| Exam Board | Edexcel |
|---|---|
| Module | S1 (Statistics 1) |
| Year | 2019 |
| Session | January |
| Marks | 15 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Binomial Distribution |
| Type | Finding binomial parameters from properties |
| Difficulty | Moderate -0.3 This is a straightforward S1 question testing basic binomial distribution calculations and comparison with data. Parts (a)-(c) involve routine mean calculation and standard binomial probability formulas with n=3, p=0.4. Parts (d)-(e) require calculating probabilities with varying p values across trials, which is slightly more involved but still mechanical multiplication of probabilities. The question is slightly easier than average because it's highly structured with 'show that' parts providing answers to check against, and requires only standard techniques without problem-solving insight. |
| Spec | 2.02f Measures of average and spread2.04a Discrete probability distributions2.04b Binomial distribution: as model B(n,p)2.04c Calculate binomial probabilities |
| \(x\) | 0 | 1 | 2 | 3 |
| Frequency | 16 | 36 | 24 | 4 |
| \(s\) | 0 | 1 | 2 | 3 |
| \(\mathrm { P } ( S = s )\) | 0.432 | 0.064 |
| \(t\) | 0 | 1 | 2 | 3 |
| \(\mathrm { P } ( T = t )\) | 0.45 | 0.055 |
| Answer | Marks | Guidance |
|---|---|---|
| \(\left[\bar{x} = \dfrac{96}{80} =\right] \mathbf{1.2}\) | B1 | For 1.2 or any exact equivalent |
| (1) |
| Answer | Marks | Guidance |
|---|---|---|
| \(P(S = 2) = 3 \times 0.4^2 \times 0.6\) | M1 | For \(k \times 0.4^2 \times 0.6\) (including \(k=1\)) where \(k \in \mathbb{Z}^+\) |
| \(= \mathbf{0.288}\) | A1 | For 0.288 or exact equivalent e.g. \(\frac{36}{125}\) |
| (2) |
| Answer | Marks | Guidance |
|---|---|---|
| \(P(S = 0) = 1 - (0.496 +\) "0.288") or \(0.6^3 = \mathbf{0.216}\) | B1ft | Based on sum of probabilities \(= 1\) |
| (1) |
| Answer | Marks | Guidance |
|---|---|---|
| \(p_1 = 0.25\) and \(p_2 = 0.4\) and \(p_3 = 0.55\) | M1 | For correct attempt to find all 3 \(p_i\) values |
| \(P(T = 3) = p_1 \times p_2 \times p_3 = 0.25 \times 0.4 \times 0.55 = \mathbf{0.055}\) | A1cso | For correct numerical product, no incorrect working seen |
| Answer | Marks |
|---|---|
| \(P(T=1) = 0.25 \times 0.6 \times 0.45 + 0.75 \times 0.4 \times 0.45 + 0.75 \times 0.6 \times 0.55\) | M1A1 |
| \(= 0.0675 + 0.135 + 0.2475\) | |
| \(= \mathbf{0.45}\) | A1 cso |
| (5) |
| Answer | Marks |
|---|---|
| \(P(T=0) = 0.75 \times 0.6 \times 0.45\) (or equivalent for \(P(T=2)\)) | M1 |
| \(= \mathbf{0.2025}\) (Allow \(\frac{81}{400}\)) | A1 |
| \(P(T=2) = 1 - (0.505 +\) "0.2025"\() = \mathbf{0.2925}\) (Allow \(\frac{117}{400}\)) | A1ft |
| (3) |
| Answer | Marks | Guidance |
|---|---|---|
| \(x\) | 0 | 1 |
| \(f/80\) | 0.2 | 0.45 |
| \(P(S=x)\) | 0.216 | 0.432 |
| \(P(T=x)\) | 0.2025 | 0.45 |
| \(f\) | 16 | 36 |
| \(S(f)\) | 17.28 | 34.56 |
| \(T(f)\) | 16.2 | 36 |
| Ting's model is always closer — so choose Ting's model | M1 | For attempt at calculating probs from data or freqs \(S(f)\) and \(T(f)\) (at least 3 correct) |
| A1 | For all figures correct (2sf) and comparison of probs (or frequencies) for the 2 models | |
| A1 | For clearly choosing Ting's model | |
| (3) [15 marks] |
# Question 5:
## Part (a):
$\left[\bar{x} = \dfrac{96}{80} =\right] \mathbf{1.2}$ | B1 | For 1.2 or any exact equivalent
| | (1)
## Part (b):
$P(S = 2) = 3 \times 0.4^2 \times 0.6$ | M1 | For $k \times 0.4^2 \times 0.6$ (including $k=1$) where $k \in \mathbb{Z}^+$
$= \mathbf{0.288}$ | A1 | For 0.288 or exact equivalent e.g. $\frac{36}{125}$
| | (2)
## Part (c):
$P(S = 0) = 1 - (0.496 +$ "0.288") or $0.6^3 = \mathbf{0.216}$ | B1ft | Based on sum of probabilities $= 1$
| | (1)
## Part (d)(i):
$p_1 = 0.25$ and $p_2 = 0.4$ and $p_3 = 0.55$ | M1 | For correct attempt to find all 3 $p_i$ values
$P(T = 3) = p_1 \times p_2 \times p_3 = 0.25 \times 0.4 \times 0.55 = \mathbf{0.055}$ | A1cso | For correct numerical product, no incorrect working seen
## Part (d)(ii):
$P(T=1) = 0.25 \times 0.6 \times 0.45 + 0.75 \times 0.4 \times 0.45 + 0.75 \times 0.6 \times 0.55$ | M1A1 |
$= 0.0675 + 0.135 + 0.2475$ | |
$= \mathbf{0.45}$ | A1 cso |
| | (5)
## Part (e):
$P(T=0) = 0.75 \times 0.6 \times 0.45$ (or equivalent for $P(T=2)$) | M1 |
$= \mathbf{0.2025}$ (Allow $\frac{81}{400}$) | A1 |
$P(T=2) = 1 - (0.505 +$ "0.2025"$) = \mathbf{0.2925}$ (Allow $\frac{117}{400}$) | A1ft |
| | (3)
## Part (f):
Estimate probs from the data:
| $x$ | 0 | 1 | 2 | 3 |
|---|---|---|---|---|
| $f/80$ | **0.2** | **0.45** | **0.3** | **0.05** |
| $P(S=x)$ | 0.216 | 0.432 | 0.288 | 0.064 |
| $P(T=x)$ | 0.2025 | 0.45 | 0.2925 | 0.055 |
| $f$ | 16 | 36 | 24 | 4 |
| $S(f)$ | 17.28 | 34.56 | 23.04 | 5.12 |
| $T(f)$ | 16.2 | 36 | 23.4 | 4.4 |
Ting's model is always closer — so **choose Ting's** model | M1 | For attempt at calculating probs from data or freqs $S(f)$ **and** $T(f)$ (at least 3 correct)
| A1 | For all figures correct (2sf) and comparison of probs (or frequencies) for the 2 models
| A1 | For clearly choosing Ting's model
| | (3) **[15 marks]**\begin{enumerate}
\item Some children are playing a game involving throwing a ball into a bucket. Each child has 3 throws and the number of times the ball lands in the bucket, $x$, is recorded. Their results are given in the table below.
\end{enumerate}
\begin{center}
\begin{tabular}{ | c | c | c | c | c | }
\hline
$x$ & 0 & 1 & 2 & 3 \\
\hline
Frequency & 16 & 36 & 24 & 4 \\
\hline
\end{tabular}
\end{center}
(a) Find $\bar { x }$\\
(1)
Sandra decides to model the game by assuming that on each throw, the probability of the ball landing in the bucket is 0.4 for every child on every throw and that the throws are all independent. The random variable $S$ represents the number of times the ball lands in the bucket for a randomly selected child.\\
(b) Find $\mathrm { P } ( S = 2 )$\\
(c) Complete the table below to show the probability distribution for $S$.
\begin{center}
\begin{tabular}{ | c | c | c | c | c | }
\hline
$s$ & 0 & 1 & 2 & 3 \\
\hline
$\mathrm { P } ( S = s )$ & & 0.432 & & 0.064 \\
\hline
\end{tabular}
\end{center}
Ting believes that the probability of the ball landing in the bucket is not the same for each throw. He suggests that the probability will increase with each throw and uses the model
$$p _ { i } = 0.15 i + 0.10$$
where $i = 1,2,3$ and $p _ { i }$ is the probability that the $i$ th throw of the ball, by any particular child, will land in the bucket.\\
The random variable $T$ represents the number of times the ball lands in the bucket for a randomly selected child using Ting's model.\\
(d) Show that\\
(i) $\mathrm { P } ( T = 3 ) = 0.055$\\
(ii) $\mathrm { P } ( T = 1 ) = 0.45$\\
(5)\\
(e) Complete the table below to show the probability distribution for $T$, stating the exact probabilities in each case.
\begin{center}
\begin{tabular}{ | c | c | c | c | c | }
\hline
$t$ & 0 & 1 & 2 & 3 \\
\hline
$\mathrm { P } ( T = t )$ & & 0.45 & & 0.055 \\
\hline
\end{tabular}
\end{center}
(f) State, giving your reasons, whether Sandra's model or Ting's model is the more appropriate for modelling this game.
\hfill \mbox{\textit{Edexcel S1 2019 Q5 [15]}}