Question 7 - A-Level Maths

OCR MEI S1 2012 June — Question 7 18 marks

Exam Board	OCR MEI
Module	S1 (Statistics 1)
Year	2012
Session	June
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 S1 question requiring basic probability calculations with independent events. Parts (i)-(ii) involve simple binomial probability computations, (iii)-(iv) are routine graphing/interpretation, (v) uses standard E(X) and Var(X) formulas from a given distribution, and (vi) requires recognizing that getting exactly 3 heads total across 3 tosses needs case enumeration (3+0+0, 0+3+0, etc.) but involves only multiplication of given probabilities. All techniques are standard textbook exercises with no novel insight required, making it slightly easier than average.
Spec	2.04a Discrete probability distributions 5.02b Expectation and variance: discrete random variables 5.02c Linear coding: effects on mean and variance

7 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, \(X\).

Show that \(\mathrm { P } ( X = 0 ) = 0.025\).
Show that \(\mathrm { P } ( X = 1 ) = 0.1375\). The table shows the probability distribution of \(X\).
\(r\) 0 1 2 3 4 5
\(\mathrm { P } ( X = r )\) 0.025 0.1375 0.3 0.325 0.175 0.0375
Draw a vertical line chart to illustrate the probability distribution.
Comment on the skewness of the distribution.
Find \(\mathrm { E } ( X )\) and \(\operatorname { Var } ( X )\).
Yasmin tosses the 5 coins three times. Find the probability that the total number of heads is 3 . \section*{THERE ARE NO QUESTIONS WRITTEN ON THIS PAGE.}

Show mark scheme Show mark scheme source

Question 7:

Part (i):

Answer	Marks	Guidance
Answer	Marks	Guidance
\(P(X=0) = 0.4\times0.5^4 = 0.025\) NB ANSWER GIVEN	M1	For \(0.5^4\)
A1

[2]

Part (ii):

Answer	Marks	Guidance
Answer	Marks	Guidance
\(P(X=1) = (0.6\times0.5^4)+(4\times0.4\times0.5\times0.5^3)\)	M1*	For \(0.6\times0.5^4\) seen as single term (not multiplied or divided by anything)
M1*	For \(4\times0.4\times0.5^4\). Allow \(4\times0.025\)
M1* dep	For sum of both
\(= 0.0375 + 0.1 = 0.1375\) NB ANSWER GIVEN	A1	0.1 MUST be justified

[4]

Part (iii):

Answer	Marks	Guidance
Answer	Marks	Guidance
Correct 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
Heights correct: last bar taller than first, fifth taller than second and fourth taller than third	G1	Visual check only. Lines must be thin (gap width > line width). Zero if vertical scale not linear

[2]

Part (iv):

Answer	Marks	Guidance
Answer	Marks	Guidance
'Negative' or 'very slight negative'	E1	E0 for symmetrical, but E1 for (very slight) negative skewness even if also mention symmetrical. Ignore any reference to unimodal

[1]

Part (v):

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	For \(\Sigma rp\) (at least 3 terms correct)
A1	CAO
\(E(X^2) = (0\times0.025)+(1\times0.1375)+(4\times0.3)+(9\times0.325)+(16\times0.175)+(25\times0.0375) = 8\)	M1*	For \(\Sigma r^2p\) (at least 3 terms correct)
\(\text{Var}(X) = 8 - 2.6^2\)	M1* dep	For their \([E(X)]^2\)
\(= 1.24\)	A1	FT their \(E(X)\) provided \(\text{Var}(X) > 0\)

[5]

Part (vi):

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 \(0.325\times0.025^2\)
\(= 3\times0.000203 + 6\times0.001031 + 0.002600\)	M1	For decimal part of second term \(0.3\times0.1375\times0.025\)
\(= 0.000609 + 0.006188 + 0.002600 = 0.00940\)	M1	For third term — ignore extra coefficient. All M marks depend on triple probability products
\((= 3\times13/64000 + 6\times33/32000 + 1331/512000)\)	A1	CAO: AWRT 0.0094. Allow 0.009 with working

[4]

# Question 7:

## Part (i):

| Answer | Marks | Guidance |
|--------|-------|----------|
| $P(X=0) = 0.4\times0.5^4 = 0.025$ **NB ANSWER GIVEN** | M1 | For $0.5^4$ |
| | A1 | |

**[2]**

## Part (ii):

| Answer | Marks | Guidance |
|--------|-------|----------|
| $P(X=1) = (0.6\times0.5^4)+(4\times0.4\times0.5\times0.5^3)$ | M1* | For $0.6\times0.5^4$ seen as single term (not multiplied or divided by anything) |
| | M1* | For $4\times0.4\times0.5^4$. Allow $4\times0.025$ |
| | M1* dep | For sum of both |
| $= 0.0375 + 0.1 = 0.1375$ **NB ANSWER GIVEN** | A1 | 0.1 MUST be justified |

**[4]**

## Part (iii):

| Answer | Marks | Guidance |
|--------|-------|----------|
| Correct 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 |
| Heights correct: last bar taller than first, fifth taller than second and fourth taller than third | G1 | Visual check only. Lines must be thin (gap width > line width). Zero if vertical scale not linear |

**[2]**

## Part (iv):

| Answer | Marks | Guidance |
|--------|-------|----------|
| 'Negative' or 'very slight negative' | E1 | E0 for symmetrical, but E1 for (very slight) negative skewness even if also mention symmetrical. Ignore any reference to unimodal |

**[1]**

## 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 | For $\Sigma rp$ (at least 3 terms correct) |
| | A1 | CAO |
| $E(X^2) = (0\times0.025)+(1\times0.1375)+(4\times0.3)+(9\times0.325)+(16\times0.175)+(25\times0.0375) = 8$ | M1* | For $\Sigma r^2p$ (at least 3 terms correct) |
| $\text{Var}(X) = 8 - 2.6^2$ | M1* dep | For their $[E(X)]^2$ |
| $= 1.24$ | A1 | FT their $E(X)$ provided $\text{Var}(X) > 0$ |

**[5]**

## 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 $0.325\times0.025^2$ |
| $= 3\times0.000203 + 6\times0.001031 + 0.002600$ | M1 | For decimal part of second term $0.3\times0.1375\times0.025$ |
| $= 0.000609 + 0.006188 + 0.002600 = 0.00940$ | M1 | For third term — ignore extra coefficient. All M marks depend on triple probability products |
| $(= 3\times13/64000 + 6\times33/32000 + 1331/512000)$ | A1 | CAO: AWRT 0.0094. Allow 0.009 with working |

**[4]**

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7 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 .

\section*{THERE ARE NO QUESTIONS WRITTEN ON THIS PAGE.}

\hfill \mbox{\textit{OCR MEI S1 2012 Q7 [18]}}

This paper (7 questions)

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Q1 6 Q2 6 Q3 8 Q4 8 Q5 8 Q6 18 Q7 18

\(r\)	0	1	2	3	4	5
\(\mathrm { P } ( X = r )\)	0.025	0.1375	0.3	0.325	0.175	0.0375