| Exam Board | Edexcel |
|---|---|
| Module | S1 (Statistics 1) |
| Year | 2024 |
| Session | October |
| Paper | Download PDF ↗ |
| Topic | Cumulative distribution functions |
| Type | Discrete CDF table with constants |
| Difficulty | Moderate -0.3 This is a straightforward S1 question testing basic understanding of CDFs, probability distributions, expectation and variance. Part (a) uses F(6)=1, part (b) finds differences, parts (c)-(d) apply standard formulas with arithmetic, part (e) uses Var(aX+b) rule, and part (f) requires listing simple cases. All techniques are routine recall with minimal problem-solving, though the multi-part structure and arithmetic make it slightly more substantial than the most trivial questions. |
| Spec | 5.02b Expectation and variance: discrete random variables5.03a Continuous random variables: pdf and cdf5.03e Find cdf: by integration5.04a Linear combinations: E(aX+bY), Var(aX+bY) |
| \(x\) | 1 | 2 | 3 | 4 | 5 | 6 |
| \(\mathrm {~F} ( x )\) | 0.1 | 0.2 | \(3 k\) | \(5 k\) | \(7 k\) | \(10 k\) |
| \(y\) | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 |
| \(\mathrm { P } ( Y = y )\) | \(a\) | \(a\) | \(a\) | \(b\) | \(b\) | \(b\) | 0.11 | 0.05 |
| Answer | Marks | Guidance |
|---|---|---|
| (a) \([10k = 1 \Rightarrow] k = 0.1\) | B1 | For \(k = 0.1\) oe |
| (b) e.g. \(P(X = 1) = 0.1\) and \(P(X = 2) \left[= F(2) - F(1)\right] = 0.1\) | B1 | For correct use of \(F(x)\) to find 2 probabilities May be implied by two correct probabilities |
| e.g. \(P(X = 3) \left[= F(3) - F(2)\right] = 0.1\) | M1 | For correct use of \(F(x)\) to find one other probability. May be implied by one other correct probability |
| X | 1 | 2 |
| P(X = x) | 0.1 | 0.1 |
| A1 | For a fully correct probability distribution. Need not be in a table but probabilities must be attached to the correct \(X\) values | |
| (c) \(a + a + a + b + b + b + 0.11 + 0.05 = 1\) \([3a + 3b = 0.84]\) | M1 | For use of the sum of the probabilities = 1 to form a linear equation in \(a\) and \(b\) (May be implied by \(3a + 3b = 0.84\)) |
| \(a + 2a + 3a + 4b + 5b + 6b + 0.77 + 0.4 = 4.02\) \([\Rightarrow 6a + 15b = 2.85]\) | M1 | For use of \(\sum xy = P(Y = y) = 4.02\) to form a linear equation in \(a\) and \(b\) (May be implied by \(6a + 15b = 2.85\)) |
| e.g. \(9a = 1.35 \Rightarrow a = 0.15\) * | A1* | Answer is given so there must be a correct line between the 2 equations and the given answer |
| (d) \(b = 0.13\) | ||
| \(E(Y^2) = 1^2 \times 0.15 + 2^2 \times 0.15 + 3^2 \times 0.15 + 4^2 \times '0.13' + 5^2 \times '0.13' + 6^2 \times '0.13'\) | M1 | For finding \(E(Y^2)\) with their \(b\) (At least 4 correct terms). Values for \(a\) and \(b\) must be substituted into their expression for \(E(Y^2)\) |
| \(+ 7^2 \times 0.11 + 8^2 \times 0.05 = \left[20.7\right]\) * | A1* | For a fully correct expression (no incorrect working seen) eg. \(14 \times 0.15 + 77 \times 0.13 + 49 \times 0.11 + 64 \times 0.05\) Allow \(0.15 + 0.6 + 1.35 + 2.08 + 3.25 + 4.68 + 5.39 + 3.2\) |
| (e) \(\left[\text{Var}(Y) \right] = 20.7 - 4.02^2 = [4.5396]\) | M1 | For a correct expression for Var(Y). May be implied by awrt 4.54 |
| \(\text{Var}(5 - 2Y) = 4\text{Var}(Y) = 4 \times '4.5396' = 18.1584\) | M1 A1 | For use of 4Var(Y) ft their Var(Y) provided Var(Y) is not 20.7 or 4.02 Do not allow \(5 + 4\text{Var}(Y)\) |
| awrt 18.2 | Allow \(\text{Var}(5 - 2Y) = \frac{137}{2} - \left(\frac{25}{22}\right)^2\) ft their \(E(5 - 2Y)\) and their \(E((5 - 2Y)^2)\) | |
| (f) '0.1'×0.15 + '0.1'×0.15 = 0.03 | M1 A1 | For use of \(P(X = 1) \times P(Y = 2) + P(X = 2) \times P(Y = 1)\) ft their \(X\) distribution probabilities |
| Cao |
**(a)** $[10k = 1 \Rightarrow] k = 0.1$ | B1 | For $k = 0.1$ oe
**(b)** e.g. $P(X = 1) = 0.1$ and $P(X = 2) \left[= F(2) - F(1)\right] = 0.1$ | B1 | For correct use of $F(x)$ to find 2 probabilities May be implied by two correct probabilities
e.g. $P(X = 3) \left[= F(3) - F(2)\right] = 0.1$ | M1 | For correct use of $F(x)$ to find one other probability. May be implied by one other correct probability
| X | 1 | 2 | 3 | 4 | 5 | 6 |
| P(X = x) | 0.1 | 0.1 | 0.1 | 0.2 | 0.2 | 0.3 |
| | | | | | | | A1 | For a fully correct probability distribution. Need not be in a table but probabilities must be attached to the correct $X$ values
**(c)** $a + a + a + b + b + b + 0.11 + 0.05 = 1$ $[3a + 3b = 0.84]$ | M1 | For use of the sum of the probabilities = 1 to form a linear equation in $a$ and $b$ (May be implied by $3a + 3b = 0.84$)
$a + 2a + 3a + 4b + 5b + 6b + 0.77 + 0.4 = 4.02$ $[\Rightarrow 6a + 15b = 2.85]$ | M1 | For use of $\sum xy = P(Y = y) = 4.02$ to form a linear equation in $a$ and $b$ (May be implied by $6a + 15b = 2.85$)
e.g. $9a = 1.35 \Rightarrow a = 0.15$ * | A1* | Answer is given so there must be a correct line between the 2 equations and the given answer
**(d)** $b = 0.13$ | |
$E(Y^2) = 1^2 \times 0.15 + 2^2 \times 0.15 + 3^2 \times 0.15 + 4^2 \times '0.13' + 5^2 \times '0.13' + 6^2 \times '0.13'$ | M1 | For finding $E(Y^2)$ with their $b$ (At least 4 correct terms). Values for $a$ and $b$ must be substituted into their expression for $E(Y^2)$
$+ 7^2 \times 0.11 + 8^2 \times 0.05 = \left[20.7\right]$ * | A1* | For a fully correct expression (no incorrect working seen) eg. $14 \times 0.15 + 77 \times 0.13 + 49 \times 0.11 + 64 \times 0.05$ Allow $0.15 + 0.6 + 1.35 + 2.08 + 3.25 + 4.68 + 5.39 + 3.2$
**(e)** $\left[\text{Var}(Y) \right] = 20.7 - 4.02^2 = [4.5396]$ | M1 | For a correct expression for Var(Y). May be implied by awrt 4.54
$\text{Var}(5 - 2Y) = 4\text{Var}(Y) = 4 \times '4.5396' = 18.1584$ | M1 A1 | For use of 4Var(Y) ft their Var(Y) provided Var(Y) is not 20.7 or 4.02 Do not allow $5 + 4\text{Var}(Y)$
awrt 18.2 | | Allow $\text{Var}(5 - 2Y) = \frac{137}{2} - \left(\frac{25}{22}\right)^2$ ft their $E(5 - 2Y)$ and their $E((5 - 2Y)^2)$
**(f)** '0.1'×0.15 + '0.1'×0.15 = 0.03 | M1 A1 | For use of $P(X = 1) \times P(Y = 2) + P(X = 2) \times P(Y = 1)$ ft their $X$ distribution probabilities
| | Cao
**Total: 14 marks**\begin{enumerate}
\item A biased die with six faces is rolled. The discrete random variable $X$ represents the score which is uppermost. The cumulative distribution function of $X$ is shown in the table below.
\end{enumerate}
\begin{center}
\begin{tabular}{ | c | c | c | c | c | c | c | }
\hline
$x$ & 1 & 2 & 3 & 4 & 5 & 6 \\
\hline
$\mathrm {~F} ( x )$ & 0.1 & 0.2 & $3 k$ & $5 k$ & $7 k$ & $10 k$ \\
\hline
\end{tabular}
\end{center}
(a) Find the value of the constant $k$\\
(b) Find the probability distribution of $X$
A biased die with eight faces is rolled. The discrete random variable $Y$ represents the score which is uppermost. The probability distribution of $Y$ is shown in the table below, where $a$ and $b$ are constants.
\begin{center}
\begin{tabular}{ | c | c | c | c | c | c | c | c | c | }
\hline
$y$ & 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 \\
\hline
$\mathrm { P } ( Y = y )$ & $a$ & $a$ & $a$ & $b$ & $b$ & $b$ & 0.11 & 0.05 \\
\hline
\end{tabular}
\end{center}
Given that $\mathrm { E } ( Y ) = 4.02$\\
(c) form and solve two equations in $a$ and $b$ to show that $a = 0.15$
You must show your working.\\
(Solutions relying on calculator technology are not acceptable.)\\
(d) Show that $\mathrm { E } \left( Y ^ { 2 } \right) = 20.7$\\
(e) Find $\operatorname { Var } ( 5 - 2 Y )$
These dice are each rolled once. The scores on the two dice are independent.\\
(f) Find the probability that the sum of these two scores is 3
\hfill \mbox{\textit{Edexcel S1 2024 Q6}}