| Exam Board | OCR MEI |
|---|---|
| Module | S1 (Statistics 1) |
| Marks | 18 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Discrete Probability Distributions |
| Type | Verify probability from independent trials |
| Difficulty | Standard +0.3 This is a straightforward probability question involving independent trials with one biased coin. Parts (i)-(ii) require basic probability calculations with binomial coefficients, (iii)-(iv) are routine graphing/interpretation, (v) uses standard expectation formulas, and (vi) applies the given distribution. All techniques are standard S1 material with no novel problem-solving required, making it slightly easier than average. |
| Spec | 2.02i Select/critique data presentation5.02a Discrete probability distributions: general5.02b Expectation and variance: discrete random variables |
| \(r\) | 0 | 1 | 2 | 3 | 4 | 5 |
| \(\mathrm { P } ( X = r )\) | 0.025 | 0.1375 | 0.3 | 0.325 | 0.175 | 0.0375 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(P(X=0) = 0.4 \times 0.5^4 = 0.025\) NB ANSWER GIVEN | M1, A1 [2] | For \(0.5^4\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(P(X=1) = (0.6 \times 0.5^4) + (4 \times 0.4 \times 0.5 \times 0.5^3)\) | M1* | For \(0.6 \times 0.5^4\) seen as a single term (not multiplied or divided by anything) |
| M1* | For \(4 \times 0.4 \times 0.5^4\); allow \(4 \times 0.025\). Watch out for incorrect methods such as \((0.4/4)\). 0.1 MUST be justified | |
| \(= 0.0375 + 0.1 = 0.1375\) NB ANSWER GIVEN | M1* dep, A1 [4] | For sum of both, dep on both M1's |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Vertical line chart with labelled linear scales on both axes | G1 | For labelled linear scales on both axes. Dep on attempt at vertical line chart. Accept P on vertical axis |
| Correct heights with last bar taller than first, fifth taller than second, fourth taller than third | G1 [2] | Lines must be thin (gap width > line width). All correct. Zero of vertical scale not linear: G0G0. Joined tops: G0G1 MAX. Frequency polygon: G0G1 MAX. Bar chart: G0G1 MAX. Curve only (no vertical lines): G0G0. Best fit line: G0G0. Allow transposed diagram |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| 'Negative' or 'very slight negative' | E1 [1] | E0 for symmetrical. E1 for (very slight) negative skewness even if symmetrical also mentioned. Ignore any reference to unimodal |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(E(X) = (0\times0.025)+(1\times0.1375)+(2\times0.3)+(3\times0.325)+(4\times0.175)+(5\times0.0375) = 2.6\) | M1, A1 | For \(\Sigma rp\) (at least 3 terms correct). CAO |
| \(E(X^2) = (0\times0.025)+(1\times0.1375)+(4\times0.3)+(9\times0.325)+(16\times0.175)+(25\times0.075) = 8\) | M1* | For \(\Sigma r^2 p\) (at least 3 terms correct) |
| \(\text{Var}(X) = 8 - 2.6^2 = 1.24\) | M1* dep, A1 [5] | For \(E(X^2) - [E(X)]^2\). FT their \(E(X)\) provided \(\text{Var}(X) > 0\). Unsupported correct answers get 5 marks |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(P(\text{Total of } 3) = (3\times0.325\times0.025^2)+(6\times0.3\times0.1375\times0.025)+0.1375^3\) | M1 | For decimal part of first term \(0.325 \times 0.025^2\) |
| \(= 3\times0.000203 + 6\times0.001031 + 0.002600\) | M1 | For decimal part of second term \(0.3\times0.1375\times0.025\) |
| \(= 0.000609 + 0.006188 + 0.002600\) | M1 | For third term – ignore extra coefficient. All M marks depend on triple probability products |
| \(= 0.00940\) \((= 3\times13/64000 + 6\times33/32000 + 1331/512000)\) | A1 [4] | CAO: AWRT 0.0094. Allow 0.009 with working |
# Question 5:
## Part (i)
| Answer | Marks | Guidance |
|--------|-------|----------|
| $P(X=0) = 0.4 \times 0.5^4 = 0.025$ **NB ANSWER GIVEN** | M1, A1 [2] | For $0.5^4$ |
---
## Part (ii)
| Answer | Marks | Guidance |
|--------|-------|----------|
| $P(X=1) = (0.6 \times 0.5^4) + (4 \times 0.4 \times 0.5 \times 0.5^3)$ | M1* | For $0.6 \times 0.5^4$ seen as a single term (not multiplied or divided by anything) |
| | M1* | For $4 \times 0.4 \times 0.5^4$; allow $4 \times 0.025$. Watch out for incorrect methods such as $(0.4/4)$. 0.1 **MUST** be justified |
| $= 0.0375 + 0.1 = 0.1375$ **NB ANSWER GIVEN** | M1* dep, A1 [4] | For sum of both, dep on both M1's |
---
## Part (iii)
| Answer | Marks | Guidance |
|--------|-------|----------|
| Vertical line chart with labelled linear scales on both axes | G1 | For labelled linear scales on both axes. Dep on attempt at vertical line chart. Accept P on vertical axis |
| Correct heights with last bar taller than first, fifth taller than second, fourth taller than third | G1 [2] | Lines must be thin (gap width > line width). All correct. Zero of vertical scale not linear: G0G0. Joined tops: G0G1 MAX. Frequency polygon: G0G1 MAX. Bar chart: G0G1 MAX. Curve only (no vertical lines): G0G0. Best fit line: G0G0. Allow transposed diagram |
---
## Part (iv)
| Answer | Marks | Guidance |
|--------|-------|----------|
| 'Negative' or 'very slight negative' | E1 [1] | E0 for symmetrical. E1 for (very slight) negative skewness even if symmetrical also mentioned. Ignore any reference to unimodal |
---
## Part (v)
| Answer | Marks | Guidance |
|--------|-------|----------|
| $E(X) = (0\times0.025)+(1\times0.1375)+(2\times0.3)+(3\times0.325)+(4\times0.175)+(5\times0.0375) = 2.6$ | M1, A1 | For $\Sigma rp$ (at least 3 terms correct). CAO |
| $E(X^2) = (0\times0.025)+(1\times0.1375)+(4\times0.3)+(9\times0.325)+(16\times0.175)+(25\times0.075) = 8$ | M1* | For $\Sigma r^2 p$ (at least 3 terms correct) |
| $\text{Var}(X) = 8 - 2.6^2 = 1.24$ | M1* dep, A1 [5] | For $E(X^2) - [E(X)]^2$. FT their $E(X)$ provided $\text{Var}(X) > 0$. Unsupported correct answers get 5 marks |
---
## Part (vi)
| Answer | Marks | Guidance |
|--------|-------|----------|
| $P(\text{Total of } 3) = (3\times0.325\times0.025^2)+(6\times0.3\times0.1375\times0.025)+0.1375^3$ | M1 | For decimal part of first term $0.325 \times 0.025^2$ |
| $= 3\times0.000203 + 6\times0.001031 + 0.002600$ | M1 | For decimal part of second term $0.3\times0.1375\times0.025$ |
| $= 0.000609 + 0.006188 + 0.002600$ | M1 | For third term – ignore extra coefficient. All M marks depend on triple probability products |
| $= 0.00940$ $(= 3\times13/64000 + 6\times33/32000 + 1331/512000)$ | A1 [4] | CAO: AWRT 0.0094. Allow 0.009 with working |5 Yasmin has 5 coins. One of these coins is biased with P (heads) $= 0.6$. The other 4 coins are fair. She tosses all 5 coins once and records the number of heads, $X$.\\
(i) Show that $\mathrm { P } ( X = 0 ) = 0.025$.\\
(ii) Show that $\mathrm { P } ( X = 1 ) = 0.1375$.
The table shows the probability distribution of $X$.
\begin{center}
\begin{tabular}{ | c | c | c | c | c | c | c | }
\hline
$r$ & 0 & 1 & 2 & 3 & 4 & 5 \\
\hline
$\mathrm { P } ( X = r )$ & 0.025 & 0.1375 & 0.3 & 0.325 & 0.175 & 0.0375 \\
\hline
\end{tabular}
\end{center}
(iii) Draw a vertical line chart to illustrate the probability distribution.\\
(iv) Comment on the skewness of the distribution.\\
(v) Find $\mathrm { E } ( X )$ and $\operatorname { Var } ( X )$.\\
(vi) Yasmin tosses the 5 coins three times. Find the probability that the total number of heads is 3 .
\hfill \mbox{\textit{OCR MEI S1 Q5 [18]}}