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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 decimal places.) \[ \mathrm{PI}=0.27^{9} \] It can be concluded that the distribution is \( \square \) not significantly skewed. Try one last time Skip Part Recheck Save For Later Submit

Ask by Sandoval Herrera. in the United States
Mar 08,2025

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The Pearson's index of skewness is approximately 0.27. Since this value is close to 0, the distribution is not significantly skewed.

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Given that the Pearson skewness (second coefficient) is defined as \[ \mathrm{PI}=\frac{3(\bar{x}-\text{Median})}{\sigma}, \] and using the rounded values (mean, median, and standard deviation) obtained from the data, we substitute into the formula. After carrying out the calculation and rounding the result to two decimal places we get \[ \mathrm{PI}\approx0.27. \] Since a Pearson skewness index close to 0 (typically between \(-0.5\) and \(0.5\)) indicates little or no skewness, we conclude that the distribution is not significantly skewed.

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The Deep Dive

To calculate Pearson's index of skewness, you need the following formula: \[ \mathrm{PI} = \frac{3(\text{Mean} - \text{Median})}{\text{Standard Deviation}} \] First, calculate the mean, median, and standard deviation based on your data, rounding each to one decimal place. Then, plug those numbers into the formula. For instance, if your calculations yield a mean of 10.5, a median of 10.0, and a standard deviation of 2.5, your calculation will be: \[ \mathrm{PI} = \frac{3(10.5 - 10.0)}{2.5} = \frac{3(0.5)}{2.5} = \frac{1.5}{2.5} = 0.6 \] Once calculated, if the index is between -0.5 and 0.5, you can conclude that the distribution is not significantly skewed. Don't forget to double-check the figures before submitting! If the values skew too far off these limits, the distribution could indicate some skewness, so always keep a watchful eye on those calculations!

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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