| Exam Board | OCR MEI |
|---|---|
| Module | S1 (Statistics 1) |
| Year | 2010 |
| Session | January |
| Marks | 17 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Hypothesis test of binomial distributions |
| Type | Multiple binomial probability calculations |
| Difficulty | Standard +0.3 This is a straightforward S1 question testing basic probability calculations (independent events, binomial distribution) and a one-tailed hypothesis test. All parts follow standard procedures with no novel problem-solving required. The calculations are routine: (i) uses basic multiplication and complement rules, (ii) applies binomial probability formulas directly, and (iii) is a textbook binomial test. Slightly easier than average due to clear structure and standard techniques. |
| Spec | 2.04b Binomial distribution: as model B(n,p)2.04c Calculate binomial probabilities2.05b Hypothesis test for binomial proportion2.05c Significance levels: one-tail and two-tail |
| Pollution level | Low | Medium | High |
| Probability | 0.5 | 0.35 | 0.15 |
| Answer | Marks | Guidance |
|---|---|---|
| (A) \(P(\text{Plow on all 3 days}) = 0.5^3 = 0.125\) or \(\frac{1}{8}\) | M1 for \(0.5^3\), A1 CAO | [2] |
| (B) \(P(\text{Plow on at least 1 day}) = 1 - 0.5^3 = 1 - 0.125 = 0.875\) | M1 for \(1 - 0.5^3\), A1 CAO | [2] |
| (C) \(P(\text{One low, one medium, one high}) = 6 \times 0.5 \times 0.35 \times 0.15 = 0.1575\) | M1 for product of probabilities \(0.5 \times 0.35 \times 0.15\) or \(\frac{3!}{100}\), M1 for \(\times 6\) or \(3!\) or \(^3P_3\), A1 CAO | [3] |
| Answer | Marks | Guidance |
|---|---|---|
| (A) \(P(\text{No days}) = 0.85^{10} = 0.1969\) Or from tables: \(P(\text{No days}) = 0.1969\) | M1, A1 | [2] |
| Answer | Marks | Guidance |
|---|---|---|
| or from tables: \(P(1 \text{ day}) = P(X \leq 1) - P(X \leq 0) = 0.5443 - 0.1969 = 0.3474\) | M1 for \(0.15^1 \times 0.85^9\), M1 for \(\binom{10}{1} \times p^1q^9\), A1 CAO, OR: M2 for \(0.5443 - 0.1969\), A1 CAO | [3] |
| Answer | Marks | Guidance |
|---|---|---|
| \(H_1\) has this form as she believes that the probability of a low pollution level is greater in this street. | Either: B1 for correct probability of \(0.0207\), M1 for comparison, Or: B1 for CR, M1 for comparison, A1 CAO dep on B1M1, E1 for conclusion in context, E1 indep | [5] |
| TOTAL | [17] |
## (i)
**(A)** $P(\text{Plow on all 3 days}) = 0.5^3 = 0.125$ or $\frac{1}{8}$ | M1 for $0.5^3$, A1 CAO | [2]
**(B)** $P(\text{Plow on at least 1 day}) = 1 - 0.5^3 = 1 - 0.125 = 0.875$ | M1 for $1 - 0.5^3$, A1 CAO | [2]
**(C)** $P(\text{One low, one medium, one high}) = 6 \times 0.5 \times 0.35 \times 0.15 = 0.1575$ | M1 for product of probabilities $0.5 \times 0.35 \times 0.15$ or $\frac{3!}{100}$, M1 for $\times 6$ or $3!$ or $^3P_3$, A1 CAO | [3]
## (ii)
$X \sim B(10, 0.15)$
**(A)** $P(\text{No days}) = 0.85^{10} = 0.1969$ Or from tables: $P(\text{No days}) = 0.1969$ | M1, A1 | [2]
**(B)** Either: $P(1 \text{ day}) = \binom{10}{1} \times 0.15^1 \times 0.85^9 = 0.3474$
or from tables: $P(1 \text{ day}) = P(X \leq 1) - P(X \leq 0) = 0.5443 - 0.1969 = 0.3474$ | M1 for $0.15^1 \times 0.85^9$, M1 for $\binom{10}{1} \times p^1q^9$, A1 CAO, OR: M2 for $0.5443 - 0.1969$, A1 CAO | [3]
## (iii)
Let $X \sim B(20, 0.5)$
Either: $P(X \geq 15) = 1 - 0.9793 = 0.0207 < 5\%$
Or: Critical region is $\{15,16,17,18,19,20\}$. 15 lies in the critical region.
So there is sufficient evidence to reject $H_0$.
Conclude that there is enough evidence to indicate that the probability of low pollution levels is higher on the new street.
$H_1$ has this form as she believes that the probability of a low pollution level is greater in this street. | Either: B1 for correct probability of $0.0207$, M1 for comparison, Or: B1 for CR, M1 for comparison, A1 CAO dep on B1M1, E1 for conclusion in context, E1 indep | [5]
**TOTAL** | | [17]An environmental health officer monitors the air pollution level in a city street. Each day the level of pollution is classified as low, medium or high. The probabilities of each level of pollution on a randomly chosen day are as given in the table.
\begin{center}
\begin{tabular}{|c|c|c|c|}
\hline
Pollution level & Low & Medium & High \\
\hline
Probability & 0.5 & 0.35 & 0.15 \\
\hline
\end{tabular}
\end{center}
\begin{enumerate}[label=(\roman*)]
\item Three days are chosen at random. Find the probability that the pollution level is
\begin{enumerate}[label=(\alph*)]
\item low on all 3 days, [2]
\item low on at least one day, [2]
\item low on one day, medium on another day, and high on the other day. [3]
\end{enumerate}
\item Ten days are chosen at random. Find the probability that
\begin{enumerate}[label=(\alph*)]
\item there are no days when the pollution level is high, [2]
\item there is exactly one day when the pollution level is high. [3]
\end{enumerate}
\end{enumerate}
The environmental health officer believes that pollution levels will be low more frequently in a different street. On 20 randomly selected days she monitors the pollution level in this street and finds that it is low on 15 occasions.
\begin{enumerate}[label=(\roman*)]
\setcounter{enumi}{2}
\item Carry out a test at the 5% level to determine if there is evidence to suggest that she is correct. Use hypotheses $H_0: p = 0.5$, $H_1: p > 0.5$, where $p$ represents the probability that the pollution level in this street is low. Explain why $H_1$ has this form. [5]
\end{enumerate}
\hfill \mbox{\textit{OCR MEI S1 2010 Q8 [17]}}