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\[ 10,18,4,15,3,21, x \] If \( x \) is the median of the 7 numbers listed above, which of the following could be the value of \( x \) ?

Ask by Lindsey Gordon. in the United States
Mar 09,2025

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Tutor-Verified Answer

Answer

\( x \) must be between 10 and 15. For example, \( x = 12 \) is a possible value.

Solution

1. First, list the six known numbers: \[ 10,\ 18,\ 4,\ 15,\ 3,\ 21 \] 2. Sort these numbers in increasing order: \[ 3,\ 4,\ 10,\ 15,\ 18,\ 21 \] 3. When we add \(x\) to the list, there will be 7 numbers. For a set of 7 numbers the median is the 4th number when the list is arranged in order. 4. For \(x\) to be the median, \(x\) must occupy the 4th position after sorting. This means that exactly 3 of the numbers must be less than \(x\) and the remaining 3 must be greater than \(x\). 5. In the sorted list of the six known numbers, the three smallest are \(3\), \(4\), and \(10\) and the three largest are \(15\), \(18\), and \(21\). 6. Therefore, for \(x\) to be placed in the 4th position, it must satisfy the inequality: \[ 10 < x < 15 \] 7. Any value of \(x\) that lies strictly between \(10\) and \(15\) will be the median. For example, if we take: \[ x = 12 \] the sorted list becomes: \[ 3,\ 4,\ 10,\ 12,\ 15,\ 18,\ 21 \] and the 4th number is indeed \(12\). Thus, one acceptable value for \(x\) is: \[ \boxed{12} \]

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The Deep Dive

To find the median of the seven numbers \( 10, 18, 4, 15, 3, 21, x \), we first need to arrange the numbers in increasing order. The median is the middle value, which means it will be the fourth number in this ordered list since there are seven numbers total. 1. Sorting the existing numbers: \( 3, 4, 10, 15, 18, 21 \) 2. We know that \( x \) will insert into this list and affect the median. The median will depend on the value of \( x \): - If \( x \leq 3 \), the sorted list becomes \( x, 3, 4, 10, 15, 18, 21 \) with the median being \( 10 \). - If \( 3 < x \leq 4 \), the list will be \( 3, x, 4, 10, 15, 18, 21 \), median remains \( 10 \). - If \( 4 < x < 10 \), the list will be \( 3, 4, x, 10, 15, 18, 21 \) with median as \( 10 \). - If \( x = 10 \), the list is \( 3, 4, 10, 10, 15, 18, 21 \) giving a median of \( 10 \). - If \( 10 < x < 15 \), it looks like \( 3, 4, 10, x, 15, 18, 21 \), keeping the median at \( 10 \). - If \( x = 15 \), it becomes \( 3, 4, 10, 15, 15, 18, 21 \) with the median still \( 15 \). - If \( 15 < x < 18 \), it becomes \( 3, 4, 10, 15, x, 18, 21 \), making the median \( 15 \). - If \( x = 18 \), it becomes \( 3, 4, 10, 15, 18, 18, 21 \) with a median of \( 15 \). - If \( 18 < x < 21 \), then \( 3, 4, 10, 15, 18, x, 21 \) keeps the median at \( 15 \). - If \( x \geq 21 \), the list will be \( 3, 4, 10, 15, 18, 21, x \) making the median \( 15 \). The possible values of \( x \) that could serve as the median are: - \( 10 \) - \( 15 \) Thus, the values \( x \) could take to meet the median requirement are \( 10 \) or any value between \( 10 \) and \( 15 \).

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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