Lesson Video: Applications of Normal Distribution | Nagwa Lesson Video: Applications of Normal Distribution | Nagwa

# Lesson Video: Applications of Normal Distribution Mathematics

In this video, we will learn how to apply the normal distribution in real-life situations.

16:13

### Video Transcript

In this video, we will learn how to apply the normal distribution in real-life situations. Many continuous variables in the real world approximately follow the normal distribution. For example, heights and weights collected from unbiased samples are expected to be normally distributed.

We should already be familiar with the theory associated with the normal distribution. If π is a normally distributed random variable with mean π and standard deviation π or variance π squared, we write that π follows a normal distribution ππ squared. We should also be familiar with the characteristic bell-shaped curve of the probability distribution for a normal random variable. We recall in particular that it is symmetrical about the mean of the distribution π. We also know that we can use the normal distribution to approximate the percentage of data points from the distribution lying within a particular range.

For example, the area under the distribution curve between the values of π and π gives the probability that a random variable taken from this distribution is between π and π, the probability that π is greater than or equal to π but less than or equal to π. Now, itβs important to remember here that normally distributed variables are continuous random variables. And so, the probability that π is actually equal to any particular value is zero. We therefore donβt need to be concerned about whether we use weak or strict inequalities for the endpoints of the interval. And we can be satisfied that, for example, the probability that π is less than π is the same as the probability that π is less than or equal to π.

We should also already be familiar with the standard normal distribution, which has a mean of zero and a standard deviation of one. If π is a normally distributed random variable with mean π and standard deviation π, then π which is equal to π minus π over π is a standard normal random variable. We can convert an observation lowercase π₯ of our random variable capital π to a standardized scale by applying the formula π equals π₯ minus π over π. And this standardized value is known as a π-score. We can then use statistical tables to look up the probabilities associated with particular π-scores.

The tables weβre going to use throughout this video are tables which correspond to the area between zero and a positive π§-score. That is, we can look up probabilities of the form the probability that π is greater than or equal to zero but less than or equal to some positive value lowercase π§. We can then use the symmetry of the normal distribution and other properties to find the probabilities associated with other regions, as weβll see during our examples. The problems weβll focus on in this video will all be real-life applications of the skills weβve just recapped. In our first problem, weβll see how we can estimate a probability from a normal distribution in context.

A crop of apples has a mean weight of 105 grams and a standard deviation of 3 grams. It is assumed that a normal distribution is an appropriate model for this data. What is the approximate probability that a randomly selected apple from the crop has a weight less than 105 grams?

Letβs use the random variable π to denote the weight of an apple. Weβre told in the question that a normal distribution is appropriate for π. And it has a mean of 105 grams and a standard deviation of 3 grams. So we can write π follows a normal distribution 105, three squared. Weβre asked to find the probability that the weight of a randomly selected apple is less than 105 grams. In probability notation, weβre looking for the probability that π is less than 105.

Now, before we jump in and try to convert this value to a standardized π-score, letβs just pause and think about this. The value of 105 grams is significant for this distribution. It is, in fact, the mean. We know that a normal distribution is symmetrical about its mean. So this normal distribution is symmetrical about the value 105. The probability weβre looking for then, the probability that π is less than 105, is the probability that an observation is in the lower half of the distribution. Without performing any calculations, we can state that this probability is equal to 0.5 or, as a percentage, 50 percent.

Now, of course, we could go through the formal process of calculating a π-score and then using our probability tables for the standard normal distribution. But if weβre smart about it and we notice the significance of this value of 105 for this distribution, then thereβs no need to. We conclude that the approximate probability that a randomly selected apple from the crop has a weight less than 105 grams is 50 percent.

Letβs now consider another example of calculating a probability for a real-life application of the normal distribution.

The monthly salaries of workers at a factory are normally distributed with mean 210 pounds and standard deviation 10 pounds. Determine the probability of choosing at random a worker with a salary between 184 and 233 pounds.

Weβll begin by introducing the random variable π to denote the monthly salary. Weβre told in the question that these salaries are normally distributed with a mean of 210 pounds and a standard deviation of 10 pounds. So we can write that π follows a normal distribution 210, 10 squared.

Weβre asked to determine the probability that a worker chosen at random has a salary between 184 and 233 pounds. Thatβs the probability that π is greater than or equal to 184 and less than or equal to 233. Letβs just consider what this would look like in terms of an area under the normal distribution curve. We know that this normal distribution is symmetrical about this mean of 210. 184 is less than 210, so this value is in the lower half of the distribution, whereas 233 is greater than 210, so this value is in the upper half. The probability weβre looking for corresponds to the area between these two values.

To find these probabilities, we first need to convert the values of 184 and 233 to standardized values from the standard normal distribution using the formula π equals π₯ minus π over π. For 184, we have π equals 184 minus 210 all over 10, which is equal to negative 2.6. And for 233, π is equal to 233 minus 210 over 10, which is equal to 2.3. On our standardized scale then, the area weβre looking for is between the values of negative 2.6 and 2.3. The probability that π is greater than or equal to 184 but less than or equal to 233 is the same as the probability that the standard normal random variable π is greater than or equal to negative 2.6 and less than or equal to 2.3.

The region weβre looking for consists of areas either side of the mean, so we need to consider how best to find it. The area to the right of the mean, which is the probability that π is between zero and 2.3, is in the correct format to be evaluated using our statistical tables. So weβll do that in a moment. But what about the area to the left of the mean? This is the probability that π is greater than or equal to negative 2.6 and less than or equal to zero.

Well, we need to recall the symmetry of the normal distribution. The probability that π is greater than or equal to negative 2.6 and less than or equal to zero is the same as the probability that π is greater than or equal to zero but less than or equal to 2.6. And this is a probability in the correct form that we can look up in our statistical tables. So weβre going to find the probabilities associated with π-scores of 2.6 and 2.3 and then add them together. Here are those statistical tables with the π-scores themselves in the first column and then the second decimal place values in the top row.

If we look carefully, we see that the probability associated with the π-score of 2.6 is 0.4953 which, remember, by symmetry is the same as the probability that π is greater than or equal to negative 2.6 and less than or equal to zero. And we see that the probability associated with the π-score of 2.3 is 0.4893. Summing these values gives 0.9846. So we can conclude that the probability of choosing at random a worker with a salary between 184 and 233 pounds is 0.9846.

Letβs now consider one final example. In this problem, one of the two parameters of the normal distribution will be unknown. And weβll be required to calculate it using other information given in the question.

The heights of a sample of flowers are normally distributed with mean π and standard deviation 12 centimeters. Given that 10.56 percent of the flowers are shorter than 47 centimeters, determine π.

Weβre told that the heights of this sample of flowers are normally distributed. Weβre given the standard deviation; itβs 12 centimeters. But we arenβt given the mean. Letβs use the random variable π to represent the height of the flower. Then we can say the π follows a normal distribution π, 12 squared. The other information given in the question is that 10.56 percent of this sample of flowers are shorter than 47 centimeters. In probability notation, we can write that the probability that π is less than 47 is 0.1056.

Letβs think about what this tells us in terms of areas under the normal curve. As this probability is less than 0.5, this tells us that the value of 47 is in the lower half of the distribution. And this probability of 0.1056 corresponds to the area to the left of 47 under our normal distribution curve. Now, this probability would have been worked out by calculating a π-score for the observation of 47 and then using statistical tables. Weβre now going to attempt this process but in reverse.

We recall that the formula for calculating the π-score for an observation π₯ is π equals π₯ minus π over π. Now, as this value of 47 is in the lower half of the distribution, the π-score associated with it will be negative. But remember, we canβt look up negative π-scores in our statistical tables. We can only look up positive ones. If we consider instead the negative of this π-score, that would be a positive value because a negative multiplied by negative gives a positive.

By the symmetry of the normal distribution curve, the area to the left of our π-score is the same as the area to the right of its negative. So this area here in the top half of the distribution is also equal to 0.1056. We can then work out the area between zero and the negative of our π-score, using the fact that the area either side of the mean is equal to 0.5. So this area, now shaded in orange, is equal to 0.3944. We finally have the probability corresponding to an area for which we can use our standard normal tables. Using our tables in reverse, we can look up the π-score associated with the probability of 0.3944.

Remember, though, that this will give us the negative of the π-score associated with 47. Looking carefully at our statistical tables, we see that the π-score associated with the probability of 0.3944 is 1.25. So this tells us that negative π is equal to 1.25, which means π is equal to negative 1.25. In practical terms, this means that the observation of 47 is 1.25 standard deviations below the mean of the distribution. To work out what the mean is, we reverse the process of calculating the π-score.

Substituting negative 1.25 for π, 47 for π₯, and 12 for π, we have negative 1.25 equals 47 minus π over 12. We can multiply both sides of the equation by 12 to give 12 multiplied by negative 1.25 equals 47 minus π. And 12 multiplied by negative 1.25 is negative 15. We can then subtract 47 from each side to give negative 15 minus 47 equals negative π, negative 15 minus 47 is negative 62. And then dividing or multiplying both sides of the equation by negative one, we have π is equal to 62. We have determined then that the mean of this distribution, which remember was the heights of a sample of flowers, is 62 centimeters.

Letβs now review some of the key points weβve seen in this video. Many continuous variables in the real world approximately follow the normal distribution. We have seen examples concerning the weights of apples, the salaries of workers, and, finally, the heights of flowers. In problems relating to the normal distribution, we usually begin by defining a letter, which is usually π, to represent the normal random variable. We then write that π follows a normal distribution ππ squared. These parameters may be known, or one of them may be unknown, depending on the problem.

We can standardize an observation lowercase π₯ over the random variable uppercase π using the formula π equals π₯ minus π over π. This will give an observation from the standard normal distribution which has a mean zero and a standard deviation one. Probabilities for the standard normal distribution can be determined using statistical tables or scientific calculators. The tables we used in this video gave probabilities of the form the probability that π is greater than or equal to zero but less than or equal to some positive value lowercase π§. But other types of tables do exist.

Finally, we can use the symmetry of the normal distribution curve to convert the probabilities from these tables into probabilities corresponding to other regions under the curve.