Approximations of the Normal Distribution | Applied Biostatistics | BIO733_Topic058

Duration: 10:29 1408 transcript words EN summary

Video summary

The video explores how the normal probability distribution serves as a powerful tool for approximating discrete distributions, specifically focusing on the binomial and Poisson distributions. For the binomial distribution, which models the number of successes in a fixed number of independent trials, the mean is calculated as $np$ and the variance as $npq$, where $q$ represents the probability of failure. The lecture explains that under certain conditions, such as when the sample size $n$ is large or when the probability of success $p$ is close to 0.5, the discrete bar chart of the binomial distribution becomes sufficiently symmetric and continuous to be modeled by a normal curve. This approximation allows statisticians to use the well-understood properties of the normal distribution to estimate probabilities for binomial scenarios without calculating complex factorials. A critical aspect of this approximation is the necessity of applying a continuity correction because one is converting from a discrete variable to a continuous one. The transcript illustrates that if we want to find the probability of a binomial random variable $X$ falling between two integers, say 4 and 7, we must adjust these boundaries by subtracting or adding 0.5 to account for the width of the bars in the discrete distribution. Consequently, the range from 4 to 7 transforms into the continuous interval from 3.5 to 7.5. The video demonstrates this process with a specific example where $n=12$ and $p=0.5$, showing how transforming the variable into a standard normal $Z$-score using these corrected limits yields a probability that closely matches the exact binomial calculation, thereby validating the accuracy of the method when conditions are met. The discussion then shifts to the Poisson distribution, which is often used to model rare events occurring over a specific interval, characterized by a single parameter $\lambda$ that represents both its mean and variance. The lecture posits that for large values of $\lambda$, typically greater than 20, the Poisson distribution can also be approximated by a normal distribution with the same mean and variance. An example involving a radioactive source emitting particles at an average rate of 25 per second is used to demonstrate this concept. By applying the continuity correction to the condition "less than 27," the upper limit becomes 26.5, which is then converted into a $Z$-score. This approach simplifies the calculation of probabilities for Poisson processes by leveraging standard normal tables, provided the rate parameter is sufficiently large to ensure the distribution's shape resembles a bell curve. In conclusion, the video emphasizes that while many discrete and continuous distributions can approximate the normal distribution under special conditions, specific criteria must be met to ensure accuracy. For binomial distributions, these rules of thumb include having a sample size greater than 10 or 30 depending on how close $p$ is to 0.5, or ensuring that both $np$ and $nq$ exceed certain thresholds like 5 or 10. Similarly, for Poisson distributions, the parameter $\lambda$ should be large enough, generally exceeding 20. The overarching message is that these approximations are highly effective tools in applied biostatistics when the underlying sample sizes are large or parameters are favorable, allowing researchers to simplify complex probability calculations while maintaining a high degree of precision through the proper application of continuity corrections.

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The normal probability distribution approximates to many distribution. But in this specific module, we will be talking about the approximation of the normal distribution to a discrete distribution, which is binomial distribution, and the Poisson distribution. Let's first look at the binomial distribution. If X is a is a is a binomial random variable, then the probability mass function will be given by n choose x p to the power of x 1 minus p n minus x. And x is a discrete random variable with its possible values to be 0, 1, 2 up till fixed n. The mean for the binomial random variable is np, and its variance is npq. Whereas the standard deviation of a binomial random variable is npq under root. So, under certain conditions, we can use the normal distribution to approximate the binomial distribution. And as we can see right here in these two figures, the figure on to the left is having the binomial distribution. It's for the random variable x that has a binomial distribution, where n is 300, and p is .5. And as we already know that when the p is equals to 0.5, the binomial distribution is is pretty symmetric. And as soon as the sample size keep on increasing, the the the bars are getting closer and closer to each other, and it's giving us a quite symmetric and quite continuous situation. Therefore, it can be approximated as the normal with mean 150 and standard deviation to be 75. So, as we already know that x if x is a binomial with n and p its parameters, then mu will be np, variance will be npq, where q will be equals to 1 - P, which is the probability of failure, and P is the probability of success. Then for large N and P not too small or not too large X will follow the normal probability distribution with mean NP and its variance NPQ. Here in this situation we want to note that when as a rule of thumb, when N is greater than 10 and P is approximately 1 by 2 or N is greater than 30 and P is moving away from 1 by 2 then normal distribution approximates well to the binomial distribution. Moreover there's another rule that if NP is greater than 5 or NPQ is greater than 10 then we can easily approximate binomial to the normal distribution. So, suppose X is a binomial random variable with N 12 and P to be 0.5 and we are interested in finding out the probability that a random variable X has a probability between 4 and 7. Then to find this distribution as we know this N and P in this situation will be equals to 6 which is more than 5. Hence, we can convert this binomial into the normal probability distribution to find out the probabilities. And in this situation, for the for the normal probability distribution, we need two parameter. One is mean, which is given by NP, and it is 6, and the other is variance, which is NPQ, and in this situation it is 3. So, hence X will follow the normal probability distribution with 6 and 3 as its parameters. But unfortunately, it's not quite so simple to convert from a discrete probability distribution to a to a continuous probability distribution. Hence, we need a continuity correction. Let's look at this figure where the bars here will will will represent the discrete situation and the curve represent the continuous situation. And we are interested in finding out the probability from 3.5 to 7.5. Hence, if we transform it into X and we are interested in finding out this probability, the first thing is that we will draw a curve and we will convert our random variable X to a standard normal variable where Z will be equals to zero, which is the mean here, and it goes from minus infinity to plus infinity. And if we have to find out the probability from minus 1.443 and 0.866, we are shading this region for which we need to find out the probability. And to find this probability, firstly, we find the cumulative probability from minus infinity to 0.866 and also the cumulative probability from minus 1.44 all the way down to the minus infinity. And subtracting this larger probability from the smaller probability, we will find out the final probability, which is 0.732. Hence, the exact answer is 0.73333. So, in this case, the approximation is very good. You might notice one more thing that we are interested in finding out the probability of X from 4 to 7, but it transforms to X to be 3.5 and 7.5. This is all due to the continuity correction. That as soon as we include the continuity correction, four will turn out to be 3.5 and seven will be 7.5. So, we are interested in this area. So, if we have to go to the right of four, we will subtract .5 from four. So, 4 - 0.5 will make it 3.5. So, it will cover all those values here. And similarly, if you want to go backward, then we will add 0.5 to it and it will be 7.5 and hence that's why our X random variable, which now follows the normal probability distribution, it's due to continuity correction, it is written as 3.5 and 7.5 and hence this probability is 0.73 3. We can also approximate the normal distribution to the Poisson distribution under certain conditions. Let's say if X is a random variable that follows the Poisson distribution with its mean, which is our only parameter, as lambda. And we also know that the for the Poisson distribution, mean and variance are all are the same, that is lambda. Then for large and large values of lambda, let's say lambda is greater than 20, then Poisson can easily be converted into the normal distribution with mean lambda and variance to lambda as well. Let's take an example that a radioactive source emits particle at an average rate of 25 particles per second. And we are interested in finding out the probability that in 1 second, the count is less than 27 particles. Here, our random variable of interest Here in this situation, our random variable of interest is which is the number of particles emitted in 1 second. Here X follows the Poisson distribution with average 25. Hence, the value of the lambda here will be 25. So, we can simply say that if X follows the Poisson with lambda to be 25, then X can be approximated to the normal with mean to be 25 and the variance to be 25 as well. But, since we are converting from a discrete probability distribution to a continuous probability distribution, we need a continuity correction. Since we are interested in finding out the probability of X less than 27, by applying the continuity correction, it will turn out to be X to be less than 26.5. So, eventually, when we have converted it into the normal random variable, we need to find out the probability of X to be less than 26.5. Since we are interested in finding out the probability of X to be less than 26.5, we draw the normal curve, and this is the area we are interested in. So, converting this normal random variable X to a standard normal, we will use Z score, which makes it 26.5 minus 25 by 5. Here, 5 is the sigma, where sigma is is under root of the variance. And in this situation, variance is equals to 25, and mean equals to 25 as well, which is also equals to the lambda of the Poisson random variable. Hence, the Z score turn out to be 0.3. So, eventually, if we have to find out the probability of X normal random variable having the probability less than 26.5, it is equivalent to finding out the probability of Z less than 0.3. And which will be the area below 0.3. And using the table for the cumulative probability, we get the value 0.6179. Many of the discrete as well as continuous distributions, they approximate well to the normal probability distribution under certain special conditions. Few of those conditions are that if the sample size is large or they have certain specified values of the parameters of that distribution.