Question

(a) Find the standard deviation of the distribution in Example 1.1.

(b) What is the probability that a photograph, selected at random, would show a distance x more than one standard deviation away from the average?

Short answer

1. Standard deviation $=0.298h$
2. Probability for a photograph, selected at random, to show a distance x more than one standard deviation away from the average would be $0.393$

Step-by-step solution

Calculating for <x2>   and  ⟨x⟩2

Refer to example 1.1, the probability distribution is,

${\mathrm{\rho }}_{\left(\mathrm{x}\right)}=\frac{1}{2\sqrt{\mathrm{hx}}}$ For$0\le x\le h$

This is the probability distribution and we will use this to calculate the probability for the photograph in step 3.

The expectation value can be calculated as,

Hence, the square of the expectation will be ${⟨x⟩}^2=\frac{h^2}{9}$.

Since, the probability density stays the same irrespective of the power ofx

Therefore,

role="math" localid="1655376313786"

Calculating for the standard deviation (a)

The equation for the standard deviation is shown below,

$\sigma =\sqrt{⟨x^2⟩-{⟨x⟩}^2}$

Substitute the calculated values in above equation,

$\begin{array}{l}\sigma =\sqrt{\frac{h^2}{5}-\frac{h^2}{9}}\\ \sigma =\sqrt{\frac{9h^2-5h^2}{45}}\\ \sigma =\sqrt{\frac{4h^2}{45}}\\ \sigma =\frac{2h}{3\sqrt{5}}\\ \sigma =0.298h\end{array}$

Thus, the standard deviation is 0.298h.

Solving for the probability for a photograph (b)

If we integrate the probability distribution over the appropriate intervals of x from $\mathbf{0}\mathbf{\le }\mathit{x}\mathbf{\le }\mathit{h}$, we can determine the probability of finding the photograph more than one standard deviation from the average.

i.e.

role="math" localid="1655376615068" $\begin{array}{l}P=\underset{0}{\overset{<x>-\sigma }{\int }}{\rho }_{\left(x\right)}dx+\underset{<x>+\sigma }{\overset{h}{\int }}{\rho }_{\left(x\right)}dx\\ P=\underset{0}{\overset{\frac{h}{3}-0.298h}{\int }}\frac{1}{2\sqrt{h}}{x}^{-1/2}dx+\underset{\frac{h}{2}+0.298h}{\overset{h}{\int }}\frac{1}{2\sqrt{h}}{x}^{-1/2}dx\\ P=\frac{1}{2\sqrt{h}}.2x^{1/2}{|}_0^{\frac{h}{3}-0.298h}+\frac{1}{2\sqrt{h}}.2x^{1/2}{|}_{\frac{h}{3}+0.298h}^h\\ P=\frac{1}{\sqrt{h}}\left(\sqrt{\frac{h}{3}-0.298h-0}\right)+\frac{1}{\sqrt{h}}\left(\sqrt{h}-\sqrt{\frac{h}{3}+0.298h}\right)\\ \end{array}$

Further solving above equation,

$\begin{array}{l}P=\sqrt{\frac{1}{3}-0.298h}+1-\sqrt{\frac{1}{3}+0.298h}\\ P=0.393\end{array}$

So, the probability for a photograph, selected at random, to show a distance x more than one standard deviation away from the average would be 0.393.