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Part 4 of 5 Find \( Q_{1}, Q_{3} \), and IQR. (Round your answers to one decimal place as needed.) \[ \begin{array}{r}Q_{1} \\ Q_{3}\end{array}=\square \] IQR \( =\square \)

Ask by Pope Cruz. in the United States
Mar 08,2025

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To find \( Q_1 \), \( Q_3 \), and the IQR, follow these steps: 1. **Sort the Data:** Arrange all numbers in order from smallest to largest. 2. **Find the Median:** Identify the middle number. If there's an odd number of data points, it's the middle one; if even, it's the average of the two middle numbers. 3. **Determine Halves:** Split the data into the lower half for \( Q_1 \) and the upper half for \( Q_3 \). 4. **Compute Quartiles:** Find the median of each half to get \( Q_1 \) and \( Q_3 \). 5. **Calculate IQR:** Subtract \( Q_1 \) from \( Q_3 \) to get the interquartile range. 6. **Round:** Round all values to one decimal place. Since the specific data isn't provided, you'll need to apply these steps to your dataset to find the exact values for \( Q_1 \), \( Q_3 \), and the IQR.

Solución

We first note that to compute the quartiles and the interquartile range (IQR) we need the raw data (or a frequency distribution) to work with. Since the dataset was not provided in this part, here is a general step‐by‐step method for finding the first quartile \(Q_1\), the third quartile \(Q_3\), and the IQR: 1. **Sort the Data:** Arrange the data values in increasing order. 2. **Find the Median:** Identify the median (the middle value). If there is an odd number of data points, the median is the middle number; if even, it is the average of the two central numbers. 3. **Determine the Lower and Upper Halves:** - The lower half of the dataset (not including the median if the number of data points is odd) is used to compute \(Q_1\). - The upper half of the dataset is used to compute \(Q_3\). 4. **Compute \(Q_1\) and \(Q_3\):** - \(Q_1\) is the median of the lower half of the data. - \(Q_3\) is the median of the upper half of the data. 5. **Calculate the IQR:** \[ \text{IQR} = Q_3 - Q_1 \] 6. **Round your Answers:** Round \(Q_1\), \(Q_3\), and the IQR to one decimal place as needed. Without the numerical data provided, we cannot compute specific numeric values for \(Q_1\), \(Q_3\), and the IQR. Please provide the dataset or the necessary values so that we can carry out the computations and fill in the blanks: \[ \begin{array}{r} Q_{1} \\[6mm] Q_{3} \end{array}=\square \] \[ \text{IQR}=\square \] If you have the dataset available from earlier parts, please share it so we may proceed with the calculation.

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Extra Insights

To determine \( Q_1 \), \( Q_3 \), and the interquartile range (IQR), you typically start with a sorted dataset. Let’s assume you have a dataset: 1. **Sorting the Data**: Arrange your dataset in ascending order. 2. **Calculating \( Q_1 \) and \( Q_3 \)**: - \( Q_1 \) is the median of the first half of the data. - \( Q_3 \) is the median of the second half of the data. 3. **Finding IQR**: The IQR is calculated as \( IQR = Q_3 - Q_1 \). After following these steps, you will find your answers to round to one decimal point as necessary. If you can provide the dataset, I can help you calculate \( Q_1 \), \( Q_3 \), and the IQR!

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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