Z-Scores and Standardization Transcript (2025)

In this video, we'll be learning about Z scores andstandardization. By learning about both of these topics, you will learn how tocalculate exact proportions using the standard normal distribution.

What is the standard normal distribution? The standard normaldistribution is a special type of normal distribution that has a mean of zeroand a standard deviation of one. Because of this, the standard normaldistribution is always centered at zero, and has intervals that increase byone. Each number on the horizontal access corresponds to Z score. A Z scoretells us how many standard deviations an observation is from the mean mu. Forexample, a Z score of -2, tells me that I am two standard deviations to theleft of the mean, and Z score of 1.5 tells me that I am one and a half standarddeviations to the right of the mean.

Most importantly, a Z score allows us to calculate how much areathat specific Z score is associated with. And we can find what that exact areausing something called Z score table, also known as the standard normal table.This table tells us the total amount of area contained to the left side of anyvalue of Z. For this table, the top row, and the first column correspond to Zvalues, and all the numbers in the middle correspond to areas.

For example, according to the table, a Z score of -1.95 has anarea of 0.0256 to the left of it. To save this in a more formal manner. We cansay that the proportion of Z less than -1.95 is equal to 0.0256. We can alsouse the standard normal table to determine the area to the right of any Zvalue. All we have to do is take one minus the area that corresponds to the Zvalue.

For example, to determine the area to the right of Z score of0.57, all we have to do is find the area that corresponds to this Z value, andthen subtract it from one. According to the table, the Z score of 0.57 has anarea of 0.7157 to the left of it. So 1-0.7157 gives us an area of 0.2843. Andthat is our answer.

The reason why we can do this is because we have to remember thatthe normal distribution is a density curve, and it always has a total areaequal to one or 100%. You can also use the Z score table to do a reverselookup, which means you can use the table to see what Z score is associatedwith a specific area. So if I wanted to know what value of Z corresponds to anarea of 0.8461 to the left of it, all we have to do is find 0.8461 on thetable, and see what value of Z it corresponds to. We see that it corresponds toa Z value of 1.02. The special thing about the standard normal distribution isthat any type of normal distribution can be transformed into it. In otherwords, any normal distribution with any value of Mu and Sigma can betransformed into the standard normal distribution, where you have a Mu of zeroand a standard deviation of one.

This conversion process is called standardization. The benefit ofstandardization is that it allows us to use the Z score table to calculateexact areas for any given normally distributed population with any value of Muor Sigma. Standardization involves using this formula. This formula says thatthe Z score is equal to an observation X minus the population mean Mu, dividedby the population of standard deviation Sigma.

So suppose that we gathered data from last year's final chemistryexam and found that it followed a normal distribution with a mean of 60 and astandard deviation of 10. If we were to draw this normal distribution, we wouldhave 60 located at the center of the distribution, because it is the value ofthe mean. And each interval would increase by 10, since that is the value ofthe standard deviation. To convert this distribution, to the standard normaldistribution, we will use the formula.

The value of Mu is equal to 60, and the value of Sigma is equalto 10. We can then take each value of X and plug it into the equation. If Iplug in 60, I will get a value of zero. If plug in 50, I will get a value of-1. If I plug in 40, I will get a value of -2. If we do this for each value,you can see that we end up with the same values as a standard normaldistribution.

When doing this conversion process, the mean of the normaldistribution will always be converted to zero. And the standard deviation willalways correspond to a value of one. It's important to remember that this willhappen with any normal distribution, no matter what value the Mu and Sigma are.Now, if I asked you what proportion of students score less than 49 on the exam,it is this area that we are interested in. However, the proportion of X lessthan 49 is unknown until we use the standardization formula.

After plugging in 49 into this formula, we end up with a value of-1.1. As a result, we will be looking for the proportion of Z less than -1.1.And finally, we can use the Z score table to determine how much area isassociated with the Z score. According to the table, there's an area of 0.1357to the left of this Z value. This means the proportion of Z less than -1.1 is0.1357. This value is in fact, the same proportion of individuals that scoredless than 49 on the exam. As a result, This is the answer.

Let's do one more example. When measuring the heights of allstudents at a local university, it was found that it was normally distributedwith a mean height to 5.5 feet and a standard deviation of 0.5 feet. Whatproportion of students are between 5.81 feet and 6.3 feet tall? Before we solvethis question, it's always a good habit to first write down importantinformation. So we have a Mu of 5.5 feet and a Sigma of 0.5 feet. We are also lookingfor the proportion of individuals between 5.81 feet and 6.3 feet tall. Thiscorresponds to this highlighted area.

To determine this area, we need to standardize the distribution.So we will use the standardization formula. Plugging in 5.81 to this formulagives us a Z score of 0.62 and plugging in 6.3 into the formula, gives us a Zscore of 1.6. According to the standard normal table, the Z score of 0.62corresponds to an area of 0.7324, and the Z score of 1.6 corresponds to an areaof 0.9452. To find the proportion of values between 0.62 and 1.6, we mustsubtract the smaller area from the bigger area. So 0.9452 minus 0.7324 gives us0.2128. As a result, the proportion of students between 5.81 feet and 6.3 feettall is 0.2128.

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Z-Scores and Standardization Transcript (2025)
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