| Exam Board | Edexcel |
|---|---|
| Module | S2 (Statistics 2) |
| Year | 2014 |
| Session | June |
| Marks | 14 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Binomial Distribution |
| Type | Direct binomial probability calculation |
| Difficulty | Standard +0.3 This is a straightforward S2 binomial distribution question requiring standard calculations: finding p from a formula, computing P(X=6) and P(X≥8) using binomial probability, solving for x using P(X≥1)=0.8, and applying normal approximation with continuity correction. All techniques are routine for S2 with no novel problem-solving required, making it slightly easier than average. |
| Spec | 5.02b Expectation and variance: discrete random variables5.02c Linear coding: effects on mean and variance5.04b Linear combinations: of normal distributions |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(X\) is the number of successes, \(X \sim B(10, 0.75)\) | B1 | Writing or using \(p=0.75\) or \(p=0.25\) anywhere in (a)(i) or (a)(ii) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(P(X=6) = (0.75)^6(0.25)^4\,^{10}C_6\) or \(P(X \leq 6)-P(X \leq 5)\) | M1 | Writing or using \((p)^6(1-p)^4\,^{10}C_6\), or for \(p=0.25\): \(P(X \leq 4)-P(X \leq 3)\), or correct answer |
| \(= 0.145998\) \(\quad\) awrt \(0.146\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| Using \(X \sim B(10, 0.75)\): \(P(X \geq 8) = P(X=8)+P(X=9)+P(X=10)\) | M1 | Writing \(B(10,0.75)\) and writing or using \(P(X=8)+P(X=9)+P(X=10)\) oe, or writing \(B(10,0.25)\) and writing or using \(P(Y \leq 2)\). Using correct Binomial must be shown by \((0.75)^n(0.25)^{10-n}\) or correct answer |
| \(=(0.75)^8(0.25)^2\,^{10}C_8+(0.75)^9(0.25)^1\,^{10}C_9+(0.75)^{10}\) | ||
| \(= 0.52559\) \(\quad\) awrt \(0.526\) | A1 | |
| Or: Using \(Y \sim B(10, 0.25)\) and \(P(Y \leq 2) = 0.5256\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(1-P(0) = 0.8\) or \(P(0) = 0.2\) | M1 | Writing or using \(1-P(0)=0.8\) or \(P(0)=0.2\) or \((1-p)^{20}=0.2\). Allow any inequality sign |
| \((1-p)^{20} = 0.2\) | ||
| \(1-p = 0.9227\) | ||
| \(p = 0.0773\) | A1 | awrt 0.0773 or awrt 0.923 |
| \(\frac{3}{200}(90-x) = 0.0773\) | M1 | Substituting \(\frac{3}{200}(90-x)\) for \(p\). Allow \(\frac{3}{200}(90-x)=k\) where \(0 < k < 1\), \(k \neq 0.8\) or \(0.2\). Allow any inequality sign |
| \(x = 84.84\) | ||
| \(x = 85\) | A1cao | Condone \(x \geq 85\). Do not allow \(x > 85\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(X\) – successes \(\sim B(100, 0.975)\); \(Y\) – not successes \(\sim B(100, 0.025)\) | B1 | Writing or using \(0.975\) or \(0.025\); may be implied by \(Po(2.5)\) |
| \(Y \sim Po(2.5)\) | M1A1 | M1: using Poisson approximation. A1: \(Po(2.5)\) |
| \(P(Y \leq 5) = 0.958\) | M1A1 | M1: writing or using \(P(Y \leq 5)\). A1: awrt 0.958. SC use of normal approximation: B1 M0 A0 M1 A0 |
# Question 4:
## Part (a)
| Answer/Working | Marks | Guidance |
|---|---|---|
| $X$ is the number of successes, $X \sim B(10, 0.75)$ | B1 | Writing or using $p=0.75$ or $p=0.25$ anywhere in (a)(i) or (a)(ii) |
### Part (a)(i)
| Answer/Working | Marks | Guidance |
|---|---|---|
| $P(X=6) = (0.75)^6(0.25)^4\,^{10}C_6$ or $P(X \leq 6)-P(X \leq 5)$ | M1 | Writing or using $(p)^6(1-p)^4\,^{10}C_6$, or for $p=0.25$: $P(X \leq 4)-P(X \leq 3)$, or correct answer |
| $= 0.145998$ $\quad$ awrt $0.146$ | A1 | |
### Part (a)(ii)
| Answer/Working | Marks | Guidance |
|---|---|---|
| Using $X \sim B(10, 0.75)$: $P(X \geq 8) = P(X=8)+P(X=9)+P(X=10)$ | M1 | Writing $B(10,0.75)$ and writing or using $P(X=8)+P(X=9)+P(X=10)$ oe, **or** writing $B(10,0.25)$ and writing or using $P(Y \leq 2)$. Using correct Binomial must be shown by $(0.75)^n(0.25)^{10-n}$ or correct answer |
| $=(0.75)^8(0.25)^2\,^{10}C_8+(0.75)^9(0.25)^1\,^{10}C_9+(0.75)^{10}$ | | |
| $= 0.52559$ $\quad$ awrt $0.526$ | A1 | |
| Or: Using $Y \sim B(10, 0.25)$ and $P(Y \leq 2) = 0.5256$ | | |
## Part (b)
| Answer/Working | Marks | Guidance |
|---|---|---|
| $1-P(0) = 0.8$ or $P(0) = 0.2$ | M1 | Writing or using $1-P(0)=0.8$ or $P(0)=0.2$ or $(1-p)^{20}=0.2$. Allow any inequality sign |
| $(1-p)^{20} = 0.2$ | | |
| $1-p = 0.9227$ | | |
| $p = 0.0773$ | A1 | awrt 0.0773 or awrt 0.923 |
| $\frac{3}{200}(90-x) = 0.0773$ | M1 | Substituting $\frac{3}{200}(90-x)$ for $p$. Allow $\frac{3}{200}(90-x)=k$ where $0 < k < 1$, $k \neq 0.8$ or $0.2$. Allow any inequality sign |
| $x = 84.84$ | | |
| $x = 85$ | A1cao | Condone $x \geq 85$. Do not allow $x > 85$ |
## Part (c)
| Answer/Working | Marks | Guidance |
|---|---|---|
| $X$ – successes $\sim B(100, 0.975)$; $Y$ – not successes $\sim B(100, 0.025)$ | B1 | Writing or using $0.975$ or $0.025$; may be implied by $Po(2.5)$ |
| $Y \sim Po(2.5)$ | M1A1 | M1: using Poisson approximation. A1: $Po(2.5)$ |
| $P(Y \leq 5) = 0.958$ | M1A1 | M1: writing or using $P(Y \leq 5)$. A1: awrt 0.958. SC use of normal approximation: B1 M0 A0 M1 A0 |
---\begin{enumerate}
\item A cadet fires shots at a target at distances ranging from 25 m to 90 m . The probability of hitting the target with a single shot is $p$. When firing from a distance $d \mathrm {~m} , p = \frac { 3 } { 200 } ( 90 - d )$. Each shot is fired independently.
\end{enumerate}
The cadet fires 10 shots from a distance of 40 m .\\
(a) (i) Find the probability that exactly 6 shots hit the target.\\
(ii) Find the probability that at least 8 shots hit the target.
The cadet fires 20 shots from a distance of $x \mathrm {~m}$.\\
(b) Find, to the nearest integer, the value of $x$ if the cadet has an $80 \%$ chance of hitting the target at least once.
The cadet fires 100 shots from 25 m .\\
(c) Using a suitable approximation, estimate the probability that at least 95 of these shots hit the target.\\
\hfill \mbox{\textit{Edexcel S2 2014 Q4 [14]}}