| Exam Board | CAIE |
|---|---|
| Module | S1 (Statistics 1) |
| Year | 2019 |
| Session | March |
| Marks | 6 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Discrete Probability Distributions |
| Type | Simple algebraic expression for P(X=x) |
| Difficulty | Moderate -0.8 This is a straightforward probability distribution question requiring basic recall: constructing a table, using ΣP(X=x)=1 to find k, then applying standard E(X) and Var(X) formulas. The arithmetic is simple (small integer values) and the method is routine textbook material with no problem-solving insight needed. |
| Spec | 2.04a Discrete probability distributions5.02b Expectation and variance: discrete random variables |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Probability distribution table: \(x = -1, 1, 2, 3\) with \(p = k, k, 4k, 9k\) | B1 | Probability distribution table with correct values of \(x\), no additional values unless with probability 0 stated, at least one correct probability including \(k\) |
| \(15k = 1\) | M1 | Equating \(\sum p = 1\), may be implied by answer |
| \(k = \frac{1}{15}\) | A1 | If 0 scored, SCB2 for probability distribution table with correct numerical probabilities |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(E(X) = 8k + 27k = 35k = \frac{35}{15} = \frac{7}{3}\) | B1FT | FT if \(0 <\) their \(k < 1\) |
| \(\text{Var}(X) = (k + k + 16k + 81k) - (35k)^2\) | M1 | Correct formula for variance, in terms of \(k\) at least — must have '\(-\text{mean}^2\)' (ft) |
| \(= 1.16,\ \frac{52}{45}\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(E(X) = \frac{8}{15} + \frac{27}{15} = \frac{35}{15} = \frac{7}{3}\) | B1FT | FT if \(0 <\) their \(k < 1\) |
| \(\text{Var}(X) = \frac{1}{15} + \frac{1}{15} + \frac{16}{15} + \frac{81}{15} - \left(\frac{7}{3}\right)^2\) | M1 | Substitute their values in correct var formula — must have '\(-\text{mean}^2\)' (ft) |
| \(= 1.16\ (= 52/45)\) | A1 | Using their values from (i) |
## Question 4:
**Part (i):**
| Answer | Marks | Guidance |
|--------|-------|----------|
| Probability distribution table: $x = -1, 1, 2, 3$ with $p = k, k, 4k, 9k$ | B1 | Probability distribution table with correct values of $x$, no additional values unless with probability 0 stated, at least one correct probability including $k$ |
| $15k = 1$ | M1 | Equating $\sum p = 1$, may be implied by answer |
| $k = \frac{1}{15}$ | A1 | If 0 scored, SCB2 for probability distribution table with correct numerical probabilities |
**Part (ii):**
| Answer | Marks | Guidance |
|--------|-------|----------|
| $E(X) = 8k + 27k = 35k = \frac{35}{15} = \frac{7}{3}$ | B1FT | FT if $0 <$ their $k < 1$ |
| $\text{Var}(X) = (k + k + 16k + 81k) - (35k)^2$ | M1 | Correct formula for variance, in terms of $k$ at least — must have '$-\text{mean}^2$' (ft) |
| $= 1.16,\ \frac{52}{45}$ | A1 | |
*Method 2:*
| Answer | Marks | Guidance |
|--------|-------|----------|
| $E(X) = \frac{8}{15} + \frac{27}{15} = \frac{35}{15} = \frac{7}{3}$ | B1FT | FT if $0 <$ their $k < 1$ |
| $\text{Var}(X) = \frac{1}{15} + \frac{1}{15} + \frac{16}{15} + \frac{81}{15} - \left(\frac{7}{3}\right)^2$ | M1 | Substitute their values in correct var formula — must have '$-\text{mean}^2$' (ft) |
| $= 1.16\ (= 52/45)$ | A1 | Using their values from (i) |
---4 The random variable $X$ takes the values $- 1,1,2,3$ only. The probability that $X$ takes the value $x$ is $k x ^ { 2 }$, where $k$ is a constant.\\
(i) Draw up the probability distribution table for $X$, in terms of $k$, and find the value of $k$.\\
(ii) Find $\mathrm { E } ( X )$ and $\operatorname { Var } ( X )$.\\
\hfill \mbox{\textit{CAIE S1 2019 Q4 [6]}}