| Exam Board | CAIE |
|---|---|
| Module | S1 (Statistics 1) |
| Year | 2009 |
| Session | November |
| Marks | 6 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Discrete Probability Distributions |
| Type | Two unknowns from sum and expectation |
| Difficulty | Moderate -0.3 This is a standard S1 probability distribution question requiring two simultaneous equations (sum of probabilities = 1, and E(X) = 1.05) to find p and q, followed by routine variance calculation using E(X²) - [E(X)]². The algebraic manipulation is straightforward and this is a textbook exercise type that appears frequently in S1 papers. |
| Spec | 5.02a Discrete probability distributions: general5.02b Expectation and variance: discrete random variables |
| \(x\) | - 2 | - 1 | 0 | 1 | 2 | 3 |
| \(\mathrm { P } ( X = x )\) | 0.08 | \(p\) | 0.12 | 0.16 | \(q\) | 0.22 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(-0.16 - p + 0.16 + 2q + 0.66 = 1.05\) | M1 | Attempt at \(\Sigma px = 1.05\) no dividing |
| \(-p + 2q = 0.39\) | A1 | Correct simplified equation |
| \(p + q = 0.42\) | B1 | Accept \(p = 0.42 - q\) oe |
| \(q = 0.27\), \(p = 0.15\) | A1 [4] | Both answers correct |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\text{Var}(X) = 4\times0.08 + p + 0.16 + 4q + 1.98 - (1.05)^2\) | M1 | Subst in \(\Sigma px^2 - \text{mean}^2\) formula, mean\(^2\) subt numerically, \(p\) +ve and \(<1\) |
| \(= 2.59\) | A1 [2] | Correct answer |
# Question 2:
## Part (i):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $-0.16 - p + 0.16 + 2q + 0.66 = 1.05$ | M1 | Attempt at $\Sigma px = 1.05$ no dividing |
| $-p + 2q = 0.39$ | A1 | Correct simplified equation |
| $p + q = 0.42$ | B1 | Accept $p = 0.42 - q$ oe |
| $q = 0.27$, $p = 0.15$ | A1 **[4]** | Both answers correct |
## Part (ii):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\text{Var}(X) = 4\times0.08 + p + 0.16 + 4q + 1.98 - (1.05)^2$ | M1 | Subst in $\Sigma px^2 - \text{mean}^2$ formula, mean$^2$ subt numerically, $p$ +ve and $<1$ |
| $= 2.59$ | A1 **[2]** | Correct answer |
---2 The probability distribution of the random variable $X$ is shown in the following table.
\begin{center}
\begin{tabular}{ | c | c | c | c | c | c | c | }
\hline
$x$ & - 2 & - 1 & 0 & 1 & 2 & 3 \\
\hline
$\mathrm { P } ( X = x )$ & 0.08 & $p$ & 0.12 & 0.16 & $q$ & 0.22 \\
\hline
\end{tabular}
\end{center}
The mean of $X$ is 1.05 .\\
(i) Write down two equations involving $p$ and $q$ and hence find the values of $p$ and $q$.\\
(ii) Find the variance of $X$.
\hfill \mbox{\textit{CAIE S1 2009 Q2 [6]}}