A continuous random variable X has probability density function

\[f(x) = \left\{ {\begin{array}{*{20}{c}}
  {0,}&{x < 0} \\
  {a{{\text{e}}^{ - ax}},}&{x \geqslant 0.}
\end{array}} \right.\]

It is known that \({\text{P}}(X < 1) = 1 - \frac{1}{{\sqrt 2 }}\).

(a)     Show that \(a = \frac{1}{2}\ln 2\).

(b)     Find the median of X.

(c)     Calculate the probability that X < 3 given that X >1.

(a)     \(\int_0^1 {a{{\text{e}}^{ - ax}}} {\text{d}}x = 1 - \frac{1}{{\sqrt 2 }}\)     M1A1

\(\left[ { - {{\text{e}}^{ - ax}}} \right]_0^1 = 1 - \frac{1}{{\sqrt 2 }}\)     M1A1

\( - {{\text{e}}^{ - a}} + 1 = 1 - \frac{1}{{\sqrt 2 }}\)     A1

Note: Accept \({{\text{e}}^0}\) instead of 1.

 

\({{\text{e}}^{ - a}} = \frac{1}{{\sqrt 2 }}\)

\({{\text{e}}^a} = \sqrt 2 \)

\(a = \ln {2^{\frac{1}{2}}}\,\,\,\,\,\left( {{\text{accept }} - a = \ln {2^{ - \frac{1}{2}}}} \right)\)     A1

\(a = \frac{1}{2}\ln 2\)     AG

[6 marks]

 

(b)     \(\int_0^M {a{{\text{e}}^{ - ax}}{\text{d}}x = \frac{1}{2}} \)     M1A1

\(\left[ { - {{\text{e}}^{ - ax}}} \right]_0^M = \frac{1}{2}\)     A1

\( - {{\text{e}}^{ - Ma}} + 1 = \frac{1}{2}\)

\({{\text{e}}^{ - Ma}} = \frac{1}{2}\)     A1

\(Ma = \ln 2\)

\(M = \frac{{\ln 2}}{a} = 2\)     A1

[5 marks]

 

(c)     \({\text{P}}(1 < X < 3) = \int_1^3 {a{{\text{e}}^{ - ax}}{\text{d}}x} \)     M1A1

\( = - {{\text{e}}^{ - 3a}} + {{\text{e}}^{ - a}}\)     A1

\({\text{P}}(X < 3|X > 1) = \frac{{{\text{P}}(1 < X < 3)}}{{{\text{P}}(X > 1)}}\)     M1A1

\( = \frac{{ - {{\text{e}}^{ - 3a}} + {{\text{e}}^{ - a}}}}{{1 - {\text{P}}(X < 1)}}\)     A1

\( = \frac{{ - {{\text{e}}^{ - 3a}} + {{\text{e}}^{ - a}}}}{{\frac{1}{{\sqrt 2 }}}}\)     A1

\( = \sqrt 2 ( - {{\text{e}}^{ - 3a}} + {{\text{e}}^{ - a}})\)

\( = \sqrt 2 \left( { - {2^{ - \frac{3}{2}}} + {2^{ - \frac{1}{2}}}} \right)\)     A1

\( = \frac{1}{2}\)     A1

Note: Award full marks for \({\text{P}}(X < 3/X > 1) = {\text{P}}(X < 2) = \frac{1}{2}\) or quoting properties of exponential distribution.

 

[9 marks]

Total [20 marks]

Many candidates did not attempt this question and many others were clearly not familiar with this topic. On the other hand, most of the candidates who were familiar with continuous random variables and knew how to start the questions were successful and scored well in parts (a) and (b). The most common errors were in the integral of \({e^{ - at}}\), having the limits from \( - \infty \) to 1, confusion over powers and signs (‘-’ sometimes just disappeared). Understanding of conditional probability was poor and marks were low in part (c). A small number of candidates from a small number of schools coped very competently with the algebra throughout the question.