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Updated on: 23 May 2017, 00:39
Originally posted by michaelkalend on 23 May 2017, 00:36.
Last edited by Bunuel on 23 May 2017, 00:39, edited 1 time in total.
Last edited by Bunuel on 23 May 2017, 00:39, edited 1 time in total.
Edited the question.
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B
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16%
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Company A has a total of n employees, where n is an odd integer, and no two employees have the same annual salary. The annual salaries of the n employees are listed in increasing order, and the 16th salary in the list is the median of the annual salaries. If the sum of the annual salaries of Company A’s employees is $942,400, what is the average (arithmetic mean) of the annual salaries of Company A’s employees?
A. $29,450
B. $30,400
C. $32,500
D. $47,120
E. $58,900
A. $29,450
B. $30,400
C. $32,500
D. $47,120
E. $58,900
23 May 2017, 01:18
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michaelkalend
Given that n is odd.
For a set with odd number of terms the median is the middle term (when arranged in ascending/descending order). Thus, 16th salary/employee is right in the middle, which implies that there are total of 31 employees.
The average salary = (total salary)/(number of employees) = 942,400/31 = 30,400.
Answer: B.
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23 May 2017, 06:05
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michaelkalend
Total number of n employee is odd integer.
16th salary in the list is the median of the annual salaries.
Therefore there are 15 employee before and after 16th. Total = 15 + 15 + 1 = 31
Sum of the annual salaries of Company A’s employees is = $942,400
Average = \(\frac{942,400}{31}\) = $30,400
Answer B...
