Question

The number of people living in six houses is $3,8,4,x,y\text{and}z,$ where $x<y<z$.
What are the values of $x,y\text{and}z$?

Given:
The median is $7.5$.
The mode is $8$.
The mean is $7$.

• A. $x=7.5,y=8,\text{and}z=7$
• B. $x=7,y=8,\text{and}z=12$
• C. $x=8,y=7,\text{and}z=12$
• D. Cannot be determined

Answer 1

The correct option is B $x=7,y=8,\text{and}z=12$

Let's assume that the ascending order of the data set can be

$3$

$4$

$8$

$x$

$y$

$z$

($\because$ $x<y<z$)

$\bullet$ It is clear that $x$, $y$ and $z$ are not the same, and the mode of the data set is $8$.
$\Rightarrow x\text{or}y\text{or}z$, one of them is $8$

Hence, the updated data set can be represented as

$3$

$4$

$8$

$8$

$\left[\right]$

$\left[\right]$

$\bullet$ Further, we know that the median is $7.5$, which is the mean of two middle-positioned values.

We can put one of the unknows before $8$, such that the mean of $\left[\right]$ and $8$ can be $7.5$.

$3$

$4$

$\left[\right]$

$8$

$8$

$\left[\right]$

$\therefore \frac{\left[\right]+8}{2}=7.5$
$\Rightarrow \frac{\left[\right]+8}{2}\times 2=7.5\times 2$
$\Rightarrow \left[\right]+8=15$
$\Rightarrow \left[\right]+8-8=15-8$
$\Rightarrow \left[\right]=7$

Hence, $x=7$ (as $x$ is the minimum value)

$\bullet$ The current data set is

$3$

$4$

$7$

$8$

$8$

$\left[\right]$

The mean of the above data set is $7$.

$\Rightarrow \frac{\text{Sum of all data points}}{\text{Total number of data points}}=7$

$\Rightarrow \frac{3+4+7+8+8+\left[\right]}{6}=7$

$\Rightarrow \frac{30+\left[\right]}{6}=7$

$\Rightarrow \frac{30+\left[\right]}{6}\times 6=7\times 6$

$\Rightarrow 30+\left[\right]=42$

$\Rightarrow 30+\left[\right]-30=42-30$

$\Rightarrow \left[\right]=12$

$\therefore$ The final ordered data set becomes

$3$

$4$

$7$

$8$

$8$

$12$

And, the unknowns are $7$, $8$ and $12$.

As $x<y<z(7<8<12)$, we can conclude
$x=7,y=8,\text{and}z=12$

$\therefore$ The final given house numbers are

$3$

$8$

$4$

$7$

$8$

$12$