\begin{enumerate}
  \item $\mathrm { E } ( a X + b Y + c ) = a \mathrm { E } ( X ) + b \mathrm { E } ( Y ) + c$,
  \item if $X$ and $Y$ are independent then $\operatorname { Var } ( a X + b Y + c ) = a ^ { 2 } \operatorname { Var } ( X ) + b ^ { 2 } \operatorname { Var } ( Y )$.
\end{enumerate}

\section*{Discrete distributions}
$X$ is a random variable taking values $x _ { i }$ in a discrete distribution with $\mathrm { P } \left( X = x _ { i } \right) = p _ { i }$\\
Expectation: $\mu = \mathrm { E } ( X ) = \sum x _ { i } p _ { i }$\\
Variance: $\sigma ^ { 2 } = \operatorname { Var } ( X ) = \sum \left( x _ { i } - \mu \right) ^ { 2 } p _ { i } = \sum x _ { i } ^ { 2 } p _ { i } - \mu ^ { 2 }$

\begin{center}
\begin{tabular}{|l|l|l|l|}
\hline
 & $P ( X = x )$ & E $( X )$ & $\operatorname { Var } ( X )$ \\
\hline
Binomial $\mathrm { B } ( n , p )$ & $\binom { n } { x } p ^ { x } ( 1 - p ) ^ { n - x }$ & $n p$ & $n p ( 1 - p )$ \\
\hline
Uniform distribution over $1,2 , \ldots , n , \mathrm { U } ( n )$ & $\frac { 1 } { n }$ & $\frac { n + 1 } { 2 }$ & $\frac { 1 } { 12 } \left( n ^ { 2 } - 1 \right)$ \\
\hline
Geometric distribution Geo(p) & $( 1 - p ) ^ { x - 1 } p$ & $\frac { 1 } { p }$ & $\frac { 1 - p } { p ^ { 2 } }$ \\
\hline
Poisson $\operatorname { Po } ( \lambda )$ & $e ^ { - \lambda } \frac { \lambda ^ { x } } { x ! }$ & $\lambda$ & $\lambda$ \\
\hline
\end{tabular}
\end{center}

\section*{Continuous distributions}
$X$ is a continuous random variable with probability density function (p.d.f.) $\mathrm { f } ( x )$\\
Expectation: $\mu = \mathrm { E } ( X ) = \int x \mathrm { f } ( x ) \mathrm { d } x$\\
Variance: $\sigma ^ { 2 } = \operatorname { Var } ( X ) = \int ( x - \mu ) ^ { 2 } \mathrm { f } ( x ) \mathrm { d } x = \int x ^ { 2 } \mathrm { f } ( x ) \mathrm { d } x - \mu ^ { 2 }$\\
Cumulative distribution function $\mathrm { F } ( x ) = \mathrm { P } ( X \leq x ) = \int _ { - \infty } ^ { x } \mathrm { f } ( t ) \mathrm { d } t$

\begin{center}
\begin{tabular}{|l|l|l|l|}
\hline
 & p.d.f. & E ( $X$ ) & $\operatorname { Var } ( X )$ \\
\hline
Continuous uniform distribution over [ $a , b$ ] & $\frac { 1 } { b - a }$ & $\frac { 1 } { 2 } ( a + b )$ & $\frac { 1 } { 12 } ( b - a ) ^ { 2 }$ \\
\hline
Exponential & $\lambda \mathrm { e } ^ { - \lambda x }$ & $\frac { 1 } { \lambda }$ & $\frac { 1 } { \lambda ^ { 2 } }$ \\
\hline
Normal $N \left( \mu , \sigma ^ { 2 } \right)$ & $\frac { 1 } { \sigma \sqrt { 2 \pi } } \mathrm { e } ^ { - \frac { 1 } { 2 } \left( \frac { x - \mu } { \sigma } \right) ^ { 2 } }$ & $\mu$ & $\sigma ^ { 2 }$ \\
\hline
\end{tabular}
\end{center}

\section*{Percentage points of the normal distribution}
If $Z$ has a normal distribution with mean 0 and variance 1 then, for each value of $p$, the table gives the value of $z$ such that $P ( Z \leq z ) = p$.

\begin{center}
\begin{tabular}{ | c | c c c | c c c | c c c | }
\hline
$p$ & 0.75 & 0.90 & 0.95 & 0.975 & 0.99 & 0.995 & 0.9975 & 0.999 & 0.9995 \\
\hline
$z$ & 0.674 & 1.282 & 1.645 & 1.960 & 2.326 & 2.576 & 2.807 & 3.090 & 3.291 \\
\hline
\end{tabular}
\end{center}

\begin{enumerate}
  \item The random variable $X$ is uniformly distributed over the interval $[ - 1,5 ]$.\\
a. Sketch the probability density function $f ( x )$ of $X$.\\
b. State the value of $\mathrm { P } ( X = 2 )$
\end{enumerate}

Find\\
c. $\mathrm { E } ( X )$\\
d. $\operatorname { Var } ( X )$\\

\hfill \mbox{\textit{SPS SPS FM Statistics 2020 Q1 [6]}}