| Exam Board | OCR |
|---|---|
| Module | H240/02 (Pure Mathematics and Statistics) |
| Year | 2018 |
| Session | March |
| Marks | 10 |
| Topic | Binomial Distribution |
| Type | Verify conditions in context |
| Difficulty | Standard +0.3 This is a straightforward Stats 1 question testing understanding of when binomial approximation is valid for sampling without replacement. Part (i) is routine hypergeometric calculation, part (ii) tests conceptual understanding (standard textbook explanation), parts (iii-iv) involve standard binomial calculations with no novel insight required. Slightly above average difficulty only due to the multi-part nature and percentage error calculation. |
| Spec | 2.01c Sampling techniques: simple random, opportunity, etc2.03a Mutually exclusive and independent events2.04b Binomial distribution: as model B(n,p)2.04c Calculate binomial probabilities2.05a Hypothesis testing language: null, alternative, p-value, significance |
| Answer | Marks | Guidance |
|---|---|---|
| \(\text{P}(X = 3) = {^6C_3} \times \frac{100}{300} \times \frac{99}{299} \times \frac{98}{298} \times \frac{200}{297} \times \frac{199}{296}\) | M1 | or equiv methods |
| \(= 0.164318883 = 0.164\) (3 sf) | A1 | |
| [2] |
| Answer | Marks | Guidance |
|---|---|---|
| \(\text{P}(\text{disc is black})\) changes each trial (because no replacement) | E1 | oe |
| But change in prob is small | E1 | oe |
| Hence bin gives approx, but not exact, probs | E1 | oe |
| [3] |
| Answer | Marks |
|---|---|
| \(\text{P}(X = 3 \text{ using bin}) = {^6C_3} \times \left(\frac{2}{3}\right)^2 \left(\frac{1}{3}\right)^3\) |
| Answer | Marks | Guidance |
|---|---|---|
| \(= 0.177\%\) | M1, A1 | ft their values for M1 |
| Answer | Marks |
|---|---|
| [2] |
| Answer | Marks | Guidance |
|---|---|---|
| \((= 137.5129127)\) | M1 | both \(np\) and \(npq\) correct method ft their 0.1643.... |
| Answer | Marks | Guidance |
|---|---|---|
| \(= 152.88\) to \(176.34\) | M1 | ft their \(\mu\) and \(\sqrt{\sigma^2}\). Allow rounding to 3 sf |
| Answer | Marks | Guidance |
|---|---|---|
| Estimated limits are 153 to 176 | A1 | Allow (150 – 155) to (174 – 180) |
| [3] |
## 9(i)
$\text{P}(X = 3) = {^6C_3} \times \frac{100}{300} \times \frac{99}{299} \times \frac{98}{298} \times \frac{200}{297} \times \frac{199}{296}$ | M1 | or equiv methods
$= 0.164318883 = 0.164$ (3 sf) | A1 |
| [2]
## 9(ii)
$\text{P}(\text{disc is black})$ changes each trial (because no replacement) | E1 | oe
But change in prob is small | E1 | oe
Hence bin gives approx, but not exact, probs | E1 | oe
| [3]
## 9(iii)
$\text{P}(X = 3 \text{ using bin}) = {^6C_3} \times \left(\frac{2}{3}\right)^2 \left(\frac{1}{3}\right)^3$ | |
$(= 0.164609053)$
$'0.164609053' - 0.164318883' \times 100$
$= 0.177\%$ | M1, A1 | ft their values for M1
cao
| [2]
## 9(iv)
$\mu = 1000 \times 0.164609053$ $(= 164.609053)$
$\sigma^2 = 1000 \times 0.164609053 \times (1 - 0.164609053)$
$(= 137.5129127)$ | M1 | both $np$ and $npq$ correct method ft their 0.1643....
Allow use of 0.1643....
$X \sim \text{Normal}$
$164.609053 \pm \sqrt{137.5129127}$
$= 152.88$ to $176.34$ | M1 | ft their $\mu$ and $\sqrt{\sigma^2}$. Allow rounding to 3 sf
$165 \pm \sqrt{138}$ or better
Estimated limits are 153 to 176 | A1 | Allow (150 – 155) to (174 – 180)
| [3]
---9 A bag contains 100 black discs and 200 white discs. Paula takes five discs at random, without replacement. She notes the number $X$ of these discs that are black.\\
(i) Find $\mathrm { P } ( X = 3 )$.
Paula decides to use the binomial distribution as a model for the distribution of $X$.\\
(ii) Explain why this model will give probabilities that are approximately, but not exactly, correct.\\
(iii) Paula uses the binomial model to find an approximate value for $\mathrm { P } ( X = 3 )$. Calculate the percentage by which her answer will differ from the answer in part (ii).
Paula now assumes that the binomial distribution is a good model for $X$. She uses a computer simulation to generate 1000 values of $X$. The number of times that $X = 3$ occurs is denoted by $Y$.\\
(iv) Calculate estimates of the limits between which two thirds of the values of $Y$ will lie.
\hfill \mbox{\textit{OCR H240/02 2018 Q9 [10]}}