a manager recorded how many hours each of 11 cashiers worked in a week. heres a histogram showing their…

a manager recorded how many hours each of 11 cashiers worked in a week. heres a histogram showing their data: which interval contains the median number of hours worked? choose 1 answer: 0 - 10 10 - 20 20 - 30 30 - 40 40 - 50

a manager recorded how many hours each of 11 cashiers worked in a week. heres a histogram showing their data: which interval contains the median number of hours worked? choose 1 answer: 0 - 10 10 - 20 20 - 30 30 - 40 40 - 50

Answer

Explanation:

Step1: Find the position of the median

For (n = 11) data points, the median is the (\frac{n + 1}{2}=\frac{11+ 1}{2}=6)th data point (when the data is ordered).

Step2: Cumulative - frequency approach

  • The first interval ((0 - 10)) has (1) data point.
  • The second interval ((10 - 20)) has (2) data points. The cumulative frequency for (10 - 20) is (1+2 = 3).
  • The third interval ((20 - 30)): we don't need to calculate its frequency for the median.
  • The fourth interval ((30 - 40)): The cumulative frequency up to the third - interval is less than (6). The cumulative frequency up to the (30 - 40) interval: The first interval ((0 - 10)) has (1) data point, the second interval ((10 - 20)) has (2) data points. Let's assume the frequencies of the intervals are (f_1=1) ((0 - 10)), (f_2 = 2) ((10 - 20)), (f_3) ((20 - 30)), (f_4=6) ((30 - 40)), (f_5 = 2) ((40 - 50)). The cumulative frequency (CF): (CF) for (0 - 10) is (1), for (10 - 20) is (1 + 2=3). Since (1+2+f_3+ \cdots), and we know that when we consider the (30 - 40) interval, the (6)th data point falls into it. Because (1+2\lt6) and (1 + 2+6\gt6) (even if (f_3 = 0) in the worst - case scenario for non - existence of the (20 - 30) interval in terms of getting to the median).

Answer:

D. (30 - 40)