In this video, we'll be learning about Z scores and standardization. By learning about both of these topics, you will learn how to calculate exact proportions using the standard normal distribution.
What is the standard normal distribution? The standard normal distribution is a special type of normal distribution that has a mean of zero and a standard deviation of one. Because of this, the standard normal distribution is always centered at zero, and has intervals that increase by one. Each number on the horizontal access corresponds to Z score. A Z score tells us how many standard deviations an observation is from the mean mu. For example, a Z score of -2, tells me that I am two standard deviations to the left of the mean, and Z score of 1.5 tells me that I am one and a half standard deviations to the right of the mean.
Most importantly, a Z score allows us to calculate how much area that specific Z score is associated with. And we can find what that exact area using 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 any value of Z. For this table, the top row, and the first column correspond to Z values, and all the numbers in the middle correspond to areas.
For example, according to the table, a Z score of -1.95 has an area of 0.0256 to the left of it. To save this in a more formal manner. We can say that the proportion of Z less than -1.95 is equal to 0.0256. We can also use the standard normal table to determine the area to the right of any Z value. All we have to do is take one minus the area that corresponds to the Z value.
For example, to determine the area to the right of Z score of 0.57, all we have to do is find the area that corresponds to this Z value, and then subtract it from one. According to the table, the Z score of 0.57 has an area of 0.7157 to the left of it. So 1-0.7157 gives us an area of 0.2843. And that is our answer.
The reason why we can do this is because we have to remember that the normal distribution is a density curve, and it always has a total area equal to one or 100%. You can also use the Z score table to do a reverse lookup, which means you can use the table to see what Z score is associated with a specific area. So if I wanted to know what value of Z corresponds to an area of 0.8461 to the left of it, all we have to do is find 0.8461 on the table, and see what value of Z it corresponds to. We see that it corresponds to a Z value of 1.02. The special thing about the standard normal distribution is that any type of normal distribution can be transformed into it. In other words, any normal distribution with any value of Mu and Sigma can be transformed into the standard normal distribution, where you have a Mu of zero and a standard deviation of one.
This conversion process is called standardization. The benefit of standardization is that it allows us to use the Z score table to calculate exact areas for any given normally distributed population with any value of Mu or Sigma. Standardization involves using this formula. This formula says that the Z score is equal to an observation X minus the population mean Mu, divided by the population of standard deviation Sigma.
So suppose that we gathered data from last year's final chemistry exam and found that it followed a normal distribution with a mean of 60 and a standard deviation of 10. If we were to draw this normal distribution, we would have 60 located at the center of the distribution, because it is the value of the mean. And each interval would increase by 10, since that is the value of the standard deviation. To convert this distribution, to the standard normal distribution, we will use the formula.
The value of Mu is equal to 60, and the value of Sigma is equal to 10. We can then take each value of X and plug it into the equation. If I plug 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 normal distribution.
When doing this conversion process, the mean of the normal distribution will always be converted to zero. And the standard deviation will always correspond to a value of one. It's important to remember that this will happen 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 less than 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 is associated with the Z score. According to the table, there's an area of 0.1357 to the left of this Z value. This means the proportion of Z less than -1.1 is 0.1357. This value is in fact, the same proportion of individuals that scored less than 49 on the exam. As a result, This is the answer.
Let's do one more example. When measuring the heights of all students at a local university, it was found that it was normally distributed with a mean height to 5.5 feet and a standard deviation of 0.5 feet. What proportion of students are between 5.81 feet and 6.3 feet tall? Before we solve this question, it's always a good habit to first write down important information. So we have a Mu of 5.5 feet and a Sigma of 0.5 feet. We are also looking for the proportion of individuals between 5.81 feet and 6.3 feet tall. This corresponds 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 formula gives us a Z score of 0.62 and plugging in 6.3 into the formula, gives us a Z score of 1.6. According to the standard normal table, the Z score of 0.62 corresponds to an area of 0.7324, and the Z score of 1.6 corresponds to an area of 0.9452. To find the proportion of values between 0.62 and 1.6, we must subtract the smaller area from the bigger area. So 0.9452 minus 0.7324 gives us 0.2128. As a result, the proportion of students between 5.81 feet and 6.3 feet tall is 0.2128.
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