Find expectation E(X)

1 A random variable \(X\) has probability density function f , where $$f ( x ) = \begin{cases} \frac { 3 } { 2 } \left( 1 - x ^ { 2 } \right) & 0 \leqslant x \leqslant 1
0 & \text { otherwise } \end{cases}$$ Find \(\mathrm { E } ( X )\).

5 A continuous random variable \(X\) has probability density function given by $$f ( x ) = \begin{cases} \frac { 1 } { 6 } x & 2 \leqslant x \leqslant 4
0 & \text { otherwise } \end{cases}$$

1. Find \(\mathrm { E } ( X )\).
2. Find the median of \(X\).
3. Two independent values of \(X\) are chosen at random. Find the probability that both these values are greater than 3 .

5 Two random variables \(S\) and \(T\) have probability density functions given by $$\begin{aligned} & f _ { S } ( x ) = \begin{cases} \frac { 3 } { a ^ { 3 } } ( x - a ) ^ { 2 } & 0 \leqslant x \leqslant a
0 & \text { otherwise } \end{cases}
& f _ { T } ( x ) = \begin{cases} c & 0 \leqslant x \leqslant a
0 & \text { otherwise } \end{cases} \end{aligned}$$ where \(a\) and \(c\) are constants.

1. On a single diagram sketch both probability density functions.
2. Calculate the mean of \(S\), in terms of \(a\).
3. Use your diagram to explain which of \(S\) or \(T\) has the bigger variance. (Answers obtained by calculation will score no marks.)

1 A continuous random variable \(X\) has probability density function given by $$f ( x ) = \begin{cases} \frac { 2 x } { 5 } & 0 \leqslant x \leqslant 1
\frac { 2 } { 5 \sqrt { x } } & 1 < x \leqslant 4
0 & \text { otherwise } \end{cases}$$ Find

1. \(\mathrm { E } ( X )\),
2. \(\mathrm { P } ( X \geqslant \mathrm { E } ( X ) )\).

2 A particular species of reed that grows up to 2 metres in length is used for thatching. The lengths in metres of the reeds when harvested are modelled by the random variable \(X\) which has the following probability density function, \(\mathrm { f } ( x )\). $$f ( x ) = \begin{cases} \frac { 3 } { 16 } \left( 4 x - x ^ { 2 } \right) & \text { for } 0 \leqslant x \leqslant 2
0 & \text { elsewhere } \end{cases}$$

1. Sketch \(\mathrm { f } ( x )\).
2. Show that \(\mathrm { E } ( X ) = \frac { 5 } { 4 }\) and find the standard deviation of the lengths of the harvested reeds.
3. Find the standard error of the mean length for a random sample of 100 reeds. Once the harvested reeds have been collected, any that are shorter than 1 metre are discarded.
4. Find the proportion of reeds that should be discarded according to the model.
5. Reeds are harvested from a large area which is divided into several reed beds. A sample of the harvested reeds is required for quality control. How might the method of cluster sampling be used to obtain it?

6. A continuous random variable \(X\) has probability density function $$f ( x ) = \begin{cases} a x ^ { 2 } + b x & 1 \leqslant x \leqslant 7
0 & \text { otherwise } \end{cases}$$ where \(a\) and \(b\) are constants.

1. Show that \(114 a + 24 b = 1\) Given that \(a = \frac { 1 } { 90 }\)
2. use algebraic integration to find \(\mathrm { E } ( X )\)
3. find the cumulative distribution function of \(X\), specifying it for all values of \(x\)
4. find \(\mathrm { P } ( X > \mathrm { E } ( X ) )\)
5. use your answer to part (d) to describe the skewness of the distribution.
   

1. The time, in minutes, spent waiting for a call to a call centre to be answered is modelled by the random variable \(T\) with probability density function

$$f ( t ) = \left\{ \begin{array} { l c } \frac { 1 } { 192 } \left( t ^ { 3 } - 48 t + 128 \right) & 0 \leqslant t \leqslant 4
0 & \text { otherwise } \end{array} \right.$$

1. Use algebraic integration to find, in minutes and seconds, the mean waiting time.
2. Show that \(\mathrm { P } ( 1 < T < 3 ) = \frac { 7 } { 16 }\) A supervisor randomly selects 256 calls to the call centre.
3. Use a suitable approximation to find the probability that more than 125 of these calls take between 1 and 3 minutes to be answered.

5 The continuous random variable \(X\) has probability density function $$f ( x ) = \begin{cases} \frac { 1 } { 6 } e ^ { \frac { x } { 3 } } & 0 \leq x \leq \ln 27
0 & \text { otherwise } \end{cases}$$ Show that the mean of \(X\) is \(\frac { 3 } { 2 } ( \ln 27 - 2 )\)