If 10⁵ particles are taken, what is the relative standard deviation of each measurement?

Mixture contains 1% KCl and 99% KNO₃ particles.

No of particles taken 10⁵

Let,

n= number of particles

p = Probability of getting the corresponding particle

q=1-p= Probability of not getting cthe orresponding particle

Now from the given data the following can be calculated

The probability of getting KCl particles p=0.01 and number of particles n=10⁵

The expected number of KCl particles will be

$\begin{array}{rcl}np& =& {10}^5\times 0.01\\ & =& {10}^3\end{array}$

The standard deviation for KCl particles will be

$\begin{array}{rcl}\sqrt{npq}& =& \sqrt{{10}^5\times 0.01\times 0.99}\\ & =& 31.46\end{array}$

Now the relative standard deviation for KCl particles will be

$\begin{array}{rcl}& & \frac{\sqrt{npq}\times 100}{np}\\ & =& \frac{31.46\times 100}{{10}^3}\\ & =& 3.15\%\end{array}$

The probability of getting KNO₃ particles p=0.99 and number of particles n=10⁵

The expected number of KNO₃ particles will be

$\begin{array}{rcl}np& =& {10}^5\times 0.99\\ & =& 99000\end{array}$

The standard deviation forKNO₃ particles will be

$\begin{array}{rcl}\sqrt{npq}& =& \sqrt{{10}^5\times 0.99\times 0.01}\\ & =& 31.46\end{array}$

Now the relative standard deviation forKNO₃ particles will be

$\begin{array}{rcl}& & \frac{\sqrt{npq}\times 100}{np}\\ & =& \frac{31.46\times 100}{99000}\\ & =& 0.032\%\end{array}$

Therefore, the relative standard deviation for KCl and KNO₃ particles will be 3.15% and 0.032% respectively.