Question

The number of computers available in different labs of a school are $35,30,40,x,y\text{and}z,$ where $x>y>z$.

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

Given:
Median= $39.5$
Mode= $40$
Mean= $38.1\overline{6}$

• A. $x=39$
  $y=40$
  $z=45$
• B. $x=45$
  $y=40$
  $z=39$
• C. $x=45$
  $y=39$
  $z=40$

Answer 1

The correct option is B $x=45$

$y=40$
$z=39$
Let's assume the ascending order of the given data set is

$30$

$35$

$40$

$z$

$y$

$x$

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

$\bullet$ The mode of the data set is $40$.
$\Rightarrow x\text{or}y\text{or}z=40$

The modified data set:

$30$

$35$

$40$

$40$

$\left[\right]$

$\left[\right]$

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

As median of the data set is $39.5$, the unknown data value can be placed before $40$, such that there is a possibility to get the mean of $\left[\right]$ and $40$ as $39.5$.

The updated data set can be

$30$

$35$

$\left[\right]$

$40$

$40$

$\left[\right]$

The mean of $\left[\right]$ and $40$ is $39.5$.

$\Rightarrow \frac{\left[\right]+40}{2}=39.5$

$\Rightarrow \frac{\left[\right]+40}{2}\times 2=39.5\times 2$

$\Rightarrow \left[\right]+40=79$

​​​​​​​$\Rightarrow \left[\right]+40-40=79-40$

​​​​​​​​​​​​​​$\Rightarrow \left[\right]=39$

$\therefore$ The data set becomes

$30$

$35$

$39$

$40$

$40$

$\left[\right]$

$\bullet$ The mean of the data set is $38.1\overline{6}$.

$\Rightarrow \frac{30+35+39+40+40+\left[\right]}{6}=38.1\overline{6}$

$\Rightarrow \frac{184+\left[\right]}{6}\times 6=38.1\overline{6}\times 6$

$\Rightarrow 184+\left[\right]=229$

$\Rightarrow 184+\left[\right]-184=229-184$

$\Rightarrow \left[\right]=45$

$\therefore$ The complete data set is

$30$

$35$

$\left[39\right]$

$40$

$\left[40\right]$

$\left[45\right]$

As $x>y>z$ ($45>40>39$),
$x=45,y=40,\text{and}z=39$, i.e., option (b.) is the correct one.