Chapter 4.3, Problem 18E

a.

To determine

Find the value of $P\left(X\le 1\right)$, if the mean decay rate is exactly 1 per second.

Answer to Problem 18E

The value of $P\left(X\le 1\right)=4.994\times {10}^{-4}$.

Explanation of Solution

Given info:

The mean decay rate of radioactive mass is at least 1 particle per second. If the mean decay rate is less than 1 per second, then the product will return for a refund. The number of decay events counted in 10 seconds is denoted as X.

Calculation:

The random variable X is defined as the number of decay events counted in 10 seconds. The X follows Poisson with mean, $\lambda =10$

The Poisson distribution formula is,

$P\left(X=x\right)=\frac{e^{-\lambda }{\lambda }^x}{x!}$

Where, x is a non-negative integer, $\lambda$ is a positive constant, and e is a constant

The required probability is $P\left(X\le 1\right)$. That is,

$P\left(X\le 1\right)=P\left(X=0\right)+P\left(X=1\right)$

Substitute x as 0, 1, $\lambda$ as 10 and $e\text{ as }2.71828$

$\begin{array}{c}P\left(X\le 1\right)=\frac{e^{-10}{\left(10\right)}^0}{0!}+\frac{e^{-10}{\left(10\right)}^1}{1!}\\ =\frac{{\left(2.71828\right)}^{-10}\times {\left(10\right)}^0}{1}+\frac{{\left(2.71828\right)}^{-10}\times {\left(10\right)}^1}{1}\\ =\frac{\left(4.5399\times {10}^{-5}\right)\left(1\right)}{1}+\frac{\left(4.5399\times {10}^{-5}\right)\left(10\right)}{1}\\ =4.5399\times {10}^{-5}+4.5399\times {10}^{-4}\end{array}$

              $=4.994\times {10}^{-4}$

Thus, the required probability is, $P\left(X\le 1\right)=4.994\times {10}^{-4}$

b.

To determine

Check whether one event in 10 seconds be an unusually small number, if the mean decay rate is 1 particle per second.

Answer to Problem 18E

Yes, one event in 10 seconds be an unusually small number.

Explanation of Solution

Justification:

From part (a), the value of $P\left(X\le 1\right)=4.994\times {10}^{-4}$. That is, about 5 out of every 10,000 would have 1 or fewer events in 10 seconds.

Thus, the one event in 10 seconds is an unusually small number.

c.

To determine

Explain whether counted one decay event in 10 seconds would be convincing evidence that the product should be returned or not.

Answer to Problem 18E

Yes. Since one event in 10 seconds is an unusually small number.

Explanation of Solution

Justification:

From part (b), it clear that one event in 10 seconds is an unusually small number. Thus, it can be concluded that one decay event in 10 seconds would be convincing evidence that the product should be returned.

d.

To determine

Find the value of $P\left(X\le 8\right)$, if the mean decay rate is exactly 1 per second.

Answer to Problem 18E

The value of $P\left(X\le 8\right)=0.3328$.

Explanation of Solution

Calculation:

The random variable X is defined as the number of decay events counted in 10 seconds. The X follows Poisson with mean, $\lambda =10$

The required probability is $P\left(X\le 8\right)$. That is,

$P\left(X\le 8\right)={\displaystyle \sum _{x=0}^{8}P\left(X=x\right)}$

Substitute x as 0, 1, $\lambda$ as 10 and $e\text{ as }2.71828$.

$\begin{array}{c}P\left(X\le 8\right)=\left(\begin{array}{c}\frac{e^{-10}{\left(10\right)}^0}{0!}+\frac{e^{-10}{\left(10\right)}^1}{1!}+\frac{e^{-10}{\left(10\right)}^2}{2!}+\frac{e^{-10}{\left(10\right)}^3}{3!}+\frac{e^{-10}{\left(10\right)}^4}{4!}+\\ \frac{e^{-10}{\left(10\right)}^5}{5!}+\frac{e^{-10}{\left(10\right)}^6}{6!}+\frac{e^{-10}{\left(10\right)}^7}{7!}+\frac{e^{-10}{\left(10\right)}^8}{8!}\end{array}\right)\\ =\left(\begin{array}{c}\frac{{\left(2.71828\right)}^{-10}\times {\left(10\right)}^0}{1}+\frac{{\left(2.71828\right)}^{-10}\times {\left(10\right)}^1}{1}+\\ \frac{{\left(2.71828\right)}^{-10}\times {\left(10\right)}^2}{2\times 1}+\frac{{\left(2.71828\right)}^{-10}\times {\left(10\right)}^3}{3\times 2\times 1}+\\ \frac{{\left(2.71828\right)}^{-10}\times {\left(10\right)}^4}{4\times 3\times 2\times 1}+\frac{{\left(2.71828\right)}^{-10}\times {\left(10\right)}^5}{5\times 4\times 3\times 2\times 1}+\\ \frac{{\left(2.71828\right)}^{-10}\times {\left(10\right)}^6}{6\times 5\times 4\times 3\times 2\times 1}+\frac{{\left(2.71828\right)}^{-10}\times {\left(10\right)}^7}{7\times 6\times 5\times 4\times 3\times 2\times 1}\\ +\frac{{\left(2.71828\right)}^{-10}\times {\left(10\right)}^8}{8\times 7\times 6\times 5\times 4\times 3\times 2\times 1}\end{array}\right)\\ =\left(\begin{array}{c}\frac{\left(4.5399\times {10}^{-5}\right)\left(1\right)}{1}+\frac{\left(4.5399\times {10}^{-5}\right)\left(10\right)}{1}+\frac{\left(4.5399\times {10}^{-5}\right)\left(100\right)}{2\times 1}+\\ \frac{\left(4.5399\times {10}^{-5}\right)\left(1,000\right)}{3\times 2\times 1}+\frac{\left(4.5399\times {10}^{-5}\right)\left(10,000\right)}{4\times 3\times 2\times 1}+\\ \frac{\left(4.5399\times {10}^{-5}\right)\left(100,000\right)}{5\times 4\times 3\times 2\times 1}+\frac{\left(4.5399\times {10}^{-5}\right)\left(1,000,000\right)}{6\times 5\times 4\times 3\times 2\times 1}+\\ \frac{\left(4.5399\times {10}^{-5}\right)\left(10,000,000\right)}{7\times 6\times 5\times 4\times 3\times 2\times 1}+\frac{\left(4.5399\times {10}^{-5}\right)\left(100,000,000\right)}{8\times 7\times 6\times 5\times 4\times 3\times 2\times 1}\end{array}\right)\\ =\left(\begin{array}{c}4.5399\times {10}^{-5}+4.5399\times {10}^{-4}+2.2699\times {10}^{-3}+7.5666\times {10}^{-3}\\ +0.0189+0.0378+0.0631+0.0901+0.1126\end{array}\right)\end{array}$

              $=0.3328$

Thus, the required probability is, $P\left(X\le 8\right)=0.3328$.

e.

To determine

Check whether 8 events in 10 seconds be an unusually small number, if the mean decay rate is 1 particle per second.

Answer to Problem 18E

No. Since about $\frac{1}{3}$ of the time would have 8 or fewer events in 10 seconds.

Explanation of Solution

Justification:

From part (d), the value of $P\left(X\le 8\right)=0.3328$. That is, about $\frac{1}{3}$ of the time would have 8 or fewer events in 10 seconds.

Thus, the 8 events in 10 seconds is not an unusually small number.

f.

To determine

Explain whether counted 8 decay events in 10 seconds would be convincing evidence that the product should be returned or not.

Answer to Problem 18E

No. Since 8 events in 10 seconds is not an unusually small number.

Explanation of Solution

Justification:

From part (e), it clear that 8 events in 10 seconds is not an unusually small number. Thus it can be concluded that 8 decay events in 10 seconds would not be convincing evidence that the product should be returned.