Easy -1.2 This is a straightforward application of the fundamental property that a PDF must integrate to 1. Students simply integrate t/8 from 0 to k, set equal to 1, and solve for k. It's a single-step calculation with no conceptual difficulty beyond knowing the basic PDF property, and the multiple-choice format further reduces difficulty.
1 The random variable \(T\) has probability density defined by
$$\mathrm { f } ( t ) = \left\{ \begin{array} { c c }
\frac { t } { 8 } & 0 \leq t \leq k \\
0 & \text { otherwise }
\end{array} \right.$$
Find the value of \(k\)
[0pt]
[1 mark]
$$\begin{array} { l l l l }
\frac { 1 } { 16 } & \frac { 1 } { 4 } & 4 & 16
\end{array}$$
1 The random variable $T$ has probability density defined by
$$\mathrm { f } ( t ) = \left\{ \begin{array} { c c }
\frac { t } { 8 } & 0 \leq t \leq k \\
0 & \text { otherwise }
\end{array} \right.$$
Find the value of $k$\\[0pt]
[1 mark]
$$\begin{array} { l l l l }
\frac { 1 } { 16 } & \frac { 1 } { 4 } & 4 & 16
\end{array}$$
\hfill \mbox{\textit{AQA Further AS Paper 2 Statistics Q1 [1]}}