Chapter 9, Problem 9.13P

(a)

To determine

Probability of observing exactly 50 counts in 1min.

Answer to Problem 9.13P

Probability of observing exactly 50 counts in 1-min is 0.056.

Explanation of Solution

Given:

True mean counting rate is 50 cpm

Mean counts in 1-min, $\overline{n}=50$

Exact counts, $n=50counts$

Formula used:

Standard deviation, $\sigma =\sqrt{\overline{n}}$

  $\begin{array}{l}\text{where,}\\ \overline{n}=\text{ mean value, and }\end{array}$

Probability in normal distribution, $p(n)=\frac{1}{\sigma \sqrt{2\pi }}{e}^{-\frac{{(n-\overline{n})}^2}{2{\sigma }^2}}$

  $\begin{array}{l}\text{where,}\\ p(n)=\text{ probability of finding exactly n}\\ \overline{n}=\text{ mean value, and }\\ \sigma =\text{ standard deviation}\end{array}$

Calculation:

Standard deviation, $\sigma =\sqrt{\overline{n}}=\sqrt{50}=7.07$

Probability for exactly 50 counts

  $\begin{array}{l}p(n)=\frac{1}{\sigma \sqrt{2\pi }}{e}^{-\frac{{(n-\overline{n})}^2}{2{\sigma }^2}}\\ p(50)=\frac{1}{7.07\sqrt{2\pi }}{e}^{-\frac{{(50-50)}^2}{2{(7.07)}^2}}=0.056\end{array}$

Conclusion:

Probabilityof observing exactly 50 count is 0.056.

(b)

To determine

Probability of measuring between 47-57 counts in 1 min.

Answer to Problem 9.13P

Probability of measuring between 47-57 cpm is 0.68

Explanation of Solution

Given:

Mean counts, $\overline{n}=50$ cpm

Measuring lie between, $47<n<50$ cpm

Formula used:

Standard deviation, $\sigma =\sqrt{\overline{n}}$

  $\begin{array}{l}\text{where,}\\ \overline{n}=\text{ mean value, and }\end{array}$

Z-score of normal distribution table, $Z=\frac{n-\overline{n}}{\sigma }$

Calculation:

Standard deviation, $\sigma =\sqrt{\overline{n}}=\sqrt{50}=7.07$

Z-score for n = 43, $Z=\frac{n-\overline{n}}{\sigma }=\frac{43-50}{7.07}=-0.99$

Z-score for n = 57, $Z=\frac{n-\overline{n}}{\sigma }=\frac{57-50}{7.07}=0.99$

Using Z-score table of normal distribution

Probability of measuring counts between 47-57 cpm

  $\begin{array}{l}P\left(43<n<57\right)=P\left(-0.99<Z<0.99\right)\\ =2\left[P\left(Z=0.99\right)-0.5\right]=2\left[0.8389-0.5\right]=0.68\end{array}$

Conclusion:

Probability of measuring between 47-57 cpm is 0.68.

(c)

To determine

Probability of measuring more than 57 counts in 1min.

Answer to Problem 9.13P

Probability of measuring more than 57 counts in 1-min is 0.16.

Explanation of Solution

Given:

True mean counting rate, $\overline{n}=50$

Exact count rate n liesbetween, $47<n<50$

Formula used:

Standard deviation, $\sigma =\sqrt{\overline{n}}$

  $\begin{array}{l}\text{where,}\\ \overline{n}=\text{ mean value, and }\end{array}$

Z-score of normal distribution table, $Z=\frac{n-\overline{n}}{\sigma }$

Calculation:

Standard deviation, $\sigma =\sqrt{\overline{n}}=\sqrt{50}=7.07$

Z-score for n = 43, $Z=\frac{n-\overline{n}}{\sigma }=\frac{57-50}{7.07}=0.99$

Using the Z-score table of normal distribution

The probability of measuring more than 57 counts in 1min.

  $\begin{array}{l}P\left(n>57\right)=P\left(Z>0.99\right)\\ =\left[1-P\left(Z=0.99\right)\right]=\left[1-0.8389\right]=0.16\end{array}$

Conclusion:

Probability of measuring more than 57 counts in 1-min is 0.16.