| Exam Board | OCR |
|---|---|
| Module | S4 (Statistics 4) |
| Year | 2018 |
| Session | June |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Chi-squared goodness of fit |
| Type | Chi-squared distribution theory and properties |
| Difficulty | Standard +0.8 This is a Further Maths S4 question requiring manipulation of moment generating functions to derive mean and variance, then applying MGF properties for sums of independent variables. While the techniques are standard for this level (differentiating MGFs, using independence property), it requires solid understanding of MGF theory and multiple careful algebraic steps, placing it moderately above average difficulty. |
| Spec | 5.02a Discrete probability distributions: general5.02b Expectation and variance: discrete random variables |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(M'_X(t) = (-\frac{1}{2}\nu)(-2)(1-2t)^{-\frac{1}{2}\nu - 1}\) | M1A1 | \(1+t\nu\) B1; \(E(X) =\) coeff of \(t{=}\nu\) M1A1 |
| \(M'_X(0) = \nu\) | B1, [3] | CWO |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(M''_X(t) = (\nu)(-2)(-\frac{1}{2}\nu - 1)(1-2t)^{-\frac{1}{2}\nu - 2}\) | B1 | \(3^{\text{rd}}\) term \((\frac{1}{2}\nu^2 + \nu)t^2\) B1 may be unsimplified |
| \(M''_X(0) = \nu^2 + 2\nu\), \(\text{Var}(X) = \nu^2 + 2\nu - \nu^2 = 2\nu\) | M1A1, [3] | \(E(X^2) = \nu^2 + 2\nu\) B1ft; \(\text{Var}(X) = 2\nu\) B1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \((1-2t)^{-3} \times (1-2t)^{-4} = (1-2t)^{-7}\) | M1A1, [2] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\chi^2\), \(\nu = 14\) | B1B1, [2] |
## Question 4(i):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $M'_X(t) = (-\frac{1}{2}\nu)(-2)(1-2t)^{-\frac{1}{2}\nu - 1}$ | M1A1 | $1+t\nu$ B1; $E(X) =$ coeff of $t{=}\nu$ M1A1 |
| $M'_X(0) = \nu$ | B1, [3] | CWO |
## Question 4(ii):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $M''_X(t) = (\nu)(-2)(-\frac{1}{2}\nu - 1)(1-2t)^{-\frac{1}{2}\nu - 2}$ | B1 | $3^{\text{rd}}$ term $(\frac{1}{2}\nu^2 + \nu)t^2$ B1 may be unsimplified |
| $M''_X(0) = \nu^2 + 2\nu$, $\text{Var}(X) = \nu^2 + 2\nu - \nu^2 = 2\nu$ | M1A1, [3] | $E(X^2) = \nu^2 + 2\nu$ B1ft; $\text{Var}(X) = 2\nu$ B1 |
## Question 4(iii):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $(1-2t)^{-3} \times (1-2t)^{-4} = (1-2t)^{-7}$ | M1A1, [2] | |
## Question 4(iv):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\chi^2$, $\nu = 14$ | B1B1, [2] | |
---4 The random variable $X$ has a $\chi ^ { 2 }$ distribution with $v$ degrees of freedom. The moment generating function of $X$ is
$$\mathrm { M } _ { X } ( t ) = ( 1 - 2 t ) ^ { - \frac { 1 } { 2 } v }$$
(i) Show that $\mathrm { E } ( X ) = v$.\\
(ii) Find $\operatorname { Var } ( X )$.\\
(iii) Obtain the moment generating function of the sum $Y$ of two independent $\chi ^ { 2 }$ random variables, one with 6 degrees of freedom and the other with 8 degrees of freedom.\\
(iv) Identify the distribution of $Y$.
\hfill \mbox{\textit{OCR S4 2018 Q4 [10]}}