Chapter 9, Problem 9.1P

(a)

To determine

Probability of observing exactly 400 counts in 1 minute.

Answer to Problem 9.1P

Probability observing exactly 400, P(n=400) = 0.02

Explanation of Solution

Given:

Mean activity, $\overline{n}=400cpm$

Number of counts per minute, n = 400 cpm

Formula used:

Standard deviation

  $\sigma =\sqrt{n}$

Where, n = number of counts

Probability

  $p(n)=\frac{1}{\sigma \sqrt{2\pi }}{e}^{-{(n-\overline{n})}^2/2{\sigma }^2}$

Where,

P(n) = probability of finding exactly n

  $\overline{n}$ = mean activity,

  $\sigma$ = standard deviation

Calculation:

Standard deviation

  $\sigma =\sqrt{n}=\sqrt{400}=20$

Probability

  $\begin{array}{l}p(n)=\frac{1}{\sigma \sqrt{2\pi }}{e}^{-{(n-\overline{n})}^2/2{\sigma }^2}\\ p(400)=\frac{1}{20\times \sqrt{2\pi }}{e}^{-{(400-400)}^2/2{(20)}^2}=0.02\end{array}$

Conclusion:

Probability observing exactly 400, P(400) = 0.02

(b)

To determine

Probability of measuring 390 − 410 counts in 1 minute?

Answer to Problem 9.1P

Probability of measuring range (390 − 410) counts is 0.383

Explanation of Solution

Given:

Mean activity, $\overline{n}=400cpm$

Count, $n_1=390,n_2=410$

Formula used:

  $t=\frac{n-\overline{n}}{\sigma }$

Where,

t = test

  $\sigma$ = standard deviation

Calculation:

Test value for 390 counts

  $t_1=\frac{n_1-\overline{n}}{\sigma }=\frac{390-400}{20}=-0.5$

Test value for 410counts

  $t_2=\frac{n_2-\overline{n}}{\sigma }=\frac{410-400}{20}=0.5$

The probability of measuring 390 − 410 counts

  $\begin{array}{l}P\left(390<n<410\right)=P\left(-0.5<t<0.5\right)=2\left[P\left(t=0.5\right)-0.5\right]\\ \Rightarrow 2\left[0.6915-0.5\right]=0.383\end{array}$

Conclusion:

The probability of measuring range (390 − 410)counts is 0.383.