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Part: \( 2 / 5 \) 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.) It can be concluded that the distribution is not significantly Skip Part skewed. Recheck

Ask by Horton Mcguire. in the United States
Mar 08,2025

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To calculate Pearson's index of skewness, first find the mean, median, and standard deviation of the data. Then, use the formula: \[ \text{Pearson's Index of Skewness} = \frac{\text{Mean} - \text{Median}}{\text{Standard Deviation}} \] After calculating, if the index is close to zero, the distribution is not significantly skewed.

Solution

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: \[ \text{Pearson's Index of Skewness} = \frac{\text{Mean} - \text{Median}}{\text{Standard Deviation}} \] Let's assume the data is \( x_1, x_2, x_3, ..., x_n \). 1. Calculate the mean: \[ \text{Mean} = \frac{\sum_{i=1}^{n} x_i}{n} \] 2. Calculate the median: To find the median, we need to arrange the data in ascending order and find the middle value. If the number of data points is odd, the median is the middle value. If the number of data points is even, the median is the average of the two middle values. 3. Calculate the standard deviation: \[ \text{Standard Deviation} = \sqrt{\frac{\sum_{i=1}^{n} (x_i - \text{Mean})^2}{n}} \] 4. Calculate the Pearson's index of skewness: \[ \text{Pearson's Index of Skewness} = \frac{\text{Mean} - \text{Median}}{\text{Standard Deviation}} \] Please provide the data points \( x_1, x_2, x_3, ..., x_n \) so that I can calculate the Pearson's index of skewness for you.

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Bonus Knowledge

To calculate Pearson's index of skewness, you will need the mean, median, and standard deviation of your dataset. The formula for Pearson’s index of skewness (commonly referred to as "Pearson's First Coefficient of Skewness") is: \[ Skewness = \frac{3(\text{Mean} - \text{Median})}{\text{Standard Deviation}} \] Once you have obtained the mean, median, and standard deviation (all rounded to one decimal place), plug those values into the formula. If your calculated skewness value is close to 0, it indicates that the distribution is not significantly skewed. A skewness value greater than 1 or less than -1 typically suggests a significant skew in the distribution. Now that you understand the calculation, make sure to double-check your calculations and rounding to ensure accurate results—rounding can sometimes lead to unexpected interpretations!

Related Questions

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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