| Exam Board | Edexcel |
|---|---|
| Module | S2 (Statistics 2) |
| Year | 2022 |
| Session | January |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Central limit theorem |
| Type | Sampling distribution theory |
| Difficulty | Standard +0.3 Part (a) is pure recall of a definition. Parts (b) and (c) require constructing a sampling distribution from given probabilities and calculating its expectation—standard S2 material involving systematic enumeration of outcomes and probability calculations, but straightforward with no conceptual surprises or complex reasoning required. |
| Spec | 2.01a Population and sample: terminology5.04a Linear combinations: E(aX+bY), Var(aX+bY)5.04b Linear combinations: of normal distributions |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| A sampling distribution is all the values of a statistic (obtained from a random sample) and the associated probabilities, or the probability distribution of the statistic (under random sampling) | B1 | A correct explanation with the words in bold |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(P(6) = \dfrac{6}{11}\), \(P(7) = \dfrac{3}{11}\), \(P(8) = \dfrac{2}{11}\) | B1 | Correct probabilities – may be seen in equation or implied by correct probability for \(T=14\) |
| Totals \((T)\): \(12, 13, 14, 15, 16\) | B1 | All 5 totals correct with no extras |
| \((6,6),(6,7),(6,8),(7,6),(7,7),(7,8),(8,6),(8,7),(8,8)\) | B1 | All 6 basic combinations correct |
| \([P(T=12)] = \left(\dfrac{6}{11}\right)^2 = \left[\dfrac{36}{121}\right]\) | M1 | Correct method for one probability ft their \(P(6), P(7), P(8)\) |
| \([P(T=13)] = 2 \times \dfrac{6}{11} \times \dfrac{3}{11} = \left[\dfrac{36}{121}\right]\) | M1 | Correct method for three of the five probabilities |
| \([P(T=14)] = 2 \times \dfrac{6}{11} \times \dfrac{2}{11} + \left(\dfrac{3}{11}\right)^2 = \left[\dfrac{33}{121}\right]\) | M1 | Correct method for all five probabilities |
| \([P(T=15)] = 2 \times \dfrac{3}{11} \times \dfrac{2}{11} = \left[\dfrac{12}{121}\right]\) | ||
| \([P(T=16)] = \left(\dfrac{2}{11}\right)^2 = \left[\dfrac{4}{121}\right]\) | ||
| Full table: \(T\): \(12, 13, 14, 15, 16\); \(P(T=t)\): \(\dfrac{36}{121}, \dfrac{36}{121}, \dfrac{33}{121}, \dfrac{12}{121}, \dfrac{4}{121}\) | A1 | cao. Need not be in a table but probabilities must be attached to correct total |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(E(T) = 12 \times \dfrac{36}{121} + 13 \times \dfrac{36}{121} + 14 \times \dfrac{33}{121} + 15 \times \dfrac{12}{121} + 16 \times \dfrac{4}{121}\) | M1 | Use of \(\sum t P(T=t)\), two or more products ft their table |
| \(= \dfrac{1606}{121} = \dfrac{146}{11} = 13.272\ldots\) | A1 | awrt \(13.3\) (Allow \(\dfrac{146}{11}\) oe) |
# Question 6(a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| A sampling distribution is all the **values** of a **statistic** (obtained from a random sample) and the associated **probabilities**, or the **probability distribution** of the **statistic** (under random sampling) | B1 | A correct explanation with the words in bold |
# Question 6(b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $P(6) = \dfrac{6}{11}$, $P(7) = \dfrac{3}{11}$, $P(8) = \dfrac{2}{11}$ | B1 | Correct probabilities – may be seen in equation or implied by correct probability for $T=14$ |
| Totals $(T)$: $12, 13, 14, 15, 16$ | B1 | All 5 totals correct with no extras |
| $(6,6),(6,7),(6,8),(7,6),(7,7),(7,8),(8,6),(8,7),(8,8)$ | B1 | All 6 basic combinations correct |
| $[P(T=12)] = \left(\dfrac{6}{11}\right)^2 = \left[\dfrac{36}{121}\right]$ | M1 | Correct method for one probability ft their $P(6), P(7), P(8)$ |
| $[P(T=13)] = 2 \times \dfrac{6}{11} \times \dfrac{3}{11} = \left[\dfrac{36}{121}\right]$ | M1 | Correct method for three of the five probabilities |
| $[P(T=14)] = 2 \times \dfrac{6}{11} \times \dfrac{2}{11} + \left(\dfrac{3}{11}\right)^2 = \left[\dfrac{33}{121}\right]$ | M1 | Correct method for all five probabilities |
| $[P(T=15)] = 2 \times \dfrac{3}{11} \times \dfrac{2}{11} = \left[\dfrac{12}{121}\right]$ | | |
| $[P(T=16)] = \left(\dfrac{2}{11}\right)^2 = \left[\dfrac{4}{121}\right]$ | | |
| Full table: $T$: $12, 13, 14, 15, 16$; $P(T=t)$: $\dfrac{36}{121}, \dfrac{36}{121}, \dfrac{33}{121}, \dfrac{12}{121}, \dfrac{4}{121}$ | A1 | cao. Need not be in a table but probabilities must be attached to correct total |
# Question 6(c):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $E(T) = 12 \times \dfrac{36}{121} + 13 \times \dfrac{36}{121} + 14 \times \dfrac{33}{121} + 15 \times \dfrac{12}{121} + 16 \times \dfrac{4}{121}$ | M1 | Use of $\sum t P(T=t)$, two or more products ft their table |
| $= \dfrac{1606}{121} = \dfrac{146}{11} = 13.272\ldots$ | A1 | awrt $13.3$ (Allow $\dfrac{146}{11}$ oe) |
---6
\begin{enumerate}[label=(\alph*)]
\item Explain what you understand by the sampling distribution of a statistic.
At Sam's cafe a standard breakfast consists of 6 breakfast items. Customers can then choose to upgrade to a medium breakfast by adding 1 extra breakfast item or they can upgrade to a large breakfast by adding 2 extra breakfast items. Standard, medium and large breakfasts are sold in the ratio $6 : 3 : 2$ respectively.
A random sample of 2 customers is taken from customers who have bought a breakfast from Sam's cafe on a particular day.
\item Find the sampling distribution for the total number, $T$, of breakfast items bought by these 2 customers. Show your working clearly.
\item Find $\mathrm { E } ( T )$
\end{enumerate}
\hfill \mbox{\textit{Edexcel S2 2022 Q6 [10]}}