Question

The number of computers available in different labs of a school are $1,4,2,x,y\text{and}z,$ where $x<y<z$.

Determine the values of $x,y\text{and}z$.

Given:
Median= $3.5$
Mode= $4$
Mean= $4.5$

• A. $x=13$
  $y=4$
  $z=3$
• B. $x=3$
  $y=4$
  $z=13$

Answer 1

The correct option is B $x=3$

$y=4$
$z=13$
Let's assume the ascending order of the given data set is

$1$

$2$

$4$

$x$

$y$

$z$

($\because x<y<z$)

$\bullet$ The mode of the data set is $4$.
$\Rightarrow x\text{or}y\text{or}z=4$

The modified data set:

$1$

$2$

$4$

$4$

$\left[\right]$

$\left[\right]$

$\bullet$ For even-numbered data set, median is the mean of two central values.

As median of the data set is $3.5$, the unknown data value can be placed before $4$, such that there is a possibility to get the mean of $\left[\right]$ and $4$ as $3.5$.

The updated data set can be

$1$

$2$

$\left[\right]$

$4$

$4$

$\left[\right]$

The mean of $\left[\right]$ and $4$ is $3.5$.

$\Rightarrow \frac{\left[\right]+4}{2}=3.5$

$\Rightarrow \frac{\left[\right]+4}{2}\times 2=3.5\times 2$

$\Rightarrow \left[\right]+4=7$

$\Rightarrow \left[\right]+4-4=7-4$

​​​​​​​​​​​​​​$\Rightarrow \left[\right]=3$

$\therefore$ The data set becomes

$1$

$2$

$3$

$4$

$4$

$\left[\right]$

$\bullet$ The mean of the data set is $4.5$.

$\Rightarrow \frac{1+2+3+4+4+\left[\right]}{6}=4.5$

$\Rightarrow \frac{14+\left[\right]}{6}\times 6=4.5\times 6$

$\Rightarrow 14+\left[\right]=27$

$\Rightarrow 14+\left[\right]-14=27-14$

$\Rightarrow \left[\right]=13$

$\therefore$ The complete data set is

$1$

$2$

$\left[3\right]$

$4$

$\left[4\right]$

$\left[13\right]$

As $x<y<z$ ($3<4<13$),
$x=3,y=4,\text{and}z=13$, i.e., option (b.) is the correct one.