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The mathematical tool of probability is used to examine unpredictability. It is concerned with the possibility (probability) of an event occurring. If you throw a fair coin four times, the results are unlikely to be two heads and two tails.

Given:

\(\lambda = \mu = 8t\)

Formula Poisson probability:

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

(a)

\(t = 1\)

\(\lambda = \mu = 8t = 8(1) = 8\)

Evaluate the formula at\(k = 0,1,2, \ldots ,10\):

\(\begin{array}{l}P(X = 0) = \frac{{{8^0}{e^{ - 8}}}}{{0!}} \approx 3.3546 \times {10^{ - 4}}\\P(X = 1) = \frac{{{8^1}{e^{ - 8}}}}{{1!}} \approx 0.0027\\P(X = 5) = \frac{{{8^5}{e^{ - 8}}}}{{5!}} \approx 0.0916\\P(X = 6) = \frac{{{8^6}{e^{ - 8}}}}{{6!}} \approx 0.1221\\.\\.\\.\\P(X = 9) = \frac{{{8^9}{e^{ - 8}}}}{{9!}} \approx 0.1241\end{array}\)

Addition rule for disjoint or mutually exclusive events:

\(P(A{\rm{ or }}B) = P(A) + P(B)\)

Add the corresponding probabilities:

\(\begin{array}{l}P(X \ge 6) = 1 - P(X < 6) = 1 - P(X = 0) - P(X = 1) - \ldots - P(X = 5)\\\;\;\;\;\;\;\;\;\;\;\;\; = 1 - 3.3546 \times {10^{ - 4}} + 0.0027 + \ldots + 0.0916\\\;\;\;\;\;\;\;\;\;\;\;\; = 0.8088\end{array}\)

Command \(TI83/84\)-calculator: \(1\)-Poissoncdf\((8,5)\)

\(\begin{array}{l}P(X \le 10) = 1 - P(X < 10) = 1 - P(X = 0) - P(X = 1) - \ldots - P(X = 9)\\\;\;\;\;\;\;\;\;\;\;\;\;\;\; = 1 - 3.3546 \times {10^{ - 4}} - 0.0027 - \ldots - 0.1241\\\;\;\;\;\;\;\;\;\;\;\;\;\;\; = 0.2834\end{array}\)

Command \(TI83/84\)-calculator: \(1\) -Poissoncdf\((8,9)\)

(b)

\(t = 90\;{\rm{min}} = 1.5{\rm{ hours }}\)

The mean (expected value) is the product of the rate per hour and the number of hours:

\(\lambda = \mu = 8t = 8(1.5) = 12\)

The standard deviation of a Poisson distribution is the square root of the mean:

\(\sigma = \sqrt \mu = \sqrt \lambda = \sqrt {12} = 2\sqrt 3 \approx 3.4641\)

(c)

\(\begin{array}{l}t = 2.5\\\lambda = \mu = 8\\t = 8(2.5) = 20\end{array}\)

Evaluate the formula at\(k = 0,1,2, \ldots ,19\):

\(\begin{array}{l}P(X = 0) = \frac{{{{20}^0}{e^{ - 20}}}}{{0!}} \approx 2.0612 \times {10^{ - 9}}\\P(X = 1) = \frac{{{{20}^1}{e^{ - 20}}}}{{1!}} \approx 4.1223 \times {10^{ - 8}} \cdots \\P(X = 10) = \frac{{{{20}^{10}}{e^{ - 20}}}}{{10!}} \approx 0.0058\\P(X = 19) = \frac{{{{20}^{19}}{e^{ - 20}}}}{{19!}} \approx 0.0888\end{array}\)

Addition rule for disjoint or mutually exclusive events:

\(P(A{\rm{ or }}B) = P(A) + P(B)\)

Add the corresponding probabilities:

\(\begin{array}{l}P(X \le 19) = P(X = 0) + P(X = 1) + \ldots + P(X = 19)\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = 2.0612 \times {10^{ - 9}} + 4.1223 \times {10^{ - 8}} + \ldots + 0.0888\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = 0.4703\end{array}\)

Command \(TI83/84\)-calculator: Poissoncdf\((20,19)\)

\(\begin{array}{l}P(X \le 10) = P(X = 0) + P(X = 1) + \ldots + P(X = 10)\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = 2.0612 \times {10^{ - 9}} + 4.1223 \times {10^{ - 8}} + \ldots + 0.0058\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = 0.0108\end{array}\)

Command \(TI83/84\)-calculator: Poissoncdf\((20,10)\)

Complement rule:

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

Use the complement rule:

\(\begin{array}{l}P(X \ge 20) = 1 - P(X < 20)\\\;\;\;\;\;\;\;\;\;\;\;\;\;\,\, = 1 - P(X \le 19)\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = 1 - 0.4703\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = 0.5297\end{array}\)