Standard +0.3 This is a straightforward CDF question requiring standard techniques: calculating a probability from F(x) (simple substitution), identifying the upper quartile, and transforming a CDF using Y=X². While the transformation requires careful consideration of the domain and the relationship P(Y≤y)=P(-√y≤X≤√y), these are well-practiced methods in Further Maths statistics with no novel insight required.
A continuous random variable \(X\) has cumulative distribution function F given by
$$\mathrm{F}(x) = \begin{cases}
0 & x < -1, \\
\frac{1}{4}(x^3 + 1) & -1 \leqslant x \leqslant 1, \\
1 & x > 1.
\end{cases}$$
Find \(\mathrm{P}\left(X \geqslant \frac{3}{4}\right)\), and state what can be deduced about the upper quartile of \(X\). [3]
Obtain the cumulative distribution function of \(Y\), where \(Y = X^2\). [5]
State deduction about upper quartile Q3 (dep *A1): Q3 > ¾ B1
Express cum. dist. fn. G of Y in terms of X: G(y) = P(Y Y y) = P(–√y Y X Y √y) M1 A1
Relate to F: = F(√y) – F(–√y) M1
Simplify: = ½[(√y) 3 + 1] – ½[(–√y) 3 + 1]
3/2
= y A1
Answer
Marks
State G in full (√ on previous result): 0 (y < 0), y 3/2 (0 Y y Y 1), 1 (y > 1) A1√
(3)
(5)
[8]
Question 7:
7 | Find P(X ≥ ¾): 1 – F(¾) = 1 – ½[(¾) 3 + 1] M1
= 37/128 or 0⋅289 *A1
State deduction about upper quartile Q3 (dep *A1): Q3 > ¾ B1
Express cum. dist. fn. G of Y in terms of X: G(y) = P(Y Y y) = P(–√y Y X Y √y) M1 A1
Relate to F: = F(√y) – F(–√y) M1
Simplify: = ½[(√y) 3 + 1] – ½[(–√y) 3 + 1]
3/2
= y A1
State G in full (√ on previous result): 0 (y < 0), y 3/2 (0 Y y Y 1), 1 (y > 1) A1√ | (3)
(5) | [8]
A continuous random variable $X$ has cumulative distribution function F given by
$$\mathrm{F}(x) = \begin{cases}
0 & x < -1, \\
\frac{1}{4}(x^3 + 1) & -1 \leqslant x \leqslant 1, \\
1 & x > 1.
\end{cases}$$
Find $\mathrm{P}\left(X \geqslant \frac{3}{4}\right)$, and state what can be deduced about the upper quartile of $X$. [3]
Obtain the cumulative distribution function of $Y$, where $Y = X^2$. [5]
\hfill \mbox{\textit{CAIE FP2 2009 Q7 [8]}}