Of the 200 numbers, 16.5% are greater than 40, and 33.3% of the number

The median occurs at term\(\frac{(n + 1)}{2}=\frac{201}{2}=100.5\), and for a set with an even count (200 in this case), it will be the average of terms \(100\) and \(101\).

With \(200*16.5% = \frac{1}{3}x\), you have established that the \(x = 200*16.5%*3=99\) highest observations are each greater than 35 and that the \(200*16.5%=33\) highest observations are each greater than 40. All other terms must be less than or equal to 35.

The average of the 101st and 100th terms is the median. The 99th term in descending order is the \(200-99+1=102\)nd term in ascending order. The average of the 101st term and the 100th term, which are each less than or equal to 35, can only be 34 or 35.