IBDP Maths AHL 4.14 Variance of random variable AA HL Paper 1- Exam Style Questions- New Syllabus
▶️ Answer/Explanation
(a)
Since probabilities must be between 0 and 1, each probability must be non-negative.
\( 0.6-2a\ge0,\qquad 3a\ge0,\qquad 0.4-a\ge0 \)
These inequalities give
\( a\le0.3,\qquad a\ge0,\qquad a\le0.4. \)
Combining them,
\( \boxed{0\le a\le0.3} \)
Notice that the probabilities always sum to 1, so only the non-negativity conditions are required.
✅ Answer: \(\boxed{0\le a\le0.3}\)
(b)
When \(a=0.2\), the distribution becomes
\( P(X=1)=0.2,\qquad P(X=2)=0.6,\qquad P(X=3)=0.2. \)
Using the property
\( \mathrm{Var}(2-X)=\mathrm{Var}(X), \)
we first calculate the mean:
\( E(X)=1(0.2)+2(0.6)+3(0.2)=2. \)
Next, calculate
\( E(X^2)=1^2(0.2)+2^2(0.6)+3^2(0.2) =0.2+2.4+1.8=4.4. \)
Therefore,
\( \mathrm{Var}(X) =E(X^2)-[E(X)]^2 =4.4-2^2 =4.4-4 =0.4. \)
Hence,
\( \boxed{\mathrm{Var}(2-X)=0.4.} \)
Since adding or subtracting a constant changes only the mean (not the spread), the variance remains unchanged.
✅ Answer: \(\boxed{0.4}\)
▶️ Answer/Explanation
\(X \sim \text{B}(5, p)\). \({\text{P}}(X = 4) = \binom{5}{4} p^4 (1 – p) = 5p^4(1 – p)\).
\(\boxed{5p^4(1 – p)}\)
(i) Maximize \({\text{P}}(X = 4) = 5p^4(1 – p)\). Derivative: \(\frac{d}{dp}[5p^4(1 – p)] = 5[4p^3(1 – p) + p^4(-1)] = 5(4p^3 – 5p^4)\).
Set to zero: \(20p^3 – 25p^4 = 0\). Factor: \(5p^3(4 – 5p) = 0\). Solutions: \(p = 0\) or \(p = \frac{4}{5}\). Maximum at \(p = \frac{4}{5}\).
(ii) \({\text{E}}(X) = np = 5 \times \frac{4}{5} = 4\).
\(\boxed{\frac{4}{5}, 4}\)
\(X \sim \text{B}(5, p)\) M1
\({\text{P}}(X = 4) = \binom{5}{4} p^4 (1 – p)\) A1
[2 marks]
(i) \(\frac{d}{dp}(5p^4 – 5p^5) = 20p^3 – 25p^4\) M1A1
\(5p^3(4 – 5p) = 0 \Rightarrow p = \frac{4}{5}\) M1A1
Note: No \(p = 0\) in final answer for A1.
(ii) \({\text{E}}(X) = np = 5 \left(\frac{4}{5}\right)\) M1
\(= 4\) A1
[6 marks]