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 $12$ $1\text{s},$ $12$ $2\text{s},$ . . . , $12$ $28\text{s},$ $11$ $29\text{s},$ $11$ $30\text{s},$ and $7$ $31\text{s}.$ 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:

$d$ must have to be less than $M$ because $M$ accounts for the larger numbers that are not modes ( $29, 30$ and $31$ ).

There are $365$ entries, so $m$ is the $183$ rd number. The first $15$ numbers take up $15 \cdot 12 = 180$ spots, so $m$ is $16.$

$\mu$ is less than $16$ since there are fewer occurrences of the larger numbers, making the distribution left skewed.

$d$ is also less than $\mu$ since $\mu$ accounts for $29, 30$ and $31.$ Therefore, we get that $d \lt \mu \lt M.$ Thus, E is the correct answer.

2019 AMC 10A Problem 12

Solution:

Problem 12 in Other Years