Each of the five divisions of a certain company sent representatives to a conference. If the numbers of representatives sent by four of the divisions were 3, 4, 5, and 5, was the range of the numbers of representatives sent by the five divisions greater than 2 ?
Give the numbers 3, 4, 5, and 5, the ONLY way for the range of the numbers to be less than 2 is if \(x < 3\) or \(x > 5\)
(1) The median of the numbers of representatives sent by the five divisions was greater than the average (arithmetic mean) of these numbers.
Scenario 1: x =1
\(\frac{1 + 3 + 4 + 5 + 5 }{ 5} = \frac{18}{5} = 3.6\)
\(median = 4\), \(average = 3.6\)
range > 2; YES
Scenario 2: x = 5
\(\frac{3 + 4 + 5 + 5 + 5 }{ 5} = \frac{22 }{ 5 }= 4.4\)
median = 5, average = 4.4
range < 2; NO
Insufficient.
(2) The median of the numbers of representatives sent by the five divisions was 4.
Scenario 1: 3, 4, 4, 5, 5
\(range < 2\)
Scenario 2: 2, 3, 4, 5, 5
\(range > 2\)
Insufficient.
(1&2) median = 4, median > mean
\(\frac{1 + 3 + 4 + 5 + 5 }{ 5} = \frac{18}{5} = 3.6\)
median = 4, average = 3.6
range > 2; YES
Answer is C.
_________________
Help me get better -- please critique my responses!