| Exam Board | OCR MEI |
|---|---|
| Module | S1 (Statistics 1) |
| Year | 2005 |
| Session | January |
| Marks | 19 |
| 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 binomial distribution question with standard calculations (direct probability, cumulative probability, and a basic hypothesis test). All parts use routine methods taught in S1 with no novel problem-solving required, though part (iv) requires systematic trial of values. Slightly easier than average due to the mechanical nature of the calculations. |
| Spec | 2.04b Binomial distribution: as model B(n,p)2.04c Calculate binomial probabilities2.05b Hypothesis test for binomial proportion |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(P(X=0) = 0.8^{16} = 0.0281\) | M1, A1 | \(0.8^{16}\) or tables |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(P(X>3) = 1 - P(X \leq 3)\) or \(P(X \leq 12)\) \(= 1 - 0.5981 = 0.4019\) | M1, M1, A1 | Manipulation; use of tables |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(X \sim B(17, 0.2) \Rightarrow P(X \geq 1) = 0.9775\) | M1, A1 | \(B(17, 0.2)\); 0.9775 |
| Greater than 0.9 so acceptable | E1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(X \sim B(18, 0.2) \Rightarrow P(X \geq 2) = 0.9009\); can make 18 appointments | M1, A1, A1 | 18 and \(\geq 2\); 0.9009; 18 ok |
| \(X \sim B(19, 0.2) \Rightarrow P(X \geq 3) = 0.7631\) | M1 | 19 and \(\geq 3\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| Now \(X \sim B(20, p)\); let \(p\) be probability of not turning up | B1 | |
| \(H_0: p = 0.2\) | B1 | |
| \(H_1: p \neq 0.2\) | B1 | |
| \(P(X \leq 1) = 0.0692 > 2.5\%\); cannot reject \(H_0\) | M1, M1, A1 | 0.0692; correct comparison; cannot reject \(H_0\) |
| Conclude that the proportion of patients not turning up is unchanged | E1 |
## Question 8:
*Number not turning up $X \sim B(16, 0.2)$*
### Part (i)
| Answer | Mark | Guidance |
|--------|------|----------|
| $P(X=0) = 0.8^{16} = 0.0281$ | M1, A1 | $0.8^{16}$ or tables |
### Part (ii)
| Answer | Mark | Guidance |
|--------|------|----------|
| $P(X>3) = 1 - P(X \leq 3)$ or $P(X \leq 12)$ $= 1 - 0.5981 = 0.4019$ | M1, M1, A1 | Manipulation; use of tables |
### Part (iii)
| Answer | Mark | Guidance |
|--------|------|----------|
| $X \sim B(17, 0.2) \Rightarrow P(X \geq 1) = 0.9775$ | M1, A1 | $B(17, 0.2)$; 0.9775 |
| Greater than 0.9 so acceptable | E1 | |
### Part (iv)
| Answer | Mark | Guidance |
|--------|------|----------|
| $X \sim B(18, 0.2) \Rightarrow P(X \geq 2) = 0.9009$; can make 18 appointments | M1, A1, A1 | 18 and $\geq 2$; 0.9009; 18 ok |
| $X \sim B(19, 0.2) \Rightarrow P(X \geq 3) = 0.7631$ | M1 | 19 and $\geq 3$ |
### Part (v)
| Answer | Mark | Guidance |
|--------|------|----------|
| Now $X \sim B(20, p)$; let $p$ be probability of not turning up | B1 | |
| $H_0: p = 0.2$ | B1 | |
| $H_1: p \neq 0.2$ | B1 | |
| $P(X \leq 1) = 0.0692 > 2.5\%$; cannot reject $H_0$ | M1, M1, A1 | 0.0692; correct comparison; cannot reject $H_0$ |
| Conclude that the proportion of patients not turning up is unchanged | E1 | |8 At a doctor's surgery, records show that $20 \%$ of patients who make an appointment fail to turn up. During afternoon surgery the doctor has time to see 16 patients.
There are 16 appointments to see the doctor one afternoon.\\
(i) Find the probability that all 16 patients turn up.\\
(ii) Find the probability that more than 3 patients do not turn up.
To improve efficiency, the doctor decides to make more than 16 appointments for afternoon surgery, although there will still only be enough time to see 16 patients. There must be a probability of at least 0.9 that the doctor will have enough time to see all the patients who turn up.\\
(iii) The doctor makes 17 appointments for afternoon surgery. Find the probability that at least one patient does not turn up. Hence show that making 17 appointments is satisfactory.\\
(iv) Now find the greatest number of appointments the doctor can make for afternoon surgery and still have a probability of at least 0.9 of having time to see all patients who turn up.
A computerised appointment system is introduced at the surgery. It is decided to test, at the 5\% level, whether the proportion of patients failing to turn up for their appointments has changed. There are always 20 appointments to see the doctor at morning surgery. On a randomly chosen morning, 1 patient does not turn up.\\
(v) Write down suitable hypotheses and carry out the test.
\hfill \mbox{\textit{OCR MEI S1 2005 Q8 [19]}}