| Exam Board | Edexcel |
|---|---|
| Module | FS2 AS (Further Statistics 2 AS) |
| Year | 2023 |
| Session | June |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Continuous Uniform Random Variables |
| Type | Find parameters from given statistics |
| Difficulty | Standard +0.3 This question tests standard formulas for continuous uniform distributions (mean, variance, probability) with straightforward algebraic manipulation. Part (a) requires using P(X) = length/total length to find k. Part (b) involves finding parameters from E(Y) and Var(Y) formulas, then a binomial probability calculation. All steps are routine applications of memorized formulas with no conceptual challenges or novel problem-solving required. |
| Spec | 5.02e Discrete uniform distribution |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(P(-4 < X < 2) = P(-3 < X < 2)\) | M1 | 1.1b – realising they need to consider \(P(-3 < X < 2)\) |
| \(\frac{2+3}{k+3} = \frac{1}{3}\) or \(k = 2 + 2\times(2--3)\) | M1 | 1.1b – correct equation for \(k\), or allow \(\frac{2+4}{k+3} = \frac{1}{3}\ [\to k=15]\) or \(k = 2+2\times(2--4)[=14]\) |
| \(k = 12\) | A1 | 1.1b – cao [NB M0M1A0 is possible, often implied by \(k=15\)] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\frac{a+b}{2} = 6\) and \(\frac{1}{12}(b-a)^2 = 192\) | M1 | 3.1a – translating the problem into 2 correct equations (using 6 and 192) |
| \(\frac{1}{12}(a-(12-a))^2 = 192\) or \(\frac{1}{12}((12-b)-b)^2 = 192\) (oe) | M1 | 1.1b – correct method to eliminate \(a\) or \(b\) (e.g. equation in \(a\) or \(b\)); also allow \(a+b=12\) and \(b-a=48\) |
| \(a = -18\) or \(b = 30\) | A1 | 1.1b – correct single value for \(a\) or \(b\) (\(a=30\) and \(b=-18\) is A0) |
| \(P(Y > 7.5) = \frac{\text{"30"}-7.5}{\text{"30"}-(\text{"-18"})} = \left[\frac{15}{32}\text{ or } 0.46875\right]\) (accept 0.469 or better) | M1 | 3.4 – using their model to find \(P(Y > 7.5)\) [\(P(Y < 7.5) = \frac{17}{32}\) is M0 unless it leads to 0.771] |
| \(R \sim B(5,\ \text{"0.46875"})\) | M1 | 3.3 – stating or using correct model i.e. \(B(5, p)\) where \(p\) is a probability based on \(Y\) and 7.5; NB use of \(B(5, \frac{17}{32})\) is OK here and typically scores M0M1A0 |
| \(P(R \geqslant 2) = 0.7710\ldots\) | A1 | 1.1b – awrt 0.771 |
## Question 4:
### Part 4(a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $P(-4 < X < 2) = P(-3 < X < 2)$ | M1 | 1.1b – realising they need to consider $P(-3 < X < 2)$ |
| $\frac{2+3}{k+3} = \frac{1}{3}$ or $k = 2 + 2\times(2--3)$ | M1 | 1.1b – correct equation for $k$, or allow $\frac{2+4}{k+3} = \frac{1}{3}\ [\to k=15]$ or $k = 2+2\times(2--4)[=14]$ |
| $k = 12$ | A1 | 1.1b – cao [NB M0M1A0 is possible, often implied by $k=15$] |
### Part 4(b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\frac{a+b}{2} = 6$ **and** $\frac{1}{12}(b-a)^2 = 192$ | M1 | 3.1a – translating the problem into 2 correct equations (using 6 and 192) |
| $\frac{1}{12}(a-(12-a))^2 = 192$ or $\frac{1}{12}((12-b)-b)^2 = 192$ (oe) | M1 | 1.1b – correct method to eliminate $a$ or $b$ (e.g. equation in $a$ or $b$); also allow $a+b=12$ and $b-a=48$ |
| $a = -18$ or $b = 30$ | A1 | 1.1b – correct single value for $a$ or $b$ ($a=30$ and $b=-18$ is A0) |
| $P(Y > 7.5) = \frac{\text{"30"}-7.5}{\text{"30"}-(\text{"-18"})} = \left[\frac{15}{32}\text{ or } 0.46875\right]$ (accept 0.469 or better) | M1 | 3.4 – using their model to find $P(Y > 7.5)$ [$P(Y < 7.5) = \frac{17}{32}$ is M0 unless it leads to 0.771] |
| $R \sim B(5,\ \text{"0.46875"})$ | M1 | 3.3 – stating or using correct model i.e. $B(5, p)$ where $p$ is a probability based on $Y$ and 7.5; NB use of $B(5, \frac{17}{32})$ is OK here and typically scores M0M1A0 |
| $P(R \geqslant 2) = 0.7710\ldots$ | A1 | 1.1b – awrt 0.771 |\begin{enumerate}
\item The random variable $X$ has a continuous uniform distribution over the interval $[ - 3 , k ]$ Given that $\mathrm { P } ( - 4 < X < 2 ) = \frac { 1 } { 3 }$\\
(a) find the value of $k$
\end{enumerate}
A computer generates a random number, $Y$, where
\begin{itemize}
\item $\quad Y$ has a continuous uniform distribution over the interval $[ a , b ]$
\item $\mathrm { E } ( Y ) = 6$
\item $\operatorname { Var } ( Y ) = 192$
\end{itemize}
The computer generates 5 random numbers.\\
(b) Calculate the probability that at least 2 of the 5 numbers generated are greater than 7.5
\hfill \mbox{\textit{Edexcel FS2 AS 2023 Q4 [9]}}