| Exam Board | CAIE |
|---|---|
| Module | S1 (Statistics 1) |
| Year | 2017 |
| Session | June |
| 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 only basic algebraic manipulation and standard formulas. Part (i) is trivial substitution showing (-2)²=(2)², part (ii) uses the fundamental property that probabilities sum to 1, and part (iii) applies the standard E(X) formula. No problem-solving insight needed, just routine application of S1 techniques. |
| Spec | 2.04a Discrete probability distributions5.02b Expectation and variance: discrete random variables |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(k(-2)^2\) is the same as \(k(2)^2 = 4k\) | B1 | Need to see \(-2^2 k\), \(2^2 k\) and \(4k\), algebraically correct expressions OE |
| Total: 1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(x\): \(-2\), \(-1\), \(2\), \(4\); Prob: \(4k\), \(k\), \(4k\), \(16k\) | B1 | \(-2, -1, 2, 4\) only seen in a table, together with at least one attempted probability involving \(k\) |
| \(4k + k + 4k + 16k = 1\) | M1 | Summing 4 probs equating to 1. Must all be positive (table not required) |
| \(k = \frac{1}{25}\ (0.04)\) | A1 | CWO |
| Total: 3 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(E(X) = -8k + -k + 8k + 64k = 63k\) | M1 | Using \(\Sigma px\) unsimplified. FT their \(k\) substituted before this stage, no inappropriate dividing |
| \(= \frac{63}{25}\ (2.52)\) | A1 | |
| Total: 2 |
## Question 3(i):
| Answer | Mark | Guidance |
|--------|------|----------|
| $k(-2)^2$ is the same as $k(2)^2 = 4k$ | B1 | Need to see $-2^2 k$, $2^2 k$ and $4k$, algebraically correct expressions OE |
| **Total: 1** | | |
---
## Question 3(ii):
| Answer | Mark | Guidance |
|--------|------|----------|
| $x$: $-2$, $-1$, $2$, $4$; Prob: $4k$, $k$, $4k$, $16k$ | B1 | $-2, -1, 2, 4$ only seen in a table, together with at least one attempted probability involving $k$ |
| $4k + k + 4k + 16k = 1$ | M1 | Summing 4 probs equating to 1. Must all be positive (table not required) |
| $k = \frac{1}{25}\ (0.04)$ | A1 | CWO |
| **Total: 3** | | |
---
## Question 3(iii):
| Answer | Mark | Guidance |
|--------|------|----------|
| $E(X) = -8k + -k + 8k + 64k = 63k$ | M1 | Using $\Sigma px$ unsimplified. FT their $k$ substituted before this stage, no inappropriate dividing |
| $= \frac{63}{25}\ (2.52)$ | A1 | |
| **Total: 2** | | |
---3 In a probability distribution the random variable $X$ takes the value $x$ with probability $k x ^ { 2 }$, where $k$ is a constant and $x$ takes values $- 2 , - 1,2,4$ only.\\
(i) Show that $\mathrm { P } ( X = - 2 )$ has the same value as $\mathrm { P } ( X = 2 )$.\\
(ii) Draw up the probability distribution table for $X$, in terms of $k$, and find the value of $k$.\\
(iii) Find $\mathrm { E } ( X )$.\\
\hfill \mbox{\textit{CAIE S1 2017 Q3 [6]}}