| Exam Board | Edexcel |
|---|---|
| Module | S2 (Statistics 2) |
| Year | 2015 |
| Session | June |
| Marks | 15 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Binomial Distribution |
| Type | Identify distribution and parameters |
| Difficulty | Moderate -0.3 This is a straightforward S2 question testing standard binomial distribution concepts: identifying parameters, stating assumptions, using tables for probability calculations, applying normal approximation, and conducting a basic hypothesis test. All parts follow routine procedures with no novel problem-solving required, making it slightly easier than average. |
| Spec | 2.04d Normal approximation to binomial5.02b Expectation and variance: discrete random variables5.02c Linear coding: effects on mean and variance5.02d Binomial: mean np and variance np(1-p)5.05b Unbiased estimates: of population mean and variance5.05c Hypothesis test: normal distribution for population mean |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(X \sim B(30, 0.4)\) | B1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Any one of: constant probability of buying insurance; customers buy insurance independently of each other | B1 | Any one of these two assumptions in context which refers to insurance |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(P(X < r) < 0.05\) | ||
| \(\{P(X\leqslant 8)=P(X<9)\}=0.0940\); \(\{P(X\leqslant 7)=P(X<8)\}=0.0435\) | M1 | For at least one of either 0.0940 or 0.0435 seen in part (c) |
| So \(r=8\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\{Y \sim B(100,0.4)\approx\}\ Y \sim N(40,24)\) | M1, A1 | Normal or N(40,24) |
| \(\{P(Y\geqslant t)\}\approx P(Y>t-0.5)\) | M1 | For either \(t-0.5\) or \(t+0.5\) |
| \(=P\left(Z>\frac{(t-0.5)-40}{\sqrt{24}}\right)=0.938\) | ||
| \(\frac{(t-0.5)-40}{\sqrt{24}}=-1.54\) | M1 | Standardising \((\pm)\) with their mean and their standard deviation and either \(t-0.5\) or \(t+0.5\) or \(t-1.5\) |
| \(-1.54\) or \(1.54\) or awrt \(-1.54\) or awrt \(1.54\) | B1 | |
| So \(t=32.955571...\Rightarrow t=33\) | A1 cao |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(H_0: p=0.4\), \(H_1: p<0.4\) | B1 | Both hypotheses stated correctly |
| Under \(H_0\), \(X\sim B(25,0.4)\) | ||
| Probability Method: \(P(X\leqslant 6)=0.0736\) | M1 | \(P(X\leqslant 6)\) |
| Critical Region Method: \(P(X\leqslant 6)=0.0736\); \(\{P(X\leqslant 7)=0.1536\}\); \(CR: X\leqslant 6\) | A1 | Either 0.0736 or \(CR: X\leqslant 6\) or \(CR: X<7\) |
| \(\{0.0736 < 0.10\}\) | ||
| Reject \(H_0\) or significant or 6 lies in the CR | dM1 | Dependent on 1st M1. See notes |
| So percentage (or proportion) who buy insurance has decreased | A1 cso |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(H_0: p = 0.4\), \(H_1: p < 0.4\) | B1 | Both hypotheses stated correctly |
| \(\{Y \sim B(25, 0.4)\} \approx Y \sim N(10, 6)\) | ||
| \(P(X \leqslant 6) = P(X < 6.5)\) | M1 | \(P(X \leqslant 6)\) or \(P(X < 6.5)\) |
| \(= P\left(Z < \frac{6.5-10}{\sqrt{6}}\right) = P(Z < -1.4288...)\) | ||
| \(\{= 1 - 0.9236\} = 0.0764\) | A0 | Award A0 here |
| \(\{0.0764 < 0.10\}\) | ||
| Reject \(H_0\) or significant | M1 | As in the main scheme |
| So percentage (or proportion) who buy insurance has decreased | A0 | Award A0 here |
# Question 6:
## Part (a)
| Answer/Working | Mark | Guidance |
|---|---|---|
| $X \sim B(30, 0.4)$ | B1 | |
**[1]**
## Part (b)
| Answer/Working | Mark | Guidance |
|---|---|---|
| Any one of: constant probability of buying insurance; customers buy insurance independently of each other | B1 | Any one of these two assumptions in context which refers to insurance |
**[1]**
## Part (c)
| Answer/Working | Mark | Guidance |
|---|---|---|
| $P(X < r) < 0.05$ | | |
| $\{P(X\leqslant 8)=P(X<9)\}=0.0940$; $\{P(X\leqslant 7)=P(X<8)\}=0.0435$ | M1 | For at least one of either 0.0940 or 0.0435 seen in part (c) |
| So $r=8$ | A1 | |
**[2]**
## Part (d)
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\{Y \sim B(100,0.4)\approx\}\ Y \sim N(40,24)$ | M1, A1 | Normal or N(40,24) |
| $\{P(Y\geqslant t)\}\approx P(Y>t-0.5)$ | M1 | For either $t-0.5$ or $t+0.5$ |
| $=P\left(Z>\frac{(t-0.5)-40}{\sqrt{24}}\right)=0.938$ | | |
| $\frac{(t-0.5)-40}{\sqrt{24}}=-1.54$ | M1 | Standardising $(\pm)$ with their mean and their standard deviation and either $t-0.5$ or $t+0.5$ or $t-1.5$ |
| $-1.54$ or $1.54$ or awrt $-1.54$ or awrt $1.54$ | B1 | |
| So $t=32.955571...\Rightarrow t=33$ | A1 cao | |
**[6]**
## Part (e)
| Answer/Working | Mark | Guidance |
|---|---|---|
| $H_0: p=0.4$, $H_1: p<0.4$ | B1 | Both hypotheses stated correctly |
| Under $H_0$, $X\sim B(25,0.4)$ | | |
| **Probability Method**: $P(X\leqslant 6)=0.0736$ | M1 | $P(X\leqslant 6)$ |
| **Critical Region Method**: $P(X\leqslant 6)=0.0736$; $\{P(X\leqslant 7)=0.1536\}$; $CR: X\leqslant 6$ | A1 | Either 0.0736 or $CR: X\leqslant 6$ or $CR: X<7$ |
| $\{0.0736 < 0.10\}$ | | |
| Reject $H_0$ or significant or 6 lies in the CR | dM1 | Dependent on 1st M1. See notes |
| So **percentage** (or **proportion**) who buy **insurance** has **decreased** | A1 cso | |
**[5]**
## Question 6(e):
**Alternative Method: Normal Approximation to Binomial**
| Answer/Working | Mark | Guidance |
|---|---|---|
| $H_0: p = 0.4$, $H_1: p < 0.4$ | B1 | Both hypotheses stated correctly |
| $\{Y \sim B(25, 0.4)\} \approx Y \sim N(10, 6)$ | | |
| $P(X \leqslant 6) = P(X < 6.5)$ | M1 | $P(X \leqslant 6)$ or $P(X < 6.5)$ |
| $= P\left(Z < \frac{6.5-10}{\sqrt{6}}\right) = P(Z < -1.4288...)$ | | |
| $\{= 1 - 0.9236\} = 0.0764$ | A0 | **Award A0 here** |
| $\{0.0764 < 0.10\}$ | | |
| Reject $H_0$ or significant | M1 | As in the main scheme |
| So **percentage** (or **proportion**) who buy **insurance** has **decreased** | A0 | **Award A0 here** |
---6. Past information at a computer shop shows that $40 \%$ of customers buy insurance when they purchase a product. In a random sample of 30 customers, $X$ buy insurance.
\begin{enumerate}[label=(\alph*)]
\item Write down a suitable model for the distribution of $X$.
\item State an assumption that has been made for the model in part (a) to be suitable.
The probability that fewer than $r$ customers buy insurance is less than 0.05
\item Find the largest possible value of $r$.
A second random sample, of 100 customers, is taken.\\
The probability that at least $t$ of these customers buy insurance is 0.938 , correct to 3 decimal places.
\item Using a suitable approximation, find the value of $t$.
The shop now offers an extended warranty on all products. Following this, a random sample of 25 customers is taken and 6 of them buy insurance.
\item Test, at the $10 \%$ level of significance, whether or not there is evidence that the proportion of customers who buy insurance has decreased. State your hypotheses clearly.
\end{enumerate}
\hfill \mbox{\textit{Edexcel S2 2015 Q6 [15]}}