# The data is arranged in descending order. The median of this data set would be

8 1 2 4.5-th term Since the median is element number 4.5 in the array, we need to calculate the average of the fourth and fifth elements. The fourth item in Table 3-11 is 43 and the fifth is 35. The average of these two items is equal to (43 35) / 2 39. Therefore, 39 is the median number of patients per day treated in the emergency room during the 8 day period.

Finding the median from pooled data Finding the median of pooled data

We often have access to the data even after we have grouped it into a frequency distribution. For example, we do not know all the observations that led to Table 3-12, which contains data on the 600 bank customers considered earlier. In this case, we have 10 class intervals and a record of the frequencies with which the observations appear in each interval. However, we can calculate the median checking account balance for these 600 customers by determining which of the 10 class intervals contains the median. To do this, we must add the frequencies that appear in the frequency column of table 3-12 until we reach element number (n 1) / 2. Since we have 600 counts, the value for (n 1) / 2 is 300.5 (the average of the numbers 300 and 301). The problem is to find the class intervals that contain elements number 300 and 301. The cumulative frequency for the first two classes is only 78 123 201. But when we take into account the third class interval and add 187 elements to the 201 accumulated, we will have a total of 388. Consequently, observations number 300 and 301 must be in this third class (the range of \$ 100.00 to \$ 149.99). The class of the median of this data set contains 187 observations. If we assume that these 187 items start at \$ 100.00 and are equally spaced throughout the inter-

\$ 150.00 \$ 100.00 \$ 0.267 wide 187 If there are 187 steps of \$ 0.267 each and we need 98 steps to get to element number 99, then this is: (\$ 0.267 98) \$ 100 \$ 126.17 and element number 100 is a further step: \$ 126.17 \$ 0.267 \$ 126.44 Therefore, we can use \$ 126.1