Finding the Mean, Variance, and Standard Deviation In Exercise...

Question

Finding the Mean, Variance, and Standard Deviation In Exercises 29–34, (a) find the mean, variance, and standard deviation of the probability distribution, and (b) interpret the results.

Machine Parts The number of defects per 1000 machine parts inspected

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Accepted Answer

Welcome back everyone. In this problem, a bakery tracks the number of burnt loaves in each batch of 200 loaves. The probability distribution for burnt loaves per batch is shown below in this table. We want to find the mean variance and standard deviation of this distribution. Now, what do we know about mean variance and standard deviation that will help us to solve in this distribution? Well, recall first of all that the mean me. Is going to be equal to the sum of the data points multiplied by the pro their probabilities, OK? Next, we also know that the variant sigma squared is equal to the sum of the squared deviations that is each value minus the mean squared multiplied by their probabilities, OK? And then our standard deviation sigma is the square root of the variant. So we're going to need to set up a few, uh, set up a table here that will help us easily find these sums for these products, OK? So let's go ahead And start with our points. So we have our data point X. Let me draw a line here. So we have our data point X and their probabilities PF X. Notice here that we need the sum of X multiplied by PF X so we can figure that out in our column. Next, we need the sum of the squared deviation so we can create a column for the deviations, and then the squared deviations. And then we're going to create one for the product of the squared deviations and the probabilities, which is what we need for our variants, OK? So now in this table we can use it to help us find each of the measures we need. So let me draw the columns here. Let me complete them, OK. And here, For the first part of our problem, OK, we are going to search for the mean. We're gonna try and find the mean, OK? Cause now, remember, the mean is equal to the sum of X multiplied by multiplied by, or sorry, of X or probabilities. Now, how many data points and probabilities do we have? Now, remember that when there are zero slows, the probability is 0.176. 1 has a probability of 0.324. 2, is a probability of 0.277, 3, 0.143. 40.065 and 5 0.015, OK? So we have all of these probabilities are all of these values and their probabilities. Now first again to find a mean, let's see if we can find a sum of X multiplied by PX. There are the probabilities. Now in this case, To do the multiplication, 0 multiplied by 0.176 is zero, and we're going to just repeat this process for the rest of the values. For instance, when X is 1 and the probability is 0.324, then their product is 0.324. For 2 and 0.277, it's 0.554, for 3 and 0.143, it's 0.429. For 4 and 0.065, it's 0.260. And for 51.015, it's 0.075. So now to find a mean, we're going to find the sum of all these values. And when we add all of them up. some of these products, we get it to be 1.642. So that means the mean mu is equal to 1.642 or approximately 1.6. Next, let's use that to find the variants, OK? Now remember, as we said earlier, the variance is equal to the sum of the square deviations. Multiplied by their probabilities. If we go back to our table, OK. Now, since we know that the mean is 1.642, for example, in our first entry where X equals 0, the deviation is going to be 0 minus 1.642, which is negative 1.642. When we square that, it's 2.6962. And when we multiply that square by its probability, which is 0.176, we get 0.474525. No, we're just going to repeat the process for the rest of values. For example, when X is 1, the deviation is negative.642, the square deviation is 0.4122, and thus, the, the product is 0.133541. When X is 2, we get 0.358. 01282 and 0.035501 respectively. When it's 3, we have 1.358. 1.8442 and point. 263-715. When it's 4, we get 2.358. 5.5602 and 0.361411. And when it's 5, we get 3.358, 11.2762 and 0.169142. Now that we have all of the products of the square deviations and the probabilities we can find their son. Because now that means the sum then is going to be equal to. 1.437836. That's what we get when we add up all of these values. So therefore, that means the variance is 1.437, 836, or approximately 1.4. Finally, we can calculate the standard deviation sigma. Remember we said earlier the standard deviation is the square root of the variants. So really, that's going to be the square root of 1.4373836, OK. Which is approximately equal to 1.2. Therefore, well, I don't have to write it again because here it is. Therefore, the mean is approximately 1.6, variance approximately 1.4, and standard deviation approximately 1.2. Thanks a lot for watching everyone. I hope this video helped.

Finding the Mean, Variance, and Standard Deviation In Exercises 29–34, (a) find the mean, variance, and standard deviation of the probability distribution, and (b) interpret the results.Machine Parts The number of defects per 1000 machine parts inspected

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Finding the Mean, Variance, and Standard Deviation In Exercises 29–34, (a) find the mean, variance, and standard deviation of the probability distribution, and (b) interpret the results.
Machine Parts The number of defects per 1000 machine parts inspected