| Exam Board | OCR MEI |
|---|---|
| Module | S1 (Statistics 1) |
| Marks | 18 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Hypothesis test of binomial distributions |
| Type | Expected value and most likely value |
| Difficulty | Standard +0.3 This is a straightforward binomial distribution question covering standard S1 content: calculating probabilities using binomial formula/tables, finding expected values, solving inequalities with binomial probabilities, and conducting a basic one-tailed hypothesis test. All parts follow textbook procedures with no novel problem-solving required, making it slightly easier than average but still requiring careful execution across multiple parts. |
| Spec | 2.04b Binomial distribution: as model B(n,p)2.04c Calculate binomial probabilities2.05a Hypothesis testing language: null, alternative, p-value, significance2.05b Hypothesis test for binomial proportion2.05c Significance levels: one-tail and two-tail |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(X \sim B(12, 0.05)\) | ||
| \((A)\): \(P(X=1) = \binom{12}{1} \times 0.05 \times 0.95^{11} = 0.3413\) | M1 | \(0.05 \times 0.95^{11}\) |
| M1 | \(\binom{12}{1} \times pq^{11}\ (p+q=1)\) | |
| A1 | cao — Total: 3 | |
| OR from tables: \(0.8816 - 0.5404 = 0.3412\) | M1 | for 0.8816 seen |
| M1 | for subtraction of 0.5404 | |
| A1 | cao — Total: 2 | |
| \((B)\): \(P(X \geq 2) = 1 - 0.8816 = 0.1184\) | M1 | for \(1 - P(X \leq 1)\) |
| A1 | cao — Total: 2 | |
| \((C)\): Expected number \(E(X) = np = 12 \times 0.05 = 0.6\) | M1 | for \(12 \times 0.05\) |
| A1 | cao (\(= 0.6\) seen) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| Either: \(1 - 0.95^n \leq \frac{1}{3}\), so \(0.95^n \geq \frac{2}{3}\) | M1 | for equation in \(n\) |
| \(n \leq \log\frac{2}{3} / \log 0.95\), so \(n \leq 7.90\) | M1 | for use of logs |
| Maximum \(n = 7\) | A1 | cao |
| Or (tables): \(n=7\): \(P(X \geq 1)=1-0.6983=0.3017 < \frac{1}{3}\) or \(0.6983 > \frac{2}{3}\) | M1indep | |
| \(n=8\): \(P(X \geq 1)=1-0.6634=0.3366 > \frac{1}{3}\) or \(0.6634 < \frac{2}{3}\) | M1indep | |
| Maximum \(n=7\) | A1 | cao dep on both M's — Total: 3 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| Let \(X \sim B(60, p)\); let \(p\) = probability of a bag being faulty | B1 | for definition of \(p\) |
| \(H_0: p = 0.05\) | B1 | for \(H_0\) |
| \(H_1: p < 0.05\) | B1 | for \(H_1\) |
| \(P(X \leq 1) = 0.95^{60} + 60 \times 0.05 \times 0.95^{59} = 0.1916 > 10\%\) | M1 A1 | for probability; M1 for comparison |
| So not enough evidence to reject \(H_0\) | A1 | |
| Conclude: not enough evidence that the new process reduces the failure rate / scientist incorrect/wrong | E1 | — Total: 8 |
## Question 4:
### Part (i):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $X \sim B(12, 0.05)$ | | |
| $(A)$: $P(X=1) = \binom{12}{1} \times 0.05 \times 0.95^{11} = 0.3413$ | M1 | $0.05 \times 0.95^{11}$ |
| | M1 | $\binom{12}{1} \times pq^{11}\ (p+q=1)$ |
| | A1 | cao — **Total: 3** |
| OR from tables: $0.8816 - 0.5404 = 0.3412$ | M1 | for 0.8816 seen |
| | M1 | for subtraction of 0.5404 |
| | A1 | cao — **Total: 2** |
| $(B)$: $P(X \geq 2) = 1 - 0.8816 = 0.1184$ | M1 | for $1 - P(X \leq 1)$ |
| | A1 | cao — **Total: 2** |
| $(C)$: Expected number $E(X) = np = 12 \times 0.05 = 0.6$ | M1 | for $12 \times 0.05$ |
| | A1 | cao ($= 0.6$ seen) |
### Part (ii):
| Answer/Working | Marks | Guidance |
|---|---|---|
| **Either:** $1 - 0.95^n \leq \frac{1}{3}$, so $0.95^n \geq \frac{2}{3}$ | M1 | for equation in $n$ |
| $n \leq \log\frac{2}{3} / \log 0.95$, so $n \leq 7.90$ | M1 | for use of logs |
| Maximum $n = 7$ | A1 | cao |
| **Or (tables):** $n=7$: $P(X \geq 1)=1-0.6983=0.3017 < \frac{1}{3}$ or $0.6983 > \frac{2}{3}$ | M1indep | |
| $n=8$: $P(X \geq 1)=1-0.6634=0.3366 > \frac{1}{3}$ or $0.6634 < \frac{2}{3}$ | M1indep | |
| Maximum $n=7$ | A1 | cao dep on both M's — **Total: 3** |
### Part (iii):
| Answer/Working | Marks | Guidance |
|---|---|---|
| Let $X \sim B(60, p)$; let $p$ = probability of a bag being faulty | B1 | for definition of $p$ |
| $H_0: p = 0.05$ | B1 | for $H_0$ |
| $H_1: p < 0.05$ | B1 | for $H_1$ |
| $P(X \leq 1) = 0.95^{60} + 60 \times 0.05 \times 0.95^{59} = 0.1916 > 10\%$ | M1 A1 | for probability; M1 for comparison |
| So not enough evidence to reject $H_0$ | A1 | |
| Conclude: not enough evidence that the new process reduces the failure rate / scientist incorrect/wrong | E1 | — **Total: 8** |4 A particular product is made from human blood given by donors. The product is stored in bags. The production process is such that, on average, $5 \%$ of bags are faulty. Each bag is carefully tested before use.
\begin{enumerate}[label=(\roman*)]
\item 12 bags are selected at random.\\
(A) Find the probability that exactly one bag is faulty.\\
(B) Find the probability that at least two bags are faulty.\\
(C) Find the expected number of faulty bags in the sample.
\item A random sample of $n$ bags is selected. The production manager wishes there to be a probability of one third or less of finding any faulty bags in the sample. Find the maximum possible value of $n$, showing your working clearly.
\item A scientist believes that a new production process will reduce the proportion of faulty bags. A random sample of 60 bags made using the new process is checked and one bag is found to be faulty. Write down suitable hypotheses and carry out a hypothesis test at the $10 \%$ level to determine whether there is evidence to suggest that the scientist is correct.
\end{enumerate}
\hfill \mbox{\textit{OCR MEI S1 Q4 [18]}}