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In this video, we'll be learning about Z scores and standardization.

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By learning about both of these topics,

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you will learn how to calculate exact

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proportions using the standard normal distribution.

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What is the standard normal distribution?

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The standard normal distribution is a special type of normal distribution that

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has a mean of 0 and a standard deviation of 1.

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Because of this,

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the standard normal distribution is always centered at

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0 and has intervals that increase by 1.

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Each number on the horizontal axis corresponds to a Z-score.

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A Z-score tells us how many standard deviations an observation is from the mean Mu.

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For example,

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a Z-score of negative 2 tells me that I

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am two standard deviations to the left of the mean

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and a Z-score of 1.5 tells me that I am one-and-a-half standard deviations to the right of the mean.

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Most importantly,

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a Z-score allows us to calculate how much

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area that specific Z score is associated with.

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And we can find out that exact area using something called a Z-score table,

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also known as the standard normal table.

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This table tells us the total amount of area

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contained to the left side of any value of Z.

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For this table,

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the top row and the first column correspond to Z values

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and all the numbers in the middle correspond to areas.

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For example, according to the table,

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a Z-score of negative 1.95 has an area of 0.0256 to the left of it.

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To say this in a more formal manner,

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we can say that the proportion of Z less than negative 1.95 is equal to 0.0256.

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We can also use the standard normal table to determine

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the area to the right of any Z value.

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All we have to do is take one minus the area that corresponds to the Z value.

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For example,

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to determine the area to the right of a Z-score of 0.57,

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all we have to do is find the area that

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corresponds to the Z value and then subtract it from 1.

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According to the table, the Z-score of 0.57 has an area of 0.7157 to the left of it.

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So, 1 minus 0.7157 gives us an area of 0.2843, and that is our answer.

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The reason why we can do this is

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because we have to remember that the normal distribution

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is a density curve and it always has a total area equal to 1 or 100%.

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You can also use the Z-score table to do a reverse look-up,

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which means you can use the table to see

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what Z score is associated with a specific area.

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So, if I wanted to know what value of Z corresponds to an area of 0.8461 to the left of it,

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all we have to do is find 0.8461 on the table and see what value of Z it corresponds to.

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We see that it corresponds to a Z value of 1.02.

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The special thing about the standard normal distribution is

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that any type of normal distribution can be transformed into

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it.

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In other words,

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any normal distribution with any value of Mu

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and Sigma can be transformed into the standard

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normal distribution where you have a Mu of 0 and a standard deviation of 1.

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This conversion process is called standardization.

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The benefit of standardization is that it allows us to use the Z-score table to

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calculate exact areas for any given normally distributed

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population with any value of Mu or Sigma.

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Standardization involves using this formula.

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This formula says that the Z-score is equal to an observation X

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minus the population mean Mu divided by the population of standard deviation Sigma.

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So, suppose that we gathered data from last year's final chemistry exam

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and found that it followed a normal distribution with a mean of 60 and

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a standard deviation of 10.

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If we were to draw this normal distribution,

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we would have 60 located at the center of the distribution because it

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is the value of the mean and each interval would increase by 10,

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since that is the value of the standard deviation.

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To convert this distribution to the standard normal distribution,

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we will use the formula.

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The value of Mu is equal to 60 and

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the value of Sigma is equal to 10.

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We can then take each value of X and plug it into the equation.

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If I plug in 60, I will get a value of 0.

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If I plug in 50, I will get a value of negative 1.

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If I plug in 40, I will get a value of negative 2.

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If we do this for each value,

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you can see that we end up with the same values as a standard normal distribution.

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When doing this conversion process,

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the mean of the normal distribution will always be converted to 0

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and the standard deviation will always correspond to a value of 1.

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It's important to remember that this will happen with any normal

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distribution no matter what value the Mu and Sigma are.

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Now, if I asked you what proportion of students score less than 49 on the exam,

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it is this area that we are interested in.

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However,

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the proportion of X less than 49 is

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unknown until we use the standardization formula.

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After plugging in 49 into this formula, we end up with a value of negative 1.1.

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As a result, we will be looking for the proportion of Z less than negative 1.1.

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And finally,

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we can use the Z score table to determine how much area is associated with the Z score.

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According to the table, there is an area of 0.1357 to the left of this Z value.

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This means that the proportion of Z less than negative 1.1 is 0.1357.

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This value is in fact the same proportion of

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individuals that scored less than 49 on the exam.

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As a result, this is the answer.

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Let's do one more example,

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when measuring the heights of all students at a local university,

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it was found that it was normally distributed with a mean

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height of 5.5 feet and a standard deviation of 0.5 feet.

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What proportion of students are between 5.81 feet and 6.3 feet tall?

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Before we solve this question,

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it's always a good habit to first write down important information.

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So, we have a Mu of 5.5 feet and a Sigma of 0.5 feet.

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We are also looking for the proportion of

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individuals between 5.81 feet and 6.3 feet tall.

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This corresponds to this highlighted area.

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To determine this area, we need to standardize the distribution,

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so we will use the standardization formula.

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Plugging in 5.81 to this formula gives us a Z-score of 0.62.

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And plugging in 6.3 into the formula gives us a Z-score of 1.6.

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According to the standard normal table,

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the Z-score of 0.62 corresponds to an area of 0.7324,

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and the Z-score of 1.6 corresponds to an area of 0.9452.

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To find the proportion of values between 0.62 and 1.6,

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we must subtract the smaller area from the bigger area.

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So, 0.9452 minus 0.7324 gives us 0.2128.

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As a result, the proportion of students between 5.81 feet and 6.3 feet tall is 0.2128.

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