[SOUND] The Standard Normal Distribution is the special case of the normal distribution. It has mean of zero and a standard deviation of one. The density function for this graph is given by this equation. Where e is the constant 2.71828, and pi is 3.14. And in the case of a Standard Normal Distribution, mu is zero and sigma is one. The shaded area in this graph represents the probability of a random variable. Falling within the two given ranges. This probability is found by solving for this shaded area. The area under this curve has been solved for the standard normal distribution, and we have access to a standard normal table, also known as Z table. Here, is an example of z. Although in this class, we will be working on large data-sets using Excel, it is still important that you understand the concept of normal table. And would know how to read it. There are many displays of normal table, to know what is being displayed, you have to look at the legend. For this table, the value represents the probability of X being less than Z. So basically, everything to the left of z. Using this table, we will use the z score to answer questions for any normal distribution. I will use this table to show you how to answer some possible research questions. To learn how to use Excel, please watch the Excel videos. The previous version of the standardized test, SAT, which high school students take has been designed, so that each of it's section, for example Math, to have a mean of 500 and standard deviation of 100. Let's that someone you know get's a score of 635. How well has this person done? First, we need to find the Z score. This will tell us how many standard deviation this person is from the mean and that is, 635 minus 500, divided by the standard deviation of 100. This person’s test score is above the mean by 1.35 standard deviation. This is the person’s Z score and we can now see in what percentile this person will fall. So, lets look at the normal table to find this value. For this particular person, who is 1.35 above the mean. To find the proportion of test takers who have scores less than 635, we are looking for P sub x less than or equal to 1.35. Now let's go to the table. For Z of 1.35, first we go down to the row that has the value of 1.3 and then go to the column which has the heading of 0.5. The value from the table is 0.9115, which is the probability of getting score less than 635. The shaded area under the curve represents the probability of being less or equal to 135 standard deviations above the mean. So this particular test taker is about 91 percentile. As you can see, the Z score, in this case 1.35 is placing this particular test taker among all the people who took this test. One might ask the question in another direction, that is, what is the probability of getting a score greater than 635? Our random variable, which is the test score, is still 635, but now we're looking for probability of finding people who score higher than this value. This is simply one minus probability of getting a score less than or equal to 635. This is because the total area under the curve is equal to 1. The blue shaded area represents the property of z being less than or equal to 135. And the white area is the complement, which is the probability of z being greater than or equal to 1.35 for all those test takers. That score higher than 635. In this case, it is 1- 0.9115, or about 0.0885. Now let's see how we can use the table if you get a negative Z-score. Using the same example, our test taker now gets a score of 458 on a test which has a mean of 500 and standard deviation of 100. The Z-score for this test taker is a -0.42. The negative sign means that this test taker is to the left of the mean, below average, as shown in the graph. The blue shaded area represents the scores below 458 and the white, the larger area it shows all those who scored higher. So what percentile would this test that could be? The table that I'm using has all positive values. But knowing the properties of normal distribution, we can deduce the negative Z values just as well. Remember that one of the properties of normal table is that it's symmetrical, which would imply that the probability of finding a test score less than or equal to -0.42 standard deviation from the mean. Is the same as finding a test scores that are greater than positive 0.42 standard deviations above the mean. Here's the probability of finding that the score less than or equal to -0.42 standard deviation from the mean. And here's the probability of finding a test score that are higher than positive 0.42 standard deviation, above the mean. The two shaded areas are exactly the same due to the symmetrical property of normal curve. We know how to use the table for positive Z, so let’s look at the table. Here’s the part of the table that pertains to our positive Z of 4 2. We first go to the row with the heading of 0.4 and then go to the column with the heading of 0.02. And where these two intersect, represents the probability of Z being less than or equal to 0.42 which is 0.6628. The probability represents the white area under the curve, which is probability of Z being less than or equal to 0.42. In other word, probability of Z being less than or equal to 0.42 is 0.6628. Thus, the probability of Z being equal or greater than 0.42 is then 1 minus 0.6628 or 0.3772 and that is the blue shaded area. So, we just found the probability of Z being equal or greater than 0.42 is 0.3772. Also because of symmetry we know that probability Z less than equal to -0.42 is the same as the probability Z being more than +0.42. Which means 37.72% of test takers have scores below 0.458. While 66.28% have higher score. So, whenever you have a negative Z value, find the probability that corresponds to the positive Z. What you get from the table is the area to the right of the positive value. To get the blue shaded area, subtract that value from 1. We did this for our example to get 0.3772. Now, let's see how we can use the table to find probabilities of finding values between two points. Here, we are going to look at probability of finding a score between 490 and 550 given that the population is 500 and the standard deviation is 100. So we first will find the Z score for the two points and thus, we are looking for probability of Z being between -0.1 and positive 0.5, and that would be this blue shaded area under the curve. To find the probability of Z being between -0.1 and positive 0.5, we first need to find the probability of z being less than or equal to 0.5. Then subtract from it the probability of z being less than or equal to -0.1. The probability of Z less than equal to 0.5 is from the table on the row with the heading 0.5 and the column of 0.00, which gives you 0.6915. Probability of Z being less than or equal to -0.1 can be solved the same way that we solve for negative Z values. Read the probability from the table as if the Z is a positive score, thus 0.1. Go to the row heading of 0.1 and column heading of 0.00. And we see that we get the value of 0.5398, but recall that in the negative Z value, the probability that we read from the table actually is for the white shaded area. Therefore, the area of interest to us, the blue shaded area, is 1 minus 0.5398, or 0.4602. Now, we are ready to calculate the probability of Z being between -0.1 and positive 0.5, and that is 0.6915 minus 0.4602, which is 0.2313. Now, let's practice. Given the average score of 500 and standard deviation of 100, what is the probability of finding someone who has scored between 600 and 650? The values you need are in this partial normal table that is displayed here. To answer this question, we will follow the same principle as we just learned. Find the Z value for 600 and 650, which will be 1 and 1.5, then using the normal table we can find the probability of Z being less than or equal to 1.5 to be 0.9332, we highlighted here in green. And probability of Z being less than or equal to 1 to be 0.8413, highlighted in pink. Subtracting these two gives us 0.0919. Thus, probability of finding someone who scores between 600 and 650 is 0.0919. Sometimes, you know the percentiles that you’re interested in, but wonder how many standard deviations from the mean would give you such percentile. Let’s say that we would like one to be the 95th percentile for the test and would want to know what score would give us that. Given what we know about the average of test is 500 and we also know that the standard deviation is 100. We're looking for a test score for which the probability of X being less than or equal to this value is 0.95, which is this. To answer this question we have to solve for Z where the area to the left of it is 0.95. To solve for Z, we can look at the normal table and look for the value that comes closest to 0.95. Here we find two values, 0.9495 and 0.9505 and the value of 0.95 is right in the middle of these two values. So we can extrapolate for the value of Z. Z value for 0.9495 is 164 and for 0.9505 is 1.65. Therefore, for 0.95 we can extrapolate that the Z is 1.645. So now we know that to be in the 95th percentile the test score must be more that 1.645 standard deviations above the mean. And that is simply the mean, 500, plus 1.645 times the standard deviation of 100, which will be a score of 664.5. Thus, anyone with this score will be at 95th percentile among all test-takers. In this lesson, you learned how to use standard normal distribution to answer questions about any normal distribution. By using the table, we don't have to actually solve for the area under the curve and that is a big time-saver.