650 --> 00:28:18,510
matter. So now, if you ask about P of X greater

311
00:28:18,510 --> 00:28:22,190
than a certain value equals a probability of, for

312
00:28:22,190 --> 00:28:27,330
example, 0.02, you have to find the probability to

313
00:28:27,330 --> 00:28:31,890
the left, which is 0.98, because our table gives

314
00:28:31,890 --> 00:28:36,470
the area to the left. Now, we have to find the

315
00:28:36,470 --> 00:28:40,820
value of A such that Probability of X is more than

316
00:28:40,820 --> 00:28:44,820
or equal to 0.98. So again, we have to look at the

317
00:28:44,820 --> 00:28:50,140
normal table, but backwards, because this value is

318
00:28:50,140 --> 00:28:53,720
given. If the probability is given, we have to

319
00:28:53,720 --> 00:28:58,900
look inside the body of the table in order to find

320
00:28:58,900 --> 00:28:59,580
the z-score.

321
00:29:03,350 --> 00:29:07,850
x equals mu plus z sigma in order to find the

322
00:29:07,850 --> 00:29:12,290
corresponding value x. So again, go back to the

323
00:29:12,290 --> 00:29:22,010
normal table, and we are looking for 98%. The

324
00:29:22,010 --> 00:29:27,930
closest value to 98%, look here, if you stop here

325
00:29:27,930 --> 00:29:30,850
at 2, go to the right,

326
00:29:33,660 --> 00:29:39,380
Here we have 9798 or

327
00:29:39,380 --> 00:29:41,480
9803.

328
00:29:42,640 --> 00:29:50,460
So the answer might be your z-score could be 2.05

329
00:29:50,460 --> 00:29:59,440
or 2.06. So again, in this case, the table does

330
00:29:59,440 --> 00:30:04,400
not give the exact. So the approximate one might

331
00:30:04,400 --> 00:30:08,920
be between them exactly. Or just take one of

332
00:30:08,920 --> 00:30:13,140
these. So either you can take 9798, which is

333
00:30:13,140 --> 00:30:19,500
closer to 98% than 9803, because it's three

334
00:30:19,500 --> 00:30:23,780
distant apart. So maybe we can take this value.

335
00:30:24,500 --> 00:30:27,640
Again, if you take the other one, you will be OK.

336
00:30:28,540 --> 00:30:36,730
So you take either 2.05. or 2.06. So let's take

337
00:30:36,730 --> 00:30:45,270
the first value, for example. So my x equals mu, z

338
00:30:45,270 --> 00:30:56,570
is 2.05, times sigma, 0.8. Multiply 2.05 by 8, 0

339
00:30:56,570 --> 00:31:03,610
.8, then add 3.2, you will get What's your answer?

340
00:31:08,390 --> 00:31:17,450
3.2 plus 2 point... So around 4.8. So your answer

341
00:31:17,450 --> 00:31:18,890
is 4.84.

342
00:31:23,810 --> 00:31:29,470
Now if you go back to the problem, and suppose you

343
00:31:29,470 --> 00:31:30,830
know the value of x.

344
00:31:34,250 --> 00:31:36,010
So the probability of X.

345
00:31:42,990 --> 00:31:45,110
Double check to the answer.

346
00:31:49,490 --> 00:31:58,010
4.84. Just check. V of X greater than this value

347
00:31:58,010 --> 00:32:03,290
should

348
00:32:03,290 --> 00:32:09,500
be Two percent. Two percent. So the probability of

349
00:32:09,500 --> 00:32:13,440
X greater than this value should be equal to one

350
00:32:13,440 --> 00:32:18,980
zero. So this problem is called backward normal

351
00:32:18,980 --> 00:32:24,960
calculation because here first step we find the

352
00:32:24,960 --> 00:32:28,040
value of this score corresponding to this

353
00:32:28,040 --> 00:32:33,420
probability. Be careful. The probability of X

354
00:32:33,420 --> 00:32:38,740
greater than 2 is 0.02. So my value here should be

355
00:32:38,740 --> 00:32:44,980
to the right. Because it says greater than A is

356
00:32:44,980 --> 00:32:51,240
just 2%. If you switch the position of A, for

357
00:32:51,240 --> 00:32:57,130
example, if A is on this side, And he asked about

358
00:32:57,130 --> 00:33:02,850
E of X greater than E is 2%. This area is not 2%.

359
00:33:02,850 --> 00:33:10,050
From A up to infinity here, this area is not 2%

360
00:33:10,050 --> 00:33:14,070
because at least it's greater than 0.5. Make

361
00:33:14,070 --> 00:33:16,610
sense? So your A should be to the right side.

362
00:33:17,590 --> 00:33:21,810
Because the value of X greater than E, 2% is on

363
00:33:21,810 --> 00:33:26,400
the other side. Let's move to the next one.

364
00:33:36,180 --> 00:33:37,660
For the same question.

365
00:33:52,790 --> 00:33:57,150
Again, the owner of Catfish Market determined the

366
00:33:57,150 --> 00:34:00,930
average weight of a catfish 3.2 with

367
00:34:00,930 --> 00:34:04,670
standardization 0.8 and we are assuming the

368
00:34:04,670 --> 00:34:08,270
weights of catfish are normally distributed, kiosk

369
00:34:08,270 --> 00:34:17,390
above. Above what weight? Do 89.8% of the weights

370
00:34:17,390 --> 00:34:18,070
care?

371
00:34:20,630 --> 00:34:27,980
Above? And above, so x greater than. X minus. And

372
00:34:27,980 --> 00:34:34,900
98, 89, sorry, 89. So this is a percentage he's

373
00:34:34,900 --> 00:34:45,860
looking for. 89.8%. Now what's the value of A? Or

374
00:34:45,860 --> 00:34:51,560
above what weight? Do 89.8% of the weights occur?

375
00:34:57,730 --> 00:35:02,670
You just make the normal curve in order to

376
00:35:02,670 --> 00:35:07,010
understand the probability. Now, A should be to

377
00:35:07,010 --> 00:35:11,550
the right or to the left side? Imagine A in the

378
00:35:11,550 --> 00:35:15,990
right side here. Do you think the area above A is

379
00:35:15,990 --> 00:35:21,670
89%? It's smaller than 0.5 for sure. So it should

380
00:35:21,670 --> 00:35:29,950
be the other side. So this is your 8. Now, this

381
00:35:29,950 --> 00:35:36,690
area makes sense that it's above 0.5. It's 0.8980.

382
00:35:38,750 --> 00:35:42,850
Now, B of X greater than equals this value. And

383
00:35:42,850 --> 00:35:46,990
again, the table gives the area to the left. So

384
00:35:46,990 --> 00:35:53,740
this is actually X less than A, 1 minus this

385
00:35:53,740 --> 00:35:56,600
value, equals 0.1020.

386
00:36:01,480 --> 00:36:08,760
Now go back to

387
00:36:08,760 --> 00:36:14,400
the normal table. Here it's 0.1020. So it should

388
00:36:14,400 --> 00:36:17,500
be negative. I mean, your z-scope should be

389
00:36:17,500 --> 00:36:17,800
negative.

390
00:36:22,000 --> 00:36:25,640
Now look at 0.102.

391
00:36:28,560 --> 00:36:38,680
Exactly this value. 0.102 is minus 1.2 up to 7. So

392
00:36:38,680 --> 00:36:39,900
minus 1.27.

393
00:36:49,120 --> 00:36:57,900
Minus 1.2. All the way up to 7 is 0.102. So the

394
00:36:57,900 --> 00:37:04,160
corresponding z-score is minus 1.17. Now x again

395
00:37:04,160 --> 00:37:05,980
equals mu plus z sigma.

396
00:37:10,280 --> 00:37:19,960
Mu is 3.2 plus z is negative 1.17 times sigma.

397
00:37:24,250 --> 00:37:34,390
So it's equal to 3.2 minus 127 times 0.3. By

398
00:37:34,390 --> 00:37:36,510
calculator, you'll get the final result.

399
00:37:51,120 --> 00:37:56,180
If the probability is smaller than 0.5, then this

400
00:37:56,180 --> 00:38:00,480
one is negative. Go to the other one. If the

401
00:38:00,480 --> 00:38:04,040
probability is above 0.5, then use the positive z

402
00:38:04,040 --> 00:38:09,600
-score. So what's the answer? 2.18.

403
00:38:12,680 --> 00:38:19,850
Be careful. In the previous one, We had a

404
00:38:19,850 --> 00:38:28,870
probability of X greater than A equal 2%. In

405
00:38:28,870 --> 00:38:33,070
this case, the value of A, for example, is located

406
00:38:33,070 --> 00:38:35,930
in the upper tail here.

407
00:38:40,210 --> 00:38:45,530
For this part, you ask about B of X greater than A

408
00:38:45,530 --> 00:38:51,090
equal 0.89. It's here more than 0.5 should be on

409
00:38:51,090 --> 00:38:54,290
the other side. So you have U of X greater than

410
00:38:54,290 --> 00:38:58,910
equal this value, which is the score in this case

411
00:38:58,910 --> 00:39:02,490
minus 1.17. So the corresponding guess score

412
00:39:02,490 --> 00:39:11,810
actually is 2.24. So this is the weight that 89.8%

413
00:39:11,810 --> 00:39:17,970
of the weights are above it. So around 90% of the

414
00:39:17,970 --> 00:39:25,350
catch fish have weights above this value. So

415
00:39:25,350 --> 00:39:31,170
around 2 pounds. So around 90% of the weights are

416
00:39:31,170 --> 00:39:37,070
above 2.18 pounds. Maybe this is one of the most

417
00:39:37,070 --> 00:39:41,690
important questions in this chapter. Any question?

418
00:39:51,660 --> 00:39:57,580
The last part here, for the same problem he asked

419
00:39:57,580 --> 00:40:01,700
about, what's the probability that a randomly

420
00:40:01,700 --> 00:40:06,120
selected fish will weigh less than 2.2 pounds? I

421
00:40:06,120 --> 00:40:12,680
think straightforward. We did similar in part A.

422
00:40:14,900 --> 00:40:23,880
So B of X less than 0.2. So he's looking for this

423
00:40:23,880 --> 00:40:28,300
probability, which is straightforward one. This

424
00:40:28,300 --> 00:40:38,800
score, 3.2 minus, I'm sorry, it's 2.2 minus minus.

425
00:40:39,280 --> 00:40:43,420
It's 2.2 minus 3.2 divided by sigma.

426
00:41:04,900 --> 00:41:10,240
So again, find the probability now of Z less than

427
00:41:10,240 --> 00:41:11,120
or equal to negative 1.5.

428
00:41:14,870 --> 00:41:19,710
Now, in this case, we have to use the negative z.

429
00:41:20,370 --> 00:41:25,010
It's negative 1.15 minus 1.2 up to 5.

430
00:41:28,150 --> 00:41:30,070
So 0.1056.

431
00:41:33,730 --> 00:41:40,660
So the answer is around 10% of the catfish will

432
00:41:40,660 --> 00:41:46,420
weigh less than 2 pounds. So the answer is 0.1056.

433
00:41:48,340 --> 00:41:49,220
Questions?

434
00:41:52,780 --> 00:41:56,100
So go back to the PowerPoint presentation we have.

435
00:41:57,780 --> 00:41:59,580
The last topic,

436
00:42:02,560 --> 00:42:03,900
evaluating normality.

437
00:42:06,930 --> 00:42:09,350
Many times we mentioned something about normality

438
00:42:09,350 --> 00:42:14,750
and outliers. For sure, if outliers exist, in this

439
00:42:14,750 --> 00:42:19,410
case, the situation is not normal. Now, how can we

440
00:42:19,410 --> 00:42:21,370
tell if a data point is an outlier?

441
00:42:24,650 --> 00:42:28,270
If you remember, we talked about outliers in

442
00:42:28,270 --> 00:42:32,510
Chapter 3. By two ways.

443
00:42:36,750 --> 00:42:38,270
By this score.

444
00:42:42,650 --> 00:42:47,390
And we mentioned that any data point.

445
00:42:54,950 --> 00:42:59,010
Below minus 3, I mean smaller than minus 3, or

446
00:42:59,010 --> 00:43:04,010
above 3, these points are suspected to be

447
00:43:04,010 --> 00:43:04,750
outliers.

448
00:43:09,230 --> 00:43:16,230
So any point, any data value smaller than minus 3

449
00:43:16,230 --> 00:43:22,330
in this form, or above plus 3 is considered to be

450
00:43:22,330 --> 00:43:30,250
an outlier. If we compute the lower limit, which

451
00:43:30,250 --> 00:43:37,310
is Q1 minus 1.5 IQR over the upper limit.

452
00:43:40,170 --> 00:43:47,490
So we

453
00:43:47,490 --> 00:43:51,690
have lower limit, upper limit. So lower limit.

454
00:43:55,400 --> 00:43:56,480
And upper limit.

455
00:43:59,980 --> 00:44:08,420
Any data point below lower limit or above upper

456
00:44:08,420 --> 00:44:13,080
limit is considered to be a type. So therefore, we

457
00:44:13,080 --> 00:44:18,320
have two methods to determine or to examine if the

458
00:44:18,320 --> 00:44:20,960
observation is enough there or not. One by using

459
00:44:20,960 --> 00:44:24,060
this score, straightforward. And the other one, we

460
00:44:24,060 --> 00:44:27,420
have to look for The lower limit and upper limit.

461
00:44:28,960 --> 00:44:34,200
The other method by using software and later on

462
00:44:34,200 --> 00:44:39,380
you will take SPSS in order to determine if the

463
00:44:39,380 --> 00:44:43,100
data is normally distributed by using something

464
00:44:43,100 --> 00:44:52,540
called QQ plot or normal probability plot. So I'm

465
00:44:52,540 --> 00:44:56,710
going to skip this part. Because data is taken by

466
00:44:56,710 --> 00:45:00,970
using software. But in general,

467
00:45:04,330 --> 00:45:11,750
you may look at this graph. Generally speaking, if

468
00:45:11,750 --> 00:45:17,730
you have a probability plot of a data, and the

469
00:45:17,730 --> 00:45:25,530
points lie on a straight line, or close to it, In

470
00:45:25,530 --> 00:45:29,650
this case, the distribution is normal. It's hard

471
00:45:29,650 --> 00:45:33,390
to make this graph manual. It's better to use

472
00:45:33,390 --> 00:45:38,510
software. But at least if we have this graph, and

473
00:45:38,510 --> 00:45:42,150
the points are close to the straight line. I mean,

474
00:45:42,250 --> 00:45:45,950
the points are either on the straight line, lies

475
00:45:45,950 --> 00:45:49,550
on the straight line, or close it. In this case,

476
00:45:50,870 --> 00:45:55,030
the data is normally distributed. If the data

477
00:45:55,030 --> 00:46:00,870
points scattered away of the straight line, then

478
00:46:00,870 --> 00:46:04,390
the distribution is not normal either skewed to

479
00:46:04,390 --> 00:46:08,690
the right or skewed to the left. So for this

480
00:46:08,690 --> 00:46:14,150
specific graph, the plot is normally distributed,

481
00:46:14,290 --> 00:46:17,550
approximately normally distributed. Because most

482
00:46:17,550 --> 00:46:23,230
of the points here lie close to the line and few

483
00:46:24,260 --> 00:46:29,660
are scattered away. Or it means that there are few

484
00:46:29,660 --> 00:46:31,940
outliers in this case, we can ignore these values.

485
00:46:33,100 --> 00:46:35,620
So here plot is approximately a straight line

486
00:46:35,620 --> 00:46:40,360
except for a few outliers at the low and the

487
00:46:40,360 --> 00:46:44,780
right, those points. So generally speaking, the

488
00:46:44,780 --> 00:46:51,200
distribution is normal distribution. That's all

489
00:46:51,200 --> 00:46:53,700
for this chapter.