Poisson Distribution and Histogram Analysis in Stat 151 Lab

Stat 151 Lab E02 TA: Brie Sawka Lab Assignment #3 Question #1

a) The shape of the poisson distribution for a randomly selected glass sheet is a extremely skewed curve to the right. Where 0 has the highest probability and then it curves down to almost 0 when x = 5. The shape suggests that the most likely number of flaws is 0 at 60% of the time. There is still some chance that there will be 1 or 2 flaws but then after that it is not likely to get 3+ flaws. b) The probability of no flaws in a randomly selected sheet of glass = 0.60653067 c) 0.60653067*0.60653067*0.60653067*0.60653067*0.60653067=0.0820850056 Question #2 a) Center: 0.4 – 0.6 Shape: slight right skew Spread: 0 to 1.4 The values are still not distributed normally but it is definitely better than question 1’s values. This slightly resembles are normally distributed graph. When compared to question 1’s histogram, we can see that the center to “typical number of flaws” has moved from 0 to 0.4 – 0.6 in this histogram. This is due to the fact that we have more random samples and not just one sheet. The graph is becoming more “normal” with more samples that are done. The shape of question 1’s histogram had an extreme skew to the right because it was just one sample whereas the histogram in this question has 5 observations, there are more samples which in turn slowly makes the shape become more normal.

The spread of question 1’s histogram had a spread of 0 to 5 whereas the spread now 0 to 1.4. It has shrunk because the mean is over an average of 50 samples whereas in question 1, x was simply about one glass sheet. c) Summary statistics: Theory: The standard deviation is 0.316227766. The standard error is 0.0447213595. The standard error has the same function as standard deviation except that it only takes into account the sample size of the mean. The standard deviation is for the first 50 averages but the standard error measures for the mean of another 50 samples done. Since the values obtained are close to the summary statistics, we can conclude that the theorem is correct and accurate at predicting values. Question #3 a) b) Center: between 0.5 – 0.6 Shape: approximately symmetric Spread: 0.2 to 0.8 Here, the histogram does appear to have a fairly normal curve except for a slight change in values. Therefore we can call it approximately symmetric. Column Mean Std. Dev. Std. Err. Mean for x=5 0.508 0.33125612 0.046846688

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When this histogram is compared to question one, we can see the center has changed from 0 to 0.5 to 0.6. Whereas when this histogram is compared to question 2, we can see that the center changed from 0.4 to 0.6 and 0.5 to 0.6 respectively. The shape of this histogram compared to question 1 is completely different. This histogram is approximately symmetric where as question 1 had an extreme skew. This question is for 30 observations so we can expect the curve to be more normal. This is true because in question 2, when compared to this question, it is more normal than question 1 but not more symmetric than this question due to the fact that question 2 is only for 5 observations. This spread is 0.2 to 0.8, the spread in question 2 is 0 to 1.4 and the spread in question 1 is 0 to 5. We can see that the spread decreases as the number of observations increase. c) Summary statistics: Theory: The standard deviation is 0.1290994449. The standard error is 0.0182574186. The values obtained for the theory are relatively close so we can conclude that the theorem is accurate at predicting the values. The values compared to question 2 are lower than for n=5. We can say this because we know that as the number of observations increases, we are going to have less error and a more normal distribution. Question #4 a) The actual number of sheets with no flaws is 4. This is really close to our calculation that we got in 1c because now we can use the percentage to calculate how many will actually have no flaws. Considering that the sheets have no flaws, the percentage was 8.2 percent. So, if we want to find out how many of those sheets are flawless, then we can calculate 0.08*50 and get 4.1 as our answer. This means that at least 4 sheets of glass should be flawless. b) The probability of the average of flaws being greater than 0.70 is 0.060667627. The actual number of sheets with average flaws being greater than 0.70 is 3. When the probability is multiplied by 50, we can a value of 3.03338135 so we can conclude that the values are close and therefore we can use the probability to estimate what the number of sheets will be with averages of flaws greater than 0.70. The two values are consistent. Column Mean Std. Dev. Std. Err. Mean for x=30 0.49 0.10843141 0.015334517

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