Jcpenny wrote:

The number of defects in the first five cars to come through a new production line are 9, 7, 10, 4, and 6, respectively. If the sixth car through the production line has either 3, 7, or 12 defects, for which of theses values does the mean number of defects per car for the first six cars equal the median?

I. 3

II. 7

III. 12

A. I only

B. II only

C. III only

D. I and III only

E. I, II, and III

D

Depending on the number of defects in 6th car,

3,4,6,7,9,10 (median 13/2 =6.5) OR

4,6,7,7,9,10 (median 14/2 = 7) OR

4,6,7,9,10,12 (median 16/2 = 8)

To find mean we can use the sum of numbers given (= 36) and add either 3, 7, or 12 to it and then divide by 6 (number of cars).

(36 + 3) /6 = 39/6 = 6.5 (== median above)

(36 + 7) /6 = 43/6 = 7.1 (does not equal median above)

(36 + 12) /6 = 48/6 = 8 (== median above)

Thus I and III are valid => D