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Part 3 of 5 Calculate the Pearson's index of skewness. (Round the mean, median and standard deviation to one decimal place and Pearson's index to two decir places.) \[ \mathrm{PI}=\square \] It can be concluded that the distribution is not significantly \( \mathbf{\nabla} \) skewed. \( \square \) Try one last time Skip Part Recheck Save For Later © 2025 McGraw Hill LLC. All Riahts Resenved. Terms of Use I

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Mar 08,2025

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The Pearson's index of skewness is 0, indicating that the distribution is not significantly skewed.

Solución

To calculate the Pearson's index of skewness, we need to follow these steps: 1. Calculate the mean of the data. 2. Calculate the median of the data. 3. Calculate the standard deviation of the data. 4. Calculate the Pearson's index of skewness using the formula: \[ \mathrm{PI} = \frac{\mathrm{Mean} - \mathrm{Median}}{\mathrm{Standard Deviation}} \] Let's assume we have a dataset: \( \{1, 2, 3, 4, 5\} \) 1. Calculate the mean: \[ \mathrm{Mean} = \frac{1 + 2 + 3 + 4 + 5}{5} = \frac{15}{5} = 3 \] 2. Calculate the median: Since the dataset is already sorted, the median is the middle value, which is 3. 3. Calculate the standard deviation: \[ \mathrm{Standard Deviation} = \sqrt{\frac{\sum_{i=1}^{n}(x_i - \mathrm{Mean})^2}{n}} \] \[ \mathrm{Standard Deviation} = \sqrt{\frac{(1-3)^2 + (2-3)^2 + (3-3)^2 + (4-3)^2 + (5-3)^2}{5}} \] \[ \mathrm{Standard Deviation} = \sqrt{\frac{4 + 1 + 0 + 1 + 4}{5}} \] \[ \mathrm{Standard Deviation} = \sqrt{\frac{10}{5}} \] \[ \mathrm{Standard Deviation} = \sqrt{2} \] 4. Calculate the Pearson's index of skewness: \[ \mathrm{PI} = \frac{3 - 3}{\sqrt{2}} = 0 \] Therefore, the Pearson's index of skewness is 0, indicating that the distribution is not significantly skewed.

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The Deep Dive

To calculate Pearson's index of skewness (PI), you'll need the mean, median, and standard deviation of your data set. The formula for Pearson's skewness is: \[ \text{PI} = \frac{3(\text{Mean} - \text{Median})}{\text{Standard Deviation}} \] After plugging in your values and rounding as indicated, you'll get your final skewness score. If the value of PI is close to 0 (typically between -0.5 and +0.5), you can conclude that the distribution is not significantly skewed. When analyzing data, keep in mind that a skewed distribution can affect statistical analyses like the t-test or ANOVA. After calculating the skewness, ensure it's properly interpreted within the context of your data, as it could suggest underlying trends or anomalies you might want to investigate further!

preguntas relacionadas

Question 12(Mulliple Choice Warth 5 points) \[ (04.06 \mathrm{HC}) \] A researcher wants to test the claim that the proportion of juniors who watch television regularly is greater than the proportion of seniors who watch television regularly She finds that 56 of 70 randomly selected juniors and 47 of 85 randomly selected seniors report watching television regularly. Construct \( 95 \% \) confidence intervals for each population proportion. Which of the statemente gives the correct outcome of the research or's tert of the dalim? The \( 95 \% \) confidence interval for juniors is (706, 894), and the \( 95 \% \) confidence interval for seniors is ( 447,659 ). Since the intervals overlap, there is not enough evidence to say the proportion of juniors who watch television regularly may be higher than that of seniors. The \( 95 \% \) confidence interval for juniors is (721, 879), and the \( 95 \% \) confidence interval for seniors is (464, 642). Since the interval for juniors is higher than the interval for seniors, there is evidence to say the proportion of juniors who watch television regularly may be higher than that of seniors. The \( 95 \% \) confidence interval for juniors is ( 706,894 ), and the \( 95 \% \) confidence interval for seniors is ( 447,659 ). Since the interval for juniors is higher than the interval for seniors, there is evidence to say the proportion of juniors who watch television regularly may be higher than that of seniors. The \( 95 \% \) confidence interval for juniors is ( \( 721, .879 \) ), and the \( 95 \% \) confidence interval for seniors is (464, 642). Since the intervals overlap, there is not enough evidence to say the proportion of juniors who watch television regularly may be higher than that of seniors.

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