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Lesson Video: Normal Distribution Mathematics

In this video, we will learn how to use the normal distribution to calculate probabilities and find unknown variables and parameters.

16:37

Video Transcript

In this video, we will learn how to use the normal distribution to calculate probabilities and find unknown variables and parameters. The normal distribution is one of the most important probability distributions because it can be used to model several types of naturally occurring phenomena, such as the heights of adults. And it can be a good approximation for other distributions when the number of data points is large.

Let’s look at this normal distribution more closely. A normal random variable is a type of continuous random variable. And if we look at a graph of its probability density function, it has a very distinctive shape. All normal distributions can be represented using a bell-shaped curve, which we call the normal curve or sometimes the Gaussian curve after the mathematician Carl Friedrich Gauss who was instrumental in developing the theory associated with the normal distribution. We use a capital letter, in this case 𝑋, to represent the normally distributed random variable. And the distribution can be described completely by two parameters, firstly the mean or expectation πœ‡ and secondly the variance 𝜎 squared. Let’s now consider some key features of this probability distribution.

First, the normal distribution is completely symmetrical about its mean value πœ‡. As with any probability distribution, the area under the entire curve is one, which means that the area either side of this vertical axis of symmetry is 0.5. The area to the left of any particular π‘₯-value on the horizontal axis gives the proportion of points from the distribution that are less than or equal to this value. We say the probability that the random variable capital 𝑋 is less than or equal to the observation lowercase π‘₯. Now it’s worth mentioning here that as we’re working with a continuous distribution, it doesn’t make any practical difference whether we talk about strictly less than or less than or equal to because the probability that our random variable is equal to any particular value is zero.

So we’ve seen that half the area lies either side of the vertical axis. But, in fact, the area below the curve can be approximately divided up further into the proportion of points that lie within certain key regions. This is called the empirical rule. If we consider one standard deviation either side of the mean first of all, this accounts for approximately 68.3 percent of the total area, which means that approximately this proportion of values from the distribution lie within one standard deviation of the mean.

In the same way, the region two standard deviations either side of the mean accounts for approximately 95 percent of the total area. So 95 percent of values from this distribution lie within two standard deviations of the mean. And if we go even further out to three standard deviations either side of the mean, this accounts for approximately 99.7 percent of the total area.

It is therefore very rare for values taken from a normal distribution to be more than three standard deviations from the mean, which has an important application in statistical process control. If a process is assumed to be normally distributed, any values which are more than three standard deviations from the mean are usually assumed to be outlying values and may indicate that an unusual event has occurred, which needs to be investigated. It’s helpful to remember the three key percentages associated with these distances from the mean as we’ll see in our first example.

For a normally distributed data set with mean 32.1 and standard deviation 2.8, between which two values would you expect 95 percent of the data set to lie?

We recall firstly that for a normally distributed random variable, approximately 95 percent of the data points lie within two standard deviations of the mean. We therefore need to calculate the values two standard deviations below and two standard deviations above the mean for this particular normal distribution.

We’re given in the question that the mean is 32.1 and the standard deviation is 2.8, so we can calculate these values fairly easily. The lower value πœ‡ minus two 𝜎 is 32.1 minus two multiplied by 2.8, which is 26.5. The upper value πœ‡ plus two 𝜎 is 32.1 plus two times 2.8, which is 37.7. And so by recalling part of the empirical rule for a normally distributed random variable, which tells us that approximately 95 percent of the data set lies within two standard deviations of the mean, we find that for this distribution, 95 percent of the data set will lie between 26.5 and 37.7.

More generally, we may want to find the proportion of points that lie in other regions under the curve. To do this, we need to consider one special case of the normal distribution, which is what we call the standard normal distribution. We usually denote this using the letter 𝑧. And it represents the normal distribution which has a mean of zero and a standard deviation, and hence variance, of one.

Values from this distribution are known as 𝑧-scores, and they represent the number of standard deviations above the mean a particular value is. For example, a 𝑧-score of 1.4 would mean a value 1.4 standard deviations above the mean, whereas a 𝑧-score of negative 2.1 would mean a value 2.1 standard deviations below the mean. These 𝑧-scores for a standard normal distribution are really useful because they allow us to view values from a normal distribution on a standardized scale.

We have a set of statistical tables which we’ll look at in detail later, in which we can look up the areas and hence the probabilities associated with particular 𝑧-scores. The type of tables we’re going to use are tables which give the probability that our random variable capital 𝑍 is between zero and an observation lowercase 𝑧. That is the proportion of points or the area between zero and a positive 𝑧-score. If we wanted to then work out the proportion of points that lie completely to the left, that is, that are completely less than a particular positive 𝑧-score, we would need to add on 0.5 to the value from our tables to account for the area to the left of the axis of symmetry. That’s the area shaded in pink.

Let’s consider a detailed example of how we can use our tables to find such a probability.

Use tables to find the normal probability corresponding to a 𝑧-score of 2.13.

We are asked to find the normal probability corresponding to a 𝑧-score of 2.13, which means the proportion of points or the area that lies to the left of this value of 2.13 under the standard normal distribution curve. So here are our statistical tables for the standard normal distribution. Now, these tables give the proportion of points or the area that lies between zero and a positive 𝑧-score. That’s only the part of the area now shaded in pink on our figure. That’s okay though because we know that the normal distribution is completely symmetrical about its mean. And so the orange part of the area is exactly 0.5. We therefore need to add 0.5 to whatever value we find in our table.

Now, looking at our tables, we can see that they have 𝑧-scores ranging from zero to three in the first column. These values increase by 0.1 each time. And then in the top row of the table, we have options for the second decimal place of our 𝑧-score. The 𝑧-score we want to look up is 2.13, so we look up 2.1 in the first column and then 0.03 because 2.1 plus 0.03 gives 2.13. We then find the value in the cell of the table where this row and this column intersect, and it is 0.4834. This tells us that the area between zero and 2.13 is 0.4834. The total area to the left of 2.13 is 0.5 plus this value, which is 0.9834. This is the normal probability corresponding to a 𝑧-score of 2.13. And it represents the total area to the left of 2.13 under the standard normal curve.

In the previous example, we saw how to use tables to find the area between zero and a positive 𝑧-score. We can also use these tables to work out the proportion of points that lie in other regions, and the symmetry of the normal distribution plays an important role. Firstly, because the curve is symmetrical about its mean, the area between zero and a positive 𝑧-score is the same as the area between the negative of that 𝑧-score and zero. We can also work out the area to the right of a particular 𝑧-score by using the fact that the total area under the curve is one. So the probability that our random variable 𝑍 is greater than or equal to a value lowercase 𝑧 is one minus the probability that it’s less than or equal to that value.

We can also work out the proportion of points that lie between two particular 𝑧-scores by subtracting one area from the other. And we’ll see some examples of the different types of problem we might encounter in our remaining examples. So this is great if the distribution we’re using is already the standard normal. But what’s even more useful is that we can use 𝑧-scores to convert values from any normal distribution with any mean and any standard deviation to a standard normal variable and therefore view them on the standard scale. We can do this using the formula 𝑧 equals π‘₯ minus πœ‡ over 𝜎. We take an observation π‘₯, subtract the mean of the distribution it’s from, and then divide by the standard deviation 𝜎. The 𝑧-score will then be an observation from the standard normal distribution.

The probability that our original random variable capital 𝑋 was between zero and lowercase π‘₯ is therefore the same as the probability that our new random variable 𝑍 is between zero and lowercase 𝑧, the 𝑧-score. And so we can use our standard normal tables to look up this probability. Let’s see an example of this.

Let 𝑋 be a random variable which is normally distributed with mean 68 and standard deviation three. Determine the probability that 𝑋 is greater than or equal to 61.7.

So we have this normally distributed random variable 𝑋, and we want to determine the probability that its value is greater than or equal to 61.7. We know that 61.7 will be in the lower half of the distribution as it’s less than the mean of 68. And so the probability we’re looking for corresponds to the area shaded in orange under our normal distribution curve. First, we need to calculate the 𝑧-score associated with this particular value using the formula 𝑧 equals π‘₯ minus πœ‡ over 𝜎. We have 𝑧 equals 61.7 minus 68 over three, which is negative 2.1, which tells us that this value of 61.7 is 2.1 standard deviations below the mean of 68.

Now, we can’t look a negative 𝑧-score up in our standard normal tables, so we need to consider instead the symmetry of the normal distribution curve. On our standardized scale, the area above a 𝑧-score of negative 2.1 will be the same as the area below a 𝑧-score of 2.1. We can look up the probability associated with a 𝑧-score of 2.1 in our standard normal tables, which will give us the area to the right of the mean. And then we can add 0.5 to account for the area to the left of the mean. Using our tables, we see that the probability associated with a 𝑧-score of 2.1 is 0.4821. So the probability that 𝑧 is less than or equal to 2.1, which is the same as the probability 𝑧 is greater than or equal to negative 2.1, which for our unstandardized random variable is the probability that 𝑋 is greater than or equal to 61.7, is 0.5 plus 0.4821, which is 0.9821.

Let’s now consider an example in which we calculate the probability between two values.

Let 𝑋 be a random variable which is normally distributed with mean 63 and variance 144. Determine the probability 𝑋 is greater than or equal to 37.56 and less than or equal to 57.36.

So we have a normally distributed variable 𝑋 with a mean of 63 and a variance of 144 β€” that’s 12 squared. We want to determine the probability that 𝑋 is between these two values, which are both in the lower half of the distribution. We begin by calculating the 𝑧-score for each value using the formula 𝑧 equals π‘₯ minus πœ‡ over 𝜎. For our first π‘₯-value, we have 37.56 minus 63 over 12, which is negative 2.12. And for our second value, the 𝑧-score is negative 0.47.

Now, we can’t look either of these values up in our standard normal tables as they’re both negative. So instead we use the symmetry of the normal distribution curve. On our standardized scale now, the probability that 𝑧 is greater than or equal to negative 2.12 but less than or equal to negative 0.47 is the same as the probability that 𝑧 is greater than or equal to positive 0.47 and less than or equal to 2.12, both of which we can look up in our standard normal tables.

Remember the tables give us the probability that 𝑧 is between zero and a positive 𝑧-score. So we can subtract the probability for 0.47 from the probability for 2.12. From our tables, the probabilities are 0.4830 and 0.1808. And then we find the difference, which is 0.3022. So, using standardized 𝑧-scores and the symmetry of the normal distribution, we found the probability that 𝑋 is greater than or equal to 37.56 and less than or equal to 57.36 is 0.3022.

Let’s now summarize the key points from this video. First, we saw the percentages associated with three key areas under the normal distribution curve using the empirical rule. To calculate the standardized 𝑧-score of an observation π‘₯, we subtract the mean πœ‡ and then divide by the standard deviation 𝜎. This will convert an observation from a normal distribution with mean πœ‡ and standard deviation 𝜎 to an observation from the standard normal distribution with mean zero and standard deviation one. We can use standard normal distribution tables to look up the area between zero and a positive 𝑧-score 𝑧. We can then use these values from the tables together with the symmetry of the normal distribution curve to calculate probabilities in a number of different formats.

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