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Randomness is studied using probability, a mathematical instrument. It is concerned with the probability of an event taking place. You might not get two heads and two tails when you throw a fair coin four times.

The cumulative probability \(P(X \le 7)\)is given in the row with \(x = 7\)and in the column with \(\mu = 10.0\)of table:

\(P(X < 8) = P(X \le 7) = 0.220\)

Complement rule:

\(P({\mathop{\rm not}\nolimits} A) = 1 - P(A)\)

Use the complement rule:

\(\begin{array}{l}P(X \ge 8) = 1 - P(X < 8)\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = 1 - 0.220\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = 0.780\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = 78.0\% \end{array}\)

The cumulative probability \(P(X \le 18)\)is given in the row with \(x = 18\)and in the column with \(\mu = 15.0\)of table:

\(P(X < 18.9) = P(X \le 18) = 0.819\)

Complement rule:

\(P({\mathop{\rm not}\nolimits} A) = 1 - P(A)\)

Use the complement rule:

\(\begin{array}{l}P(X > \mu + \sigma ) = P(X > 18.9)\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = 1 - P(X < 18.9)\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = 1 - 0.819\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = 0.181 = 18.1\% \end{array}\)

Given: Poisson distribution with

\(\mu = 10{\rm{ organisms }}/{{\rm{m}}^3}\)

Let \(x\;{{\rm{m}}^3}\)be the amount of discharge in. The mean is then the ratio multiplied by the number of \({m^3}\):

\(\mu = 10{\rm{ organisms }}/{{\rm{m}}^3} \times x\;{{\rm{m}}^3} = 10x{\rm{ organisms }}\)

We need to determine x such that:

\(P(X \ge 1) = 0.999\)

Formula Poisson probability:

\(P(X = k) = \frac{{{\mu ^k}{e^{ - \mu }}}}{{k!}}\)

Evaluate at \(k = 0\):

\(P(X = 0) = \frac{{{{(10x)}^0}{e^{ - 10x}}}}{{0!}} = {e^{ - 10x}}\)

Complement rule:

\(P({\mathop{\rm not}\nolimits} {\rm{A}}) = 1 - P(A)\)

Use the complement rule:

\(P(X \ge 1) = 1 - P(X = 0) = 1 - {e^{ - 10x}}\)

We want this probability to be equal to\(0.999\).

\(1 - {e^{ - 10x}} = 0.999\)

Subtract \(1\) from each side of the equation:

\( - {e^{ - 10x}} = - 0.001\)

Multiply each side of the equation by\( - 1\):

\({e^{ - 10x}} = 0.001\)

Take the natural logarithm of each side:

\(\ln {e^{ - 10x}} = \ln 0.001\)

Use the power property of logarithms \(\left( {\ln {a^b} = b\ln a} \right)\):

\( - 10x\ln e = \ln 0.001\)

The natural logarithm of e is l:

\( - 10x = \ln 0.001\)

Divide each side by\( - 10\):

\(x = \frac{{\ln 0.001}}{{ - 10}}\)

Evaluate:

\(x = - \frac{{\ln 0.001}}{{10}} \approx 0.6907755\)