Question

Melanie computes the mean $\mu,$ the median $M,$ and the modes of the $365$ values that are the dates in the months of $2019.$ Thus her data consist of $12$ $1$ s, $12$ $2$ s, $\ldots,$ $12$ $28$ s, $11$ $29$ s, $11$ $30$ s, and $7$ $31$ s. Let $d$ be the median of the modes. Which of the following statements is true?

$\mu \lt d \lt M$

$M \lt d \lt \mu$

$d = M = \mu$

$d \lt M \lt \mu$

$d \lt \mu \lt M$

Solution:

The values $1$ through $28$ each appear $12$ times and are the modes, so $d = \dfrac{14 + 15}{2} = 14.5.$

The $183$ rd of the $365$ ordered values is the median. Values $1$ through $15$ fill the first $180$ positions, so position $183$ is $16;$ thus $M = 16.$

The total of all values is $12(1 + \cdots + 28) + 11(29 + 30) + 7 \cdot 31 = 5738,$ so $\mu = \dfrac{5738}{365} \approx 15.7.$

Therefore $d \lt \mu \lt M.$

Thus, the correct answer is E.

2019 AMC 12A Problem 7

Solution:

Problem 7 in Other Years