| Exam Board | OCR MEI |
|---|---|
| Module | S4 (Statistics 4) |
| Year | 2012 |
| Session | June |
| Marks | 24 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Probability Generating Functions |
| Type | Find PGF from probability distribution |
| Difficulty | Standard +0.3 This is a structured, multi-part question on PGFs with clear guidance at each step. Parts (i)-(v) involve standard calculations (mean/variance of discrete uniform, geometric distribution PGF derivation). Part (vi) uses a given result K(t)=H(G(t)) with an algebraic hint provided. While PGFs are a Further Maths topic, the question requires mainly routine manipulation rather than novel insight, making it slightly easier than average. |
| Spec | 2.04b Binomial distribution: as model B(n,p)2.04c Calculate binomial probabilities5.02a Discrete probability distributions: general5.02b Expectation and variance: discrete random variables5.02g Geometric probabilities: P(X=r) = p(1-p)^(r-1)5.02h Geometric: mean 1/p and variance (1-p)/p^2 |
| Answer | Marks | Guidance |
|---|---|---|
| Mean of \(X = 3.5\) (immediately by symmetry) \(E(X^2) = \frac{1}{6}(1 + 4 + \ldots + 36) = \frac{91}{6}\) \(\therefore \text{Var}(X) = \frac{91}{6} - \left(\frac{7}{2}\right)^2 = \frac{35}{12}\) | B1, M1, A1 | Answer given. |
| Answer | Marks | Guidance |
|---|---|---|
| \(G(t) = E(t^X) = \left(t \cdot \frac{1}{6}\right) + \left(t^2 \cdot \frac{1}{6}\right) + \ldots + \left(t^6 \cdot \frac{1}{6}\right) = \frac{1}{6}(t + t^2 + \ldots + t^6) = \frac{t(1-t^6)}{6(1-t)}\) | M1, A1 | Answer given. |
| Answer | Marks | Guidance |
|---|---|---|
| \(P(N = 0) = \frac{1}{2}\), \(P(N = 1) = (\frac{1}{2})(\frac{1}{2})\), \(P(N = r) = (\frac{1}{2})^r \cdot (\frac{1}{2})\) | B1 [1] | Answer given; must be convincing. |
| Answer | Marks | Guidance |
|---|---|---|
| \(H(t) = E(t^N) = \left(t^0 \cdot \frac{1}{2}\right) + \left(t^1 \cdot \frac{1}{2}\right) + \left(t^2 \cdot \frac{1}{2}\right) + \ldots = \frac{1}{1 - \frac{t}{2}} = \frac{1}{2-t} = (2-t)^{-1}\) | M1, A1 | Answer given. |
| Answer | Marks | Guidance |
|---|---|---|
| Mean \(= H'(1)\), variance \(= H''(1) + \text{mean} - \text{mean}^2\) \(H'(t) = (-1)(2-t)^{-2}(-1) = (2-t)^{-2}\) \(\therefore\) mean \(= 1\) \(H''(t) = (-2)(2-t)^{-3}(-1) = 2(2-t)^{-3}\) \(\therefore\) variance \(= 2 + 1 - 1 = 2\) | M1, A1, M1, A1 | For use of 1st derivative. For use of 2nd derivative. For variance. |
| Answer | Marks | Guidance |
|---|---|---|
| \(K(t) = H(G(t)) = [2 - G(t)]^{-1}\) \(= \left[2 - \frac{t(1-t^6)}{6(1-t)}\right]^{-1} = \left[\frac{12(1-t) - t(1-t)(1 + t + t^2 + \ldots + t^5)}{6(1-t)}\right]^{-1} = \left[\frac{12 - t - t^2 - t^3 - \ldots - t^6}{6}\right]^{-1} = 6(12 - t - t^2 - \ldots - t^6)^{-1}\) | M1, M1, M1, A1 | Inserting \(G(t)\). Use of hint given. Answer given. |
| Answer | Marks | Guidance |
|---|---|---|
| \(K'(t) = 6(12 - t - t^2 - \ldots - t^6)^{-2}(1 + 2t + 3t^2 + 4t^3 + 5t^4 + 6t^5)\) \(K''(t) = 12(12 - t - t^2 - \ldots - t^6)^{-3}(1 + 2t + 3t^2 + 4t^3 + 5t^4 + 6t^5)^2 + 6(12 - t - t^2 - \ldots - t^6)^{-2}(2 + 6t + 12t^2 + 20t^3 + 30t^4)\) \(\therefore \text{mean} = K'(1) = 6(12 - 6)^{-2}(21) = 21/6 = 7/2\) \(\therefore K''(1) = (12 \times 6^{-3} \times 21^2) + (6 \times 6^{-2} \times 70) = (49/2) + (70/6)\) \(\therefore \text{variance} = \frac{49}{2} + \frac{70}{6} + \frac{7}{2} - \frac{49}{4} = \frac{294 + 140 + 42 - 147}{12} = \frac{329}{12}\) | M1, M1, M1, A1, A1, A1 | Reasonable attempt to differentiate \(K(t)\). Reasonable attempt at 2nd derivative. For use of derivatives. Substitution shown. Ft c's \(K'(1)\) and/or \(K''(1)\) provided variance positive. Exact. |
| Answer | Marks | Guidance |
|---|---|---|
| We have: \(\mu_X = 7/2\), \(\sigma_X^2 = 35/12\), \(\mu_N = 1\), \(\sigma_N^2 = 2\), \(\sigma_O^2 = 329/12\) Inserting in the quoted formula gives \(2 \times \left[\left(\frac{7}{2}\right)^2\right] + \left[1 \times \frac{35}{12}\right] = \frac{294 + 35}{12} = \frac{329}{12}\) as required. | M1, A1 | For correct use of candidate's values for means and variances. Answer honestly obtained (common denominator shown). A0 if different from (vii). |
**Part (i)**
| Mean of $X = 3.5$ (immediately by symmetry) $E(X^2) = \frac{1}{6}(1 + 4 + \ldots + 36) = \frac{91}{6}$ $\therefore \text{Var}(X) = \frac{91}{6} - \left(\frac{7}{2}\right)^2 = \frac{35}{12}$ | B1, M1, A1 | Answer given. |
**Part (ii)**
| $G(t) = E(t^X) = \left(t \cdot \frac{1}{6}\right) + \left(t^2 \cdot \frac{1}{6}\right) + \ldots + \left(t^6 \cdot \frac{1}{6}\right) = \frac{1}{6}(t + t^2 + \ldots + t^6) = \frac{t(1-t^6)}{6(1-t)}$ | M1, A1 | Answer given. |
**Part (iii)**
| $P(N = 0) = \frac{1}{2}$, $P(N = 1) = (\frac{1}{2})(\frac{1}{2})$, $P(N = r) = (\frac{1}{2})^r \cdot (\frac{1}{2})$ | B1 [1] | Answer given; must be convincing. |
**Part (iv)**
| $H(t) = E(t^N) = \left(t^0 \cdot \frac{1}{2}\right) + \left(t^1 \cdot \frac{1}{2}\right) + \left(t^2 \cdot \frac{1}{2}\right) + \ldots = \frac{1}{1 - \frac{t}{2}} = \frac{1}{2-t} = (2-t)^{-1}$ | M1, A1 | Answer given. |
**Part (v)**
| Mean $= H'(1)$, variance $= H''(1) + \text{mean} - \text{mean}^2$ $H'(t) = (-1)(2-t)^{-2}(-1) = (2-t)^{-2}$ $\therefore$ mean $= 1$ $H''(t) = (-2)(2-t)^{-3}(-1) = 2(2-t)^{-3}$ $\therefore$ variance $= 2 + 1 - 1 = 2$ | M1, A1, M1, A1 | For use of 1st derivative. For use of 2nd derivative. For variance. |
**Part (vi)**
| $K(t) = H(G(t)) = [2 - G(t)]^{-1}$ $= \left[2 - \frac{t(1-t^6)}{6(1-t)}\right]^{-1} = \left[\frac{12(1-t) - t(1-t)(1 + t + t^2 + \ldots + t^5)}{6(1-t)}\right]^{-1} = \left[\frac{12 - t - t^2 - t^3 - \ldots - t^6}{6}\right]^{-1} = 6(12 - t - t^2 - \ldots - t^6)^{-1}$ | M1, M1, M1, A1 | Inserting $G(t)$. Use of hint given. Answer given. |
**Part (vii)**
| $K'(t) = 6(12 - t - t^2 - \ldots - t^6)^{-2}(1 + 2t + 3t^2 + 4t^3 + 5t^4 + 6t^5)$ $K''(t) = 12(12 - t - t^2 - \ldots - t^6)^{-3}(1 + 2t + 3t^2 + 4t^3 + 5t^4 + 6t^5)^2 + 6(12 - t - t^2 - \ldots - t^6)^{-2}(2 + 6t + 12t^2 + 20t^3 + 30t^4)$ $\therefore \text{mean} = K'(1) = 6(12 - 6)^{-2}(21) = 21/6 = 7/2$ $\therefore K''(1) = (12 \times 6^{-3} \times 21^2) + (6 \times 6^{-2} \times 70) = (49/2) + (70/6)$ $\therefore \text{variance} = \frac{49}{2} + \frac{70}{6} + \frac{7}{2} - \frac{49}{4} = \frac{294 + 140 + 42 - 147}{12} = \frac{329}{12}$ | M1, M1, M1, A1, A1, A1 | Reasonable attempt to differentiate $K(t)$. Reasonable attempt at 2nd derivative. For use of derivatives. Substitution shown. Ft c's $K'(1)$ and/or $K''(1)$ provided variance positive. Exact. |
**Part (viii)**
| We have: $\mu_X = 7/2$, $\sigma_X^2 = 35/12$, $\mu_N = 1$, $\sigma_N^2 = 2$, $\sigma_O^2 = 329/12$ Inserting in the quoted formula gives $2 \times \left[\left(\frac{7}{2}\right)^2\right] + \left[1 \times \frac{35}{12}\right] = \frac{294 + 35}{12} = \frac{329}{12}$ as required. | M1, A1 | For correct use of candidate's values for means and variances. Answer honestly obtained (common denominator shown). A0 if different from (vii). |
---2 The random variable $X ( X = 1,2,3,4,5,6 )$ denotes the score when a fair six-sided die is rolled.\\
(i) Write down the mean of $X$ and show that $\operatorname { Var } ( X ) = \frac { 35 } { 12 }$.\\
(ii) Show that $\mathrm { G } ( t )$, the probability generating function (pgf) of $X$, is given by
$$\mathrm { G } ( t ) = \frac { t \left( 1 - t ^ { 6 } \right) } { 6 ( 1 - t ) }$$
The random variable $N ( N = 0,1,2 , \ldots )$ denotes the number of heads obtained when an unbiased coin is tossed repeatedly until a tail is first obtained.\\
(iii) Show that $\mathrm { P } ( N = r ) = \left( \frac { 1 } { 2 } \right) ^ { r + 1 }$ for $r = 0,1,2 , \ldots$.\\
(iv) Hence show that $\mathrm { H } ( t )$, the pgf of $N$, is given by $\mathrm { H } ( t ) = ( 2 - t ) ^ { - 1 }$.\\
(v) Use $\mathrm { H } ( t )$ to find the mean and variance of $N$.
A game consists of tossing an unbiased coin repeatedly until a tail is first obtained and, each time a head is obtained in this sequence of tosses, rolling a fair six-sided die. The die is not rolled on the first occasion that a tail is obtained and the game ends at that point. The random variable $Q ( Q = 0,1,2 , \ldots )$ denotes the total score on all the rolls of the die. Thus, in the notation above, $Q = X _ { 1 } + X _ { 2 } + \ldots + X _ { N }$ where the $X _ { i }$ are independent random variables each distributed as $X$, with $Q = 0$ if $N = 0$. The pgf of $Q$ is denoted by $\mathrm { K } ( t )$. The familiar result that the pgf of a sum of independent random variables is the product of their pgfs does not apply to $\mathrm { K } ( t )$ because $N$ is a random variable and not a fixed number; you should instead use without proof the result that $\mathrm { K } ( t ) = \mathrm { H } ( \mathrm { G } ( t ) )$.\\
(vi) Show that $\mathrm { K } ( t ) = 6 \left( 12 - t - t ^ { 2 } - \ldots - t ^ { 6 } \right) ^ { - 1 }$.\\[0pt]
[Hint. $\left. \left( 1 - t ^ { 6 } \right) = ( 1 - t ) \left( 1 + t + t ^ { 2 } + \ldots + t ^ { 5 } \right) .\right]$\\
(vii) Use $\mathrm { K } ( t )$ to find the mean and variance of $Q$.\\
(viii) Using your results from parts (i), (v) and (vii), verify the result that (in the usual notation for means and variances)
$$\sigma _ { Q } { } ^ { 2 } = \sigma _ { N } { } ^ { 2 } \mu _ { X } { } ^ { 2 } + \mu _ { N } \sigma _ { X } { } ^ { 2 } .$$
\hfill \mbox{\textit{OCR MEI S4 2012 Q2 [24]}}