Step 1

Given:

\(\displaystyle\lambda={1.7}\)

Formula Poisson probability:

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

Complement rule:

\(\displaystyle{P}{\left(\neg\ {A}\right)}={1}\ -\ {P}{\left({A}\right)}\)

Addition rule for mutually exclusive events (special addition rule):

\(\displaystyle{P}{\left({A}\ {\quad\text{or}\quad}\ {B}\right)}={P}{\left({A}\right)}\ +\ {P}{\left({B}\right)}\)

Step 2

SOLUTION

a) Evaluate the formula of Poisson probability at \(\displaystyle{k}={1}:\)

\(\displaystyle{P}{\left({X}={1}\right)}={\frac{{{1.7}^{{{1}}}{\mid}{e}^{{-{1.7}}}}}{{{1}!}}}\approx\ {0.3106}\)

Step 3

b) Evaluate the formula of Poisson probability at \(\displaystyle{k}={0},\ {1},\ {2}:\)

\(\displaystyle{P}{\left({X}={0}\right)}={\frac{{{1.7}^{{{0}}}{e}^{{-{1.7}}}}}{{{0}!}}}\approx\ {0.1827}\)

\(\displaystyle{P}{\left({X}={1}\right)}={\frac{{{1.7}^{{{1}}}{e}^{{-{1.7}}}}}{{{1}!}}}\approx\ {0.3106}\)

\(\displaystyle{P}{\left({X}={2}\right)}={\frac{{{1.7}^{{{2}}}{e}^{{-{1.7}}}}}{{{2}!}}}\approx\ {0.2640}\)

Use the special addition rule:

\(\displaystyle{P}{\left({X}\leq{2}\right)}={P}{\left({X}={0}\right)}+{P}{\left({X}={1}\right)}+{P}{\left({X}={2}\right)}\)

\(\displaystyle={0.1827}+{0.3106}+{0.2640}\)

\(\displaystyle={0.7572}\)

Step 4

c) Evaluate the formula of Poisson probability at \(\displaystyle{k}={0},{1},{2}:\)

\(\displaystyle{P}{\left({X}={0}\right)}={\frac{{{1.7}^{{{0}}}{e}^{{-{1.7}}}}}{{{0}!}}}\approx\ {0.1827}\)

\(\displaystyle{P}{\left({X}={1}\right)}={\frac{{{1.7}^{{{1}}}{e}^{{-{1.7}}}}}{{{1}!}}}\approx\ {0.3106}\)

Use the special addition rule:

\(\displaystyle{P}{\left({X}{<}{2}\right)}={P}{\left({X}={0}\right)}+{P}{\left({X}={1}\right)}\)

\(\displaystyle={0.1827}+{0.3106}\)

\(\displaystyle={0.4932}\)

Use the complement rule:

\(\displaystyle{P}{\left({X}\geq{2}\right)}={1}-{P}{\left({X}{<}{2}\right)}={1}-{0.4932}={0.5068}\)

Step 5

d) The meanof the Poisson distribition is equal to the value of the parameter \(\displaystyle\lambda\)

\(\displaystyle\mu_{{{X}}}=\lambda={1.7}\)

On average, there are 1.7 Type 1 calls are made from this motel during a period of 1 hour.

Step 6 e) The variance of the Poisson distribution is equal to the value of the parameter \(\displaystyle\lambda\)

\(\displaystyle{\sigma_{{{X}}}^{{{2}}}}=\lambda={1.7}\)

The standard deviation is the square root of the variance:

\(\displaystyle\sigma_{{{X}}}=\sqrt{{{\sigma_{{{X}}}^{{{2}}}}}}=\sqrt{{{1.7}}}\approx\ {1.3038}\)