Derive or verify variance formula

4. The continuous random variable \(X\) is uniformly distributed over the interval \([ - 4,6 ]\).

1. Write down the mean of \(X\).
2. Find \(\mathrm { P } ( X \leqslant 2.4 )\)
3. Find \(\mathrm { P } ( - 3 < X - 5 < 3 )\) The continuous random variable \(Y\) is uniformly distributed over the interval \([ a , 4 a ]\).
4. Use integration to show that \(\mathrm { E } \left( Y ^ { 2 } \right) = 7 a ^ { 2 }\)
5. Find \(\operatorname { Var } ( Y )\).
6. Given that \(\mathrm { P } \left( X < \frac { 8 } { 3 } \right) = \mathrm { P } \left( Y < \frac { 8 } { 3 } \right)\), find the value of \(a\).

3

1. The continuous random variable \(T\) follows a rectangular distribution with probability density function given by $$\mathrm { f } ( t ) = \left\{ \begin{array} { l c } k & - a \leqslant t \leqslant b
   0 & \text { otherwise } \end{array} \right.$$
   1. Express \(k\) in terms of \(a\) and \(b\).
   2. Prove, using integration, that \(\mathrm { E } ( T ) = \frac { 1 } { 2 } ( b - a )\).
2. The error, in minutes, made by a commuter when estimating the journey time by train into London may be modelled by the random variable \(T\) with probability density function $$\mathrm { f } ( t ) = \left\{ \begin{array} { c c } \frac { 1 } { 10 } & - 4 \leqslant t \leqslant 6
   0 & \text { otherwise } \end{array} \right.$$
   1. Write down the value of \(\mathrm { E } ( T )\).
   2. Calculate \(\mathrm { P } ( T < - 3\) or \(T > 3 )\).

5

1. The continuous random variable \(X\) follows a rectangular distribution with probability density function defined by $$f ( x ) = \begin{cases} \frac { 1 } { b } & 0 \leqslant x \leqslant b
   0 & \text { otherwise } \end{cases}$$
   1. Write down \(\mathrm { E } ( X )\).
   2. Prove, using integration, that $$\operatorname { Var } ( X ) = \frac { 1 } { 12 } b ^ { 2 }$$
2. At an athletics meeting, the error, in seconds, made in recording the time taken to complete the 10000 metres race may be modelled by the random variable \(T\), having the probability density function $$f ( t ) = \left\{ \begin{array} { c c } 5 & - 0.1 \leqslant t \leqslant 0.1
   0 & \text { otherwise } \end{array} \right.$$ Calculate \(\mathrm { P } ( | T | > 0.02 )\).

4 A continuous random variable \(X\) has a probability density function defined by $$f ( x ) = \begin{cases} \frac { 1 } { k } & a \leqslant x \leqslant b
0 & \text { otherwise } \end{cases}$$ where \(b > a > 0\).

1. 1. Prove that \(k = b - a\).
   2. Write down the value of \(\mathrm { E } ( X )\).
   3. Show, by integration, that \(\mathrm { E } \left( X ^ { 2 } \right) = \frac { 1 } { 3 } \left( b ^ { 2 } + a b + a ^ { 2 } \right)\).
   4. Hence derive a simplified formula for \(\operatorname { Var } ( X )\).
2. Given that \(a = 4\) and \(\operatorname { Var } ( X ) = 3\), find the numerical value of \(\mathrm { E } ( X )\).
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5. The random variable \(X\) has a continuous uniform distribution on the interval \(a \leq X \leq 3 a\).

1. Without assuming any standard results, prove that \(\mu\), the mean value of \(X\), is equal to \(2 a\) and derive an expression for \(\sigma ^ { 2 }\), the variance of \(X\), in terms of \(a\).
2. Find the probability that \(| X - \mu | < \sigma\) and compare this with the same probability when \(x\) is modelled by a Normal distribution with the same mean and variance. \section*{STATISTICS 2 (A) TEST PAPER 8 Page 2}

7. The random variable \(X\) follows a continuous uniform distribution over the interval [2,11].

1. Write down the mean of \(X\).
2. Find \(\mathrm { P } ( X \geq 8.6 )\).
3. Find \(\mathrm { P } ( | X - 5 | < 2 )\). The random variable \(Y\) follows a continuous uniform distribution over the interval \([ a , b ]\).
4. Show by integration that $$\mathrm { E } \left( Y ^ { 2 } \right) = \frac { 1 } { 3 } \left( b ^ { 2 } + a b + a ^ { 2 } \right)$$
5. Hence, prove that $$\operatorname { Var } ( Y ) = \frac { 1 } { 12 } ( b - a ) ^ { 2 }$$ You may assume that \(\mathrm { E } ( Y ) = \frac { 1 } { 2 } ( a + b )\).

3 The continuous random variable \(R\) follows a rectangular distribution with probability density function given by $$f ( r ) = \begin{cases} k & - a \leq r \leq b
0 & \text { otherwise } \end{cases}$$ Prove, using integration, that \(\mathrm { E } ( R ) = \frac { 1 } { 2 } ( b - a )\)
[0pt] [4 marks]