| Exam Board | OCR |
|---|---|
| Module | S4 (Statistics 4) |
| Year | 2015 |
| Session | June |
| Marks | 6 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Hypothesis test of binomial distributions |
| Type | Multi-stage or conditional testing |
| Difficulty | Challenging +1.2 This is a multi-stage hypothesis testing problem requiring careful probability tree analysis and conditional probability. Part (i) demands systematic enumeration of acceptance paths and algebraic manipulation to reach the given form. Part (ii) adds conditional probability. While requiring multiple steps and careful bookkeeping, the techniques are standard S4 material with no novel insights needed—harder than routine single-sample tests but well within expected Further Maths Statistics scope. |
| Spec | 2.04b Binomial distribution: as model B(n,p)2.04c Calculate binomial probabilities |
| Answer | Marks | Guidance |
|---|---|---|
| \(P(\text{batch accepted}) = q^{20} + 20pq^{38}(q + 20p)\) | M1 M1 M1 A1 | Batch accepted if: (1) No defectives in first sample: probability \(q^{20}\), OR (2) Exactly one defective in first sample AND fewer than 2 defectives in second sample. Probability of exactly 1 defective in first: \(\binom{20}{1}pq^{19} = 20pq^{19}\). Probability of 0 or 1 defectives in second: \(q^{20} + 20pq^{19}\). Combined: \(20pq^{19}(q^{20} + 20pq^{19}) = 20pq^{39} + 400p^2q^{38}\). Total: \(q^{20} + 20pq^{39} + 400p^2q^{38} = q^{20} + 20pq^{38}(q + 20p)\) |
| Answer | Marks | Guidance |
|---|---|---|
| \(P(\text{accepted from first sample} | \text{batch accepted}) = 0.5276\) (or \(\frac{q^{20}}{q^{20} + 20pq^{38}(q + 20p)}\)) | M1 M1 A1 |
**(i) Show that probability batch is accepted is $q^{20} + 20pq^{38}(q + 20p)$**
| $P(\text{batch accepted}) = q^{20} + 20pq^{38}(q + 20p)$ | M1 M1 M1 A1 | Batch accepted if: (1) No defectives in first sample: probability $q^{20}$, OR (2) Exactly one defective in first sample AND fewer than 2 defectives in second sample. Probability of exactly 1 defective in first: $\binom{20}{1}pq^{19} = 20pq^{19}$. Probability of 0 or 1 defectives in second: $q^{20} + 20pq^{19}$. Combined: $20pq^{19}(q^{20} + 20pq^{19}) = 20pq^{39} + 400p^2q^{38}$. Total: $q^{20} + 20pq^{39} + 400p^2q^{38} = q^{20} + 20pq^{38}(q + 20p)$ |
**(ii) Given batch accepted with $p = 0.01$, find probability accepted from first sample**
| $P(\text{accepted from first sample} | \text{batch accepted}) = 0.5276$ (or $\frac{q^{20}}{q^{20} + 20pq^{38}(q + 20p)}$) | M1 M1 A1 | With $p = 0.01$, $q = 0.99$. $P(\text{no defectives}) = 0.99^{20} = 0.8179$. $P(\text{accepted from first}) = 0.8179$. $P(\text{batch accepted}) = 0.99^{20} + 20(0.01)(0.99)^{38}(0.99 + 0.2) = 0.8179 + 20(0.01)(0.6676)(1.19) = 0.8179 + 0.1590 = 0.9769$. Required probability: $\frac{0.8179}{0.9769} = 0.8371$ or similar calculation. |3 The manufacturer of electronic components uses the following process to test the proportion of defective items produced.
A random sample of 20 is taken from a large batch of components.
\begin{itemize}
\item If no defective item is found, the batch is accepted.
\item If two or more defective items are found, the batch is rejected.
\item If one defective item is found, a second random sample of 20 is taken. If two or more defective items are found in this second sample, the batch is rejected, otherwise the batch is accepted.
\end{itemize}
The proportion of defective items in the batch is denoted by $p$, and $q = 1 - p$.\\
(i) Show that the probability that a batch is accepted is $q ^ { 20 } + 20 p q ^ { 38 } ( q + 20 p )$.
For a particular component, $p = 0.01$.\\
(ii) Given that a batch is accepted, find the probability that it is accepted as a result of the first sample.
\hfill \mbox{\textit{OCR S4 2015 Q3 [6]}}