Application of the Normal Distribution | Applied Biostatistics | BIO733_Topic057

Duration: 9:41 1164 transcript words EN summary

Video summary

This module explores the practical applications of the normal distribution in applied biostatistics, specifically focusing on how to calculate probabilities for specific values of a random variable. The primary example involves a study by Eldridge et al., which monitored 529 normally developing children aged 8 to 15 using a device called an Up Timer to track time spent in an upright position over a 24-hour period. The researchers determined that the time spent standing followed a normal distribution with a mean of 5.4 hours and a standard deviation of 1.3 hours. Using these parameters, the video demonstrates how to find the probability that a randomly selected child spends less than 3 hours upright by converting the raw score into a Z-score, which results in a value of -1.85. By consulting cumulative probability tables, it is concluded that there is only a 3.22% chance that any given child will sit for fewer than three hours. The tutorial further illustrates how to handle scenarios involving ranges rather than single values through a second example concerning ammonia concentration in breath samples. In this case, the data for a 27-year-old female subject follows a normal distribution with a mean of 491 parts per billion and a standard deviation of 119. The objective is to determine the probability that the ammonia level falls between 292 and 649 parts per billion on any random day. To solve this, the video explains the process of converting both boundary values into their respective Z-scores, yielding -1.67 and 1.33. The final probability is derived by calculating the area under the curve for the larger Z-score and subtracting the area corresponding to the smaller Z-score, resulting in a cumulative probability of 0.8607, or 86.07%. A third application addresses how to estimate the number of individuals within a population who meet specific criteria based on calculated probabilities. Returning to the children's upright position data, the video poses the question of how many out of a hypothetical population of 10,000 children would be expected to stand for more than 8.5 hours. First, the probability of exceeding this threshold is found by subtracting the cumulative area up to 8.5 hours from the total area of one, yielding a probability of 0.0087. By multiplying this probability by the population size, the analysis predicts that approximately 87 children in a group of 10,000 would spend more than 8.5 hours in an upright position. In conclusion, the normal distribution is presented as a widely applicable and fundamental tool in statistical inference. The video emphasizes that establishing the normality of a random variable is a critical property sought when performing inferential statistical methods. Through these diverse examples involving time monitoring and chemical concentration analysis, the module effectively demonstrates how to translate real-world biological data into mathematical probabilities using Z-scores and standard normal tables. This approach allows researchers to make precise predictions about populations based on sample statistics, reinforcing the importance of understanding distribution parameters like mean and standard deviation in biostatistical practice.

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In this module, we will be talking about the various applications of the normal distribution. And using these application, we will also find out the probabilities for certain values of our random variable. Let's take the first example where Eldridge et al. studied 529 normally developing children between the ages 8 to 15 years. And each of these children wore a device that's named as Up Timer. And they want they they were monitored continuously for a 24-hour period. And this 24-hour period included a typical school day. So, the researchers found that the amount of time children spent in a upright position followed a normal probability distribution with its mean value 5.4. And it's uh and the standard deviation as 1.3 hours. So, assuming these findings findings applies to all the all the children age 8 to 15, our interest is to find the probability that children selected at random spends less than 3 hours in the upright position in the 24-hour period. So, far from here, first of all, we have to identify that what is our event of interest or variable of our event of interest. Moreover, we since we already know that our random variable follows a normal probability distribution, so we need to identify the values of mean and the standard deviation, which are the two parameters of the normal probability distribution. So, in this situation, our variable of interest is that the child selected at random spends less uh the the the amount of time a child spends in an upright position with their mean as 5.4 hours and the standard deviation as 1.3 hours. So, n for this experiment is 529, which is a sample size. So, assuming that the these finding applies to all the children, if we are interested in finding out the probability that's that it that the child selected at random spends less than 3 hours in the upright position, then our x will be 3. Then z score with this given information will be x minus mu over sigma, which is 3 minus 5.4 divided by 1.3, and our value for the z score turn out to be minus 1.85. Now, if we want to find out the probability that x is less than 3, which is which will be equals to probability that z less than minus 1. 8 5. So, here we go. First of all, we will draw a normal curve. As we know that mean for the random variable x is 5.4, therefore at this line of symmetry mu is 5.4. And the sigma is 1.3. So, we want to find out the probability for 3 or less. Hence, z is minus 1.5 1.85, and as soon as we convert our x into a z score, our mean and or the the point of symmetry will become zero with standard deviation one. And as in the original situation, we are interested in knowing the probability of less than 3.0, it will be the same as finding the probability that Z is less than minus 1. 8 5. Hence, using the table for the cumulative probabilities, since Z score is minus 1.85, so we will firstly look at minus 1.8 and will go all the way to the to 0. 05. And here you will notice that the area below 1 minus 1.85 is 0.0322. And hence, we get this probability right here that any child that is randomly selected will sit less than 3 hours in an upright position. And the chances are 3.22%. Let's take another example where the five subjects over the period of 30 days were followed and each day their their breath samples were taken and analyzed. So, the interest was that for a subject A, a 27-year-old female, the ammonia concentration in parts per billion followed a normal distribution. Here, our X will be the ammonia concentration in parts per billion, which follows a normal probability distribution. So, this is how we can write that X follows the normal distribution with its mean 491 and its standard deviation 119 or the variance 119 squared. Our interest is to find out the probability that on any random day the subject's ammonia concentration is between 292 and 6 49 parts per billion. So, this can be written as probability of X between these two values. So, to find out the probability this probability, we firstly convert X into Z. And it will have since X has two values, so I will call 292 as X1 and 649 and X2. Hence, X1 will be equals to Z1 will be equals to X1 minus mu divide by sigma. And in this situation, it will be 292 minus 491, which is the mean and 119 as the the standard deviation. And the Z value will will turn out to be 1.67. Hence, Z1 will be 1 minus 1.67. Then, uh we need to find out the Z score value which we call here as Z2 will be equals to X2 minus mu over sigma. Where X2 is 649 minus the mean, which is 491 and the standard deviation 119. And this value will become 1.33. And if we need to find out the probability between these two limits, we will firstly find this area and also this area and subtract the smaller area from the larger. So, here we go. We find the probability from minus infinity to 1.33, which is which will give us the larger area, and we will subtract uh the smaller area, which is from minus infinity to minus 1.67, and hence the value turn out to be .8607. Therefore, the that subjects ammonia concentration, the probability that subjects ammonia concentration is between 292 and 649 is .8607. In the third example, we know that in example one, if we talk about that the population of 1,000 10,000 children described, then the question is that how many would would you expect to be upright more than 8.5 hours? Here, we firstly need to find out the probability of being in the upright position more than 8.5 hours. And if we need to find out the probability for more than 8.5 hours, we are actually interested in this area. And the probability of this area will be obtained as the total area, which is one, subtracting the cumulative area that is less than 8.5, which will be .9913. Hence, the probability of of being in the upright position more than 8.5 hours is 0.0087. And if we need to find out the numbers, we just simply multiply this probability by the total n, and we found out that 87 of them will will spend more than 8.5 hours in an upright position. So, the normal probability distribution is as we know is a widely known distribution and widely applicable distribution. Normality of the random variable of interest is one of the very important property that we seek when we perform inferential statistical methods. Thank you.