OCR MEI S1 — Question 1 18 marks

Exam BoardOCR MEI
ModuleS1 (Statistics 1)
Marks18
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicDiscrete Probability Distributions
TypeVerify probability from independent trials
DifficultyStandard +0.3 This is a straightforward probability question involving independent trials with one biased coin. Parts (i)-(ii) require basic probability calculations with clear guidance ('show that'), part (v) uses standard expectation formulas, and part (vi) applies the given distribution. While multi-part, each step follows routine procedures without requiring novel insight or complex problem-solving.
Spec2.04a Discrete probability distributions2.04b Binomial distribution: as model B(n,p)

1 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, \(X\).
  1. Show that \(\mathrm { P } ( X = 0 ) = 0.025\).
  2. Show that \(\mathrm { P } ( X = 1 ) = 0.1375\). The table shows the probability distribution of \(X\).
    \(r\)01
    \(\mathrm { P } ( X = r )\)0.0250.13750.30.3250.1750.0375
  3. Draw a vertical line chart to illustrate the probability distribution.
  4. Comment on the skewness of the distribution.
  5. Find \(\mathrm { E } ( X )\) and \(\operatorname { Var } ( X )\).
  6. Yasmin tosses the 5 coins three times. Find the probability that the total number of heads is 3 .

Question 1:
Part (iv)
AnswerMarks Guidance
AnswerMarks Guidance
'Negative' or 'very slight negative'E1 [1] E0 for symmetrical; E1 for (very slight) negative skewness even if also mention symmetrical; ignore any reference to unimodal
Part (v)
AnswerMarks Guidance
AnswerMarks 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 M1 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 \(\Sigma r^2p\) (at least 3 terms correct)
\(\text{Var}(X) = 8 - 2.6^2 = 1.24\)M1* dep, A1 [5] M1* for their \(E(X)^2\); A1 FT their \(E(X)\) provided \(\text{Var}(X) > 0\); USE of \(E(X-\mu)^2\) gets M1 for attempt at \((x-\mu)^2\), should see \((-2.6)^2, (-1.6)^2, (-0.6)^2, 0.4^2, 1.4^2, 2.4^2\)
Part (vi)
AnswerMarks Guidance
AnswerMarks Guidance
\(P(\text{Total of } 3) = (3\times0.325\times0.025^2) + (6\times0.3\times0.1375\times0.025)\)M1 For decimal part of first term \(0.325\times0.025^2\)
\(0.1375^3 = 3\times0.000203 + 6\times0.001031 + 0.002600\)M1 For decimal part of second term \(0.3\times0.1375\times0.025\)
\(= 0.000609 + 0.006188 + 0.002600 = 0.00940\)M1, A1 [4] M1 for third term – ignore extra coefficient; CAO: AWRT 0.0094; allow 0.009 with working
# Question 1:

## Part (iv)
| Answer | Marks | Guidance |
|--------|-------|----------|
| 'Negative' or 'very slight negative' | E1 [1] | E0 for symmetrical; E1 for (very slight) negative skewness even if also mention symmetrical; 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 | M1 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 $\Sigma r^2p$ (at least 3 terms correct) |
| $\text{Var}(X) = 8 - 2.6^2 = 1.24$ | M1* dep, A1 [5] | M1* for their $E(X)^2$; A1 FT their $E(X)$ provided $\text{Var}(X) > 0$; USE of $E(X-\mu)^2$ gets M1 for attempt at $(x-\mu)^2$, should see $(-2.6)^2, (-1.6)^2, (-0.6)^2, 0.4^2, 1.4^2, 2.4^2$ |

## Part (vi)
| Answer | Marks | Guidance |
|--------|-------|----------|
| $P(\text{Total of } 3) = (3\times0.325\times0.025^2) + (6\times0.3\times0.1375\times0.025)$ | M1 | For decimal part of first term $0.325\times0.025^2$ |
| $0.1375^3 = 3\times0.000203 + 6\times0.001031 + 0.002600$ | M1 | For decimal part of second term $0.3\times0.1375\times0.025$ |
| $= 0.000609 + 0.006188 + 0.002600 = 0.00940$ | M1, A1 [4] | M1 for third term – ignore extra coefficient; CAO: AWRT 0.0094; allow 0.009 with working |

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1 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 &  &  &  &  \\
\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  Q1 [18]}}