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Data Set: \( \begin{array}{llllllll}41 & 72 & 96 & 103 & 103 & 105 & 106 & 113\end{array} 118 \) Find the five number summary and IQR for the data set above. Leave answers as decimals if needed. Don't round. The minimum is

Ask by Harmon Ruiz. in the United States
Mar 09,2025

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The minimum is 41.

Solución

- The data set (in order) is: \(41,\; 72,\; 96,\; 103,\; 103,\; 105,\; 106,\; 113,\; 118\). - **Minimum:** The smallest number is the first element: \[ \text{Minimum} = 41 \] - **Median:** There are \(9\) data points. The median is the \(5^\text{th}\) value (since \(\frac{9+1}{2} = 5\)): \[ \text{Median} = 103 \] - **First Quartile (Q1):** Exclude the median and consider the lower half: \[ 41,\; 72,\; 96,\; 103 \] With \(4\) data points, Q1 is the average of the \(2^\text{nd}\) and \(3^\text{rd}\) values: \[ Q1 = \frac{72 + 96}{2} = \frac{168}{2} = 84 \] - **Third Quartile (Q3):** Exclude the median and consider the upper half: \[ 105,\; 106,\; 113,\; 118 \] With \(4\) data points, Q3 is the average of the \(2^\text{nd}\) and \(3^\text{rd}\) values: \[ Q3 = \frac{106 + 113}{2} = \frac{219}{2} = 109.5 \] - **Maximum:** The largest number is the last element: \[ \text{Maximum} = 118 \] - **Interquartile Range (IQR):** IQR is the difference between Q3 and Q1: \[ IQR = Q3 - Q1 = 109.5 - 84 = 25.5 \] Thus, the minimum is \(41\).

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Bonus Knowledge

The minimum is 41. The five-number summary includes the minimum, Q1, median, Q3, and maximum. For this data set: - Minimum: 41 - Maximum: 118 - Median (middle value): 103 - Q1 (the median of the lower half): 96 - Q3 (the median of the upper half): 106 The Interquartile Range (IQR) is calculated as Q3 - Q1, which equals 106 - 96 = 10.

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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