Discussion
Consider the following probability density function :
P(x) = 0 \(x \lt -2\) and \(x \gt 3\)
P(x) = a \(-2 \lt x \lt 2\)
P(x) = \(\frac{1}{12}\) \(2 \lt x \lt 3\)
The value of a is :
Answer: C
The constant is \(a = \frac{11}{48}\).
A probability density must integrate to 1 over all x. The density is zero outside -2 to 3, equal to a over the 4 units from -2 to 2, and 1/12 over the 1 unit from 2 to 3:
\[ \begin{aligned} 4a + \frac{1}{12} &= 1 \\ 4a &= \frac{11}{12} \\ a &= \frac{11}{48} \end{aligned} \]
The distractor 11/12 forgets to divide by the width 4 of the interval, and 1/12 and 1/48 do not satisfy the normalisation at all.
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