Stop Searching: The One Lake Forest Lawyer Saving Parents Thousands

Stop Searching: The One Lake Forest Lawyer Saving Parents Thousands

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Original dataset has median 15, so the average of 4th and 5th values is 15.After adding 18, the new median (average of 5th and 6th values) is 16.The new median implies the 5th and 6th values average to 16, meaning values greater than 15 and less than 16 must bridge them.After adding 18, new sorted list: 5th and 6th values average 16.So values between 16 and 18 could be at 5th and 6th? No — median is average of 5th and 6th.But since previous median was 15, the 4th and 5th were around 15, so after adding 18, new 5th and 6th values must straddle higher values.Since its discrete values, assume integers for simplicity (common in such problems unless otherwise).So the 5th value rose. The values greater than 15 but less than 16 are those >15 and <16.Since only one new value (18) was added, and median moved, but without exact values, minimal inferable: at most one value can be in (15,16) unless repeated.Best: use logic — after adding a value of 18 (>16), and median becomes 16 (average), it implies that the 5th and 6th values in 9 sorted values average to 16.