a)

1. $X \sim \operatorname{Poisson}(\var{thismany})$, so $\lambda = \var{thismany}$.

2. The expectation is given by $\operatorname{E}[X]=\lambda=\var{thismany}$

3. $\operatorname{stdev}(X)=\sqrt{\lambda}=\sqrt{\var{thismany}}=\var{sd}$ to 3 decimal places.

b)

1. \[ \begin{eqnarray*}\operatorname{P}(X = \var{thisnumber}) &=& \frac{e ^ { -\var{thismany}}\var{thismany} ^ {\var{thisnumber}}} {\var{thisnumber}!}\\& =& \var{prob1} \end{eqnarray*} \] to 3 decimal places.

2. If an employee receives a warning then he or she must have sold less than 3.

Hence we need to find:

\[ \begin{eqnarray*}\operatorname{P}(X < \var{number1})& =& \simplify[all,!collectNumbers]{P(X = 0) + P(X = 1) + {v}*P(X = 2)}\\& =& \simplify[all,!collectNumbers]{e ^ { -thismany} + {thismany} * e ^ { -thismany} + {v} * (({thismany} ^ 2 * e ^ { -thismany}) / 2)} \\&=& \var{prob2} \end{eqnarray*} \]

to 3 decimal places.

a)

1. $X \sim \operatorname{bin}(\var{number1},\var{prob})$, so $n= \var{number1},\;\;p=\var{prob}$.

2. The expectation is given by $\operatorname{E}[X]=n\times p=\var{number1}\times \var{prob}=\var{number1*prob}$

3. $\operatorname{stdev}(X)=\sqrt{n\times p \times (1-p)}=\sqrt{\var{number1}\times \var{prob} \times \var{1-prob}}=\var{sd}$ to 3 decimal places.

b)

1. \[ \begin{eqnarray*}\operatorname{P}(X = \var{thisnumber}) &=& \dbinom{\var{number1}}{\var{thisnumber}}\times\var{prob}^{\var{thisnumber}}\times(1-\var{prob})^{\var{number1-thisnumber}}\\& =& \var{comb(number1,thisnumber)} \times\var{prob}^{\var{thisnumber}}\times\var{1-prob}^{\var{number1-thisnumber}}\\&=&\var{prob1}\end{eqnarray*} \] to 3 decimal places.

2. 

\[ \begin{eqnarray*}\operatorname{P}(X \leq \var{thatnumber})& =& \simplify[all,!collectNumbers]{P(X = 0) + P(X = 1) + {v}*P(X = 2)}\\& =& \simplify[zeroFactor,zeroTerm,unitFactor]{{1 -prob} ^ {number1}+ {number1} *{prob} *{1 -prob} ^ {number1 -1} + {v} * ({number1} * {number1 -1}/2)* {prob} ^ 2 *( {1 -prob} ^ {number1 -2})}\\& =& \var{prob2}\end{eqnarray*} \]

to 3 decimal places.

1. Converting to $\operatorname{N}(0,1)$

$\simplify[all,!collectNumbers]{P(X < {lower}) = P(Z < ({lower} -{m}) / {s}) = P(Z < {lower-m}/{s}) = 1 -P(Z < {m-lower}/{s})} = 1 -\var{p} = \var{precround(1 -p,4)}$ to 4 decimal places.

2. Converting to $\operatorname{N}(0,1)$

$\simplify[all,!collectNumbers]{P(X > {upper}) = P(Z > ({upper} -{m}) / {s}) = P(Z > {upper-m}/{s}) = 1 -P(Z < {upper-m}/{s})} = 1-\var{p1} = \var{precround(1 -p1,4)}$ to 4 decimal places.

a)

For a uniform distribution \[X \sim \operatorname{U}(\var{lower},\var{upper})\] we have:

$\displaystyle \operatorname{E}[X] = \frac{\var{lower}+\var{upper}}{2}=\var{ans1}$m

$\displaystyle \operatorname{Var}[X] = \frac{(\var{upper}-\var{lower})^2}{12}=\frac{(\var{upper-lower})^2}{12}=\var{ans2}$ to 3 decimal places.

b)

$\displaystyle P(X \le \var{thisdis}\textrm{km})=\frac{\var{thisdis}\times 1000 -\var{lower}}{\var{upper}-\var{lower}}=\var{ans3}$ to 3 decimal places.

If $X \sim \operatorname{exp}(\lambda)$ then $\displaystyle \operatorname{E}[X] =\frac{1}{\lambda}$ and  $\displaystyle \operatorname{Var}(X)=\frac{1}{\lambda^2}$.

Also $P(X \lt a)=1-e^{-\lambda a}$.

a)

If $X \sim \operatorname{exp}(\var{ra})$ then:

$\displaystyle \operatorname{E}[X] =\frac{1}{\lambda}=\frac{1}{\var{ra}}=\var{ans1}$ to 3 decimal places.

$\displaystyle \operatorname{Var}(X) =\frac{1}{\lambda^2}=\frac{1}{\var{ra}^2}=\var{ans2}$ to 3 decimal places.

b)

$P(X \lt \var{thistime}) = 1 -(e ^ {-\var{ ra} \times \var{thistime}}) = 1 -(e ^ { -\var{ra * thistime}}) = \var{ans3}$ to 3 decimal places.