| Exam Board | Edexcel |
|---|---|
| Module | S2 (Statistics 2) |
| Year | 2020 |
| Session | October |
| Marks | 11 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Central limit theorem |
| Type | Deriving sampling distribution |
| Difficulty | Standard +0.8 This question requires understanding of sampling distributions, systematic enumeration of all possible outcomes with their probabilities (10 cases for part b), and solving an inequality involving binomial probability P(Y=0) = (1-p)^n < 0.2. While conceptually accessible, the multi-step combinatorial reasoning and algebraic manipulation exceed typical S2 questions. |
| Spec | 2.04b Binomial distribution: as model B(n,p)5.01a Permutations and combinations: evaluate probabilities |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| A sampling distribution is all the values of a statistic and the associated probabilities / or the probability distribution of the statistic | B1 | Correct explanation with bold words; allow equivalent words e.g. outcomes for values |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(P(\text{small}(40)) = 0.5\), \(P(\text{medium}(80)) = 0.3\), \(P(\text{large}(150)) = 0.2\) | B1 | Correct probabilities; may be seen in equation or implied by correct probability for \(R=0\) or 2 correct probabilities from \(R=40, 70, 110\) |
| Range \(R\): 0, 40, 70, 110 | B1 | All four ranges correct with no extra |
| \(P(R=0) = 0.5^3 + 0.3^3 + 0.2^3 = 0.16\) | M1 | Correct method for finding \(P(R=0)\) |
| \((40,40,80)\), \((40,80,80)\), \((80,80,150)\), \((80,150,150)\), \((40,40,150)\), \((40,80,150)\), \((40,150,150)\) | B1 | All correct combinations for \(R=40,70,110\); \(R=0\) combinations not required but no incorrect combinations |
| \(P(R=40) = 3\times(0.5^2\times0.3^2) + 3\times(0.5^2\times0.3)\) | M1 | Correct method for one of \(P(R=40)\), \(P(R=70)\), \(P(R=110)\) |
| \(P(R=70) = 3\times(0.3^2\times0.2) + 3\times(0.3\times0.2^2) = 0.09\) | M1 | Correct method for a second probability, or the 4 probabilities add to 1 |
| \(P(R=110) = 3\times(0.5^2\times0.2) + 3\times(0.5\times0.2^2) + 6\times(0.5\times0.3\times0.2) = 0.39\) | — | |
| \(R\): 0, 40, 70, 110 with \(r\): 0.16, 0.36, 0.09, 0.39 | A1cao | Correct answer only; allow as fractions; probabilities must be attached to correct range |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \((1 - \text{"0.09"})^n < 0.2\) or \((\text{"0.91"})^n < 0.2\) | M1 | Setting up correct inequality using their 0.09; allow written as equation |
| \(n > 17.065...\) using \(\frac{\log 0.2}{\log\text{"0.91"}}\) | M1 | For 17.1 or better; allow \(\log_{0.91} 0.2\) oe; if inequality/equation incorrect but of form \((p)^n < 0.2\) where \(0 < p < 1\), mark can be awarded if working shown |
| \(n = 18\) | A1 | Do not accept \(n > 18\) or \(n < 18\) as final answer |
# Question 6:
## Part (a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| A sampling distribution is **all** the **values** of a **statistic** and the associated **probabilities** / or the **probability distribution** of the **statistic** | B1 | Correct explanation with bold words; allow equivalent words e.g. outcomes for values |
## Part (b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $P(\text{small}(40)) = 0.5$, $P(\text{medium}(80)) = 0.3$, $P(\text{large}(150)) = 0.2$ | B1 | Correct probabilities; may be seen in equation or implied by correct probability for $R=0$ or 2 correct probabilities from $R=40, 70, 110$ |
| Range $R$: 0, 40, 70, 110 | B1 | All four ranges correct with no extra |
| $P(R=0) = 0.5^3 + 0.3^3 + 0.2^3 = 0.16$ | M1 | Correct method for finding $P(R=0)$ |
| $(40,40,80)$, $(40,80,80)$, $(80,80,150)$, $(80,150,150)$, $(40,40,150)$, $(40,80,150)$, $(40,150,150)$ | B1 | All correct combinations for $R=40,70,110$; $R=0$ combinations not required but no incorrect combinations |
| $P(R=40) = 3\times(0.5^2\times0.3^2) + 3\times(0.5^2\times0.3)$ | M1 | Correct method for one of $P(R=40)$, $P(R=70)$, $P(R=110)$ |
| $P(R=70) = 3\times(0.3^2\times0.2) + 3\times(0.3\times0.2^2) = 0.09$ | M1 | Correct method for a second probability, or the 4 probabilities add to 1 |
| $P(R=110) = 3\times(0.5^2\times0.2) + 3\times(0.5\times0.2^2) + 6\times(0.5\times0.3\times0.2) = 0.39$ | — | |
| $R$: 0, 40, 70, 110 with $r$: 0.16, 0.36, 0.09, 0.39 | A1cao | Correct answer only; allow as fractions; probabilities must be attached to correct range |
## Part (c):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $(1 - \text{"0.09"})^n < 0.2$ or $(\text{"0.91"})^n < 0.2$ | M1 | Setting up correct inequality using their 0.09; allow written as equation |
| $n > 17.065...$ using $\frac{\log 0.2}{\log\text{"0.91"}}$ | M1 | For 17.1 or better; allow $\log_{0.91} 0.2$ oe; if inequality/equation incorrect but of form $(p)^n < 0.2$ where $0 < p < 1$, mark can be awarded if working shown |
| $n = 18$ | A1 | Do not accept $n > 18$ or $n < 18$ as final answer |6. (a) Explain what you understand by the sampling distribution of a statistic.
A factory produces beads in bags for craft shops. A small bag contains 40 beads, a medium bag contains 80 beads and a large bag contains 150 beads. The factory produces small, medium and large bags in the ratio 5:3:2 respectively.
A random sample of 3 bags is taken from the factory.\\
(b) Find the sampling distribution for the range of the number of beads in the 3 bags in the sample.
A random sample of $n$ sets of 3 bags is taken. The random variable $Y$ represents the number of these $n$ sets of 3 bags that have a range of 70\\
(c) Calculate the minimum value of $n$ such that $\mathrm { P } ( Y = 0 ) < 0.2$\\
\hfill \mbox{\textit{Edexcel S2 2020 Q6 [11]}}