Edexcel FS2 AS 2024 June — Question 4 8 marks

Exam BoardEdexcel
ModuleFS2 AS (Further Statistics 2 AS)
Year2024
SessionJune
Marks8
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicContinuous Uniform Random Variables
TypeP(g(X) > c) using inverse transformation
DifficultyStandard +0.3 This is a straightforward Further Statistics question testing standard uniform distribution properties. Parts (a)-(b) are routine recall/application, part (c) requires factoring a quadratic and finding where it's positive (standard A-level technique), and part (d) involves a simple integration. While it's Further Maths content, the techniques are all standard with no novel insight required, making it slightly easier than average overall.
Spec5.02e Discrete uniform distribution
  1. The continuous random variable \(X\) is uniformly distributed over the interval [2, 7]
    1. Write down the value of \(\mathrm { E } ( X )\)
    2. Find \(\mathrm { P } ( 1 < X < 4 )\)
    3. Find \(\mathrm { P } \left( 2 X ^ { 2 } - 15 X + 27 > 0 \right)\)
    4. Find \(\mathrm { E } \left( \frac { 3 } { X ^ { 2 } } \right)\)
Question 4:
Part (a):
AnswerMarks Guidance
Answer/WorkingMark Guidance
\(E(X) = \mathbf{4.5}\)B1 4.5 oe
Part (b):
AnswerMarks Guidance
Answer/WorkingMark Guidance
\([P(1 < X < 4) = P(2 < X < 4)] = \frac{2}{5}\)B1 \(\frac{2}{5}\) oe
Part (c):
AnswerMarks Guidance
Answer/WorkingMark Guidance
\(2X^2 - 15X + 27 > 0 \to \left[\{X<3\} \cup \{X>4.5\}\right]\)M1 Attempting to find roots. Can be implied by sight of 3 and 4.5
\(P(\{X<3\} \cup \{X>4.5\}) = \frac{3-2}{7-2} + \frac{7-4.5}{7-2} = 1 - \frac{4.5-3}{7-2}\)M1 Attempt to find probability for "outside" region from \(U[2,7]\). Must have both correct ft probability statements and at least one correct ft probability
\(= \frac{7}{10}\) or 0.7A1 \(\frac{7}{10}\) oe
Part (d):
AnswerMarks Guidance
Answer/WorkingMark Guidance
\(E\!\left(\frac{3}{X^2}\right) = \int_2^7 \frac{3}{x^2} \times \frac{1}{(7-2)}\,dx\)M1 Use of \(E(g(x))\) by setting up correct integral (ignore limits)
\(= \left[-0.6x^{-1}\right]_2^7\)M1 Integration with limits
\(= \frac{3}{14}\)A1 \(\frac{3}{14}\) oe from correct working 
## Question 4:

### Part (a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $E(X) = \mathbf{4.5}$ | B1 | 4.5 oe |

### Part (b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $[P(1 < X < 4) = P(2 < X < 4)] = \frac{2}{5}$ | B1 | $\frac{2}{5}$ oe |

### Part (c):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $2X^2 - 15X + 27 > 0 \to \left[\{X<3\} \cup \{X>4.5\}\right]$ | M1 | Attempting to find roots. Can be implied by sight of 3 and 4.5 |
| $P(\{X<3\} \cup \{X>4.5\}) = \frac{3-2}{7-2} + \frac{7-4.5}{7-2} = 1 - \frac{4.5-3}{7-2}$ | M1 | Attempt to find probability for "outside" region from $U[2,7]$. Must have both correct ft probability statements and at least one correct ft probability |
| $= \frac{7}{10}$ or **0.7** | A1 | $\frac{7}{10}$ oe |

### Part (d):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $E\!\left(\frac{3}{X^2}\right) = \int_2^7 \frac{3}{x^2} \times \frac{1}{(7-2)}\,dx$ | M1 | Use of $E(g(x))$ by setting up correct integral (ignore limits) |
| $= \left[-0.6x^{-1}\right]_2^7$ | M1 | Integration with limits |
| $= \frac{3}{14}$ | A1 | $\frac{3}{14}$ oe from correct working |

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\begin{enumerate}
  \item The continuous random variable $X$ is uniformly distributed over the interval [2, 7]\\
(a) Write down the value of $\mathrm { E } ( X )$\\
(b) Find $\mathrm { P } ( 1 < X < 4 )$\\
(c) Find $\mathrm { P } \left( 2 X ^ { 2 } - 15 X + 27 > 0 \right)$\\
(d) Find $\mathrm { E } \left( \frac { 3 } { X ^ { 2 } } \right)$
\end{enumerate}

\hfill \mbox{\textit{Edexcel FS2 AS 2024 Q4 [8]}}
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