Edexcel S1 2011 January — Question 6 14 marks

Exam BoardEdexcel
ModuleS1 (Statistics 1)
Year2011
SessionJanuary
Marks14
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicDiscrete Probability Distributions
TypeProbabilities in table form with k
DifficultyModerate -0.8 This is a straightforward S1 question testing basic probability distribution concepts. Part (a) uses the fundamental property that probabilities sum to 1 (simple algebra). Parts (b)-(d) are standard expectation and variance calculations using formulas from the formula booklet. Parts (e)-(g) involve routine enumeration of outcomes for sums of independent random variables. All parts follow textbook procedures with no novel insight required, making this easier than average for A-level.
Spec2.04a Discrete probability distributions2.04b Binomial distribution: as model B(n,p)5.02b Expectation and variance: discrete random variables5.02c Linear coding: effects on mean and variance5.04a Linear combinations: E(aX+bY), Var(aX+bY)
  1. The discrete random variable \(X\) has the probability distribution
\(x\)1234
\(\mathrm { P } ( X = x )\)\(k\)\(2 k\)\(3 k\)\(4 k\)
  1. Show that \(k = 0.1\) Find
  2. \(\mathrm { E } ( X )\)
  3. \(\mathrm { E } \left( X ^ { 2 } \right)\)
  4. \(\operatorname { Var } ( 2 - 5 X )\) Two independent observations \(X _ { 1 }\) and \(X _ { 2 }\) are made of \(X\).
  5. Show that \(\mathrm { P } \left( X _ { 1 } + X _ { 2 } = 4 \right) = 0.1\)
  6. Complete the probability distribution table for \(X _ { 1 } + X _ { 2 }\)
    \(y\)2345678
    \(\mathrm { P } \left( X _ { 1 } + X _ { 2 } = y \right)\)0.010.040.100.250.24
  7. Find \(\mathrm { P } \left( 1.5 < X _ { 1 } + X _ { 2 } \leqslant 3.5 \right)\)
Question 6:
Part (a):
AnswerMarks Guidance
Answer/WorkingMarks Guidance
\(k + 2k + 3k + 4k = 1\) or \(10k = 1\), \(k = 0.1\)B1cso B1 for clear attempt to use sum of probabilities \(= 1\). Must see previous line as well as \(k = 0.1\). Allow verification with comment e.g. "so \(k = 0.1\)"
Part (b):
AnswerMarks Guidance
Answer/WorkingMarks Guidance
\(E(X) = 1\times0.1 + 2\times0.2 + 3\times0.3 + 4\times0.4 = 3\)M1 A1 M1 for completely correct expression (may be implied by correct answer of 3 or \(30k\)). A1 for 3 only
Part (c):
AnswerMarks Guidance
Answer/WorkingMarks Guidance
\(E(X^2) = 1\times0.1 + 4\times0.2 + 9\times0.3 + 16\times0.4 = 10\)M1 A1 M1 for completely correct expression (may be implied by correct answer of 10 or \(100k\)). A1 for 10 only. [\(E(X^2) = 0.1+0.8+2.7+6.4-9 = 1\) scores M0A0 but accept as \(\text{Var}(X)\) in (d)]
Part (d):
AnswerMarks Guidance
Answer/WorkingMarks Guidance
\(\text{Var}(X) = 10 - 9 (= 1)\)M1 1st M1 for using \(\text{Var}(X) = E(X^2) - E(X)^2\), ft their values from (b) and (c). Allow for \(\text{Var}(X) = 10 - 9\) or better
\(\text{Var}(2-5X) = 5^2\text{Var}(X) = 25\)M1 A1 2nd M1 for \(5^2\text{Var}(X)\) or \(25\text{Var}(X)\), ft their \(\text{Var}(X)\). Allow \(-5^2\) if it later becomes \(+25\). A1 for 25 only (dependent on both M marks). Forming distribution for \(Y = 2-5X\) gets M1 for \(E(Y^2) = 194\) then M1A1 for \(194 - 169 = 25\)
Part (e):
AnswerMarks Guidance
Answer/WorkingMarks Guidance
\(P(1,3) + P(2,2) = 2\times0.1\times0.3 + 0.2\times0.2 = 0.1\)M1 A1cso M1 for correctly identifying \((1,3)\) or \((3,1)\) and \((2,2)\) as required cases (\(3k^2 + 4k^2\) or better). A1cso for 0.1 only but must see evidence for M1
Part (f):
AnswerMarks Guidance
Answer/WorkingMarks Guidance
\(X_1 + X_2\): 2, 3, 4, 5, 6, 7, 8
\(p\): 0.01, 0.04, 0.1, 0.2, 0.25, 0.24, 0.16B1 B1 1st B1 for 0.2 correctly assigned (may be in table). 2nd B1 for 0.16 correctly assigned (may be in table)
Part (g):
AnswerMarks Guidance
Answer/WorkingMarks Guidance
\(P(2) + P(3) = 0.05\)M1 A1 M1 for \(P(2) + P(3)\) (may be implied by correct answer of 0.05). A1 for 0.05 only. Correct answer only can score full marks in parts (b), (c), (f) and (g) 
# Question 6:

## Part (a):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $k + 2k + 3k + 4k = 1$ or $10k = 1$, $k = 0.1$ | B1cso | B1 for clear attempt to use sum of probabilities $= 1$. Must see previous line as well as $k = 0.1$. Allow verification with comment e.g. "so $k = 0.1$" |

## Part (b):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $E(X) = 1\times0.1 + 2\times0.2 + 3\times0.3 + 4\times0.4 = 3$ | M1 A1 | M1 for completely correct expression (may be implied by correct answer of 3 or $30k$). A1 for 3 only |

## Part (c):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $E(X^2) = 1\times0.1 + 4\times0.2 + 9\times0.3 + 16\times0.4 = 10$ | M1 A1 | M1 for completely correct expression (may be implied by correct answer of 10 or $100k$). A1 for 10 only. [$E(X^2) = 0.1+0.8+2.7+6.4-9 = 1$ scores M0A0 but accept as $\text{Var}(X)$ in (d)] |

## Part (d):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $\text{Var}(X) = 10 - 9 (= 1)$ | M1 | 1st M1 for using $\text{Var}(X) = E(X^2) - E(X)^2$, ft their values from (b) and (c). Allow for $\text{Var}(X) = 10 - 9$ or better |
| $\text{Var}(2-5X) = 5^2\text{Var}(X) = 25$ | M1 A1 | 2nd M1 for $5^2\text{Var}(X)$ or $25\text{Var}(X)$, ft their $\text{Var}(X)$. Allow $-5^2$ if it later becomes $+25$. A1 for 25 only (dependent on both M marks). Forming distribution for $Y = 2-5X$ gets M1 for $E(Y^2) = 194$ then M1A1 for $194 - 169 = 25$ |

## Part (e):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $P(1,3) + P(2,2) = 2\times0.1\times0.3 + 0.2\times0.2 = 0.1$ | M1 A1cso | M1 for correctly identifying $(1,3)$ or $(3,1)$ and $(2,2)$ as required cases ($3k^2 + 4k^2$ or better). A1cso for 0.1 only but must see evidence for M1 |

## Part (f):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $X_1 + X_2$: 2, 3, 4, 5, **6**, 7, **8** | | |
| $p$: 0.01, 0.04, 0.1, **0.2**, 0.25, 0.24, **0.16** | B1 B1 | 1st B1 for 0.2 correctly assigned (may be in table). 2nd B1 for 0.16 correctly assigned (may be in table) |

## Part (g):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $P(2) + P(3) = 0.05$ | M1 A1 | M1 for $P(2) + P(3)$ (may be implied by correct answer of 0.05). A1 for 0.05 only. Correct answer only can score full marks in parts (b), (c), (f) and (g) |

---
\begin{enumerate}
  \item The discrete random variable $X$ has the probability distribution
\end{enumerate}

\begin{center}
\begin{tabular}{ | c | c | c | c | c | }
\hline
$x$ & 1 & 2 & 3 & 4 \\
\hline
$\mathrm { P } ( X = x )$ & $k$ & $2 k$ & $3 k$ & $4 k$ \\
\hline
\end{tabular}
\end{center}

(a) Show that $k = 0.1$

Find\\
(b) $\mathrm { E } ( X )$\\
(c) $\mathrm { E } \left( X ^ { 2 } \right)$\\
(d) $\operatorname { Var } ( 2 - 5 X )$

Two independent observations $X _ { 1 }$ and $X _ { 2 }$ are made of $X$.\\
(e) Show that $\mathrm { P } \left( X _ { 1 } + X _ { 2 } = 4 \right) = 0.1$\\
(f) Complete the probability distribution table for $X _ { 1 } + X _ { 2 }$

\begin{center}
\begin{tabular}{ | c | c | c | c | c | c | c | c | }
\hline
$y$ & 2 & 3 & 4 & 5 & 6 & 7 & 8 \\
\hline
$\mathrm { P } \left( X _ { 1 } + X _ { 2 } = y \right)$ & 0.01 & 0.04 & 0.10 &  & 0.25 & 0.24 &  \\
\hline
\end{tabular}
\end{center}

(g) Find $\mathrm { P } \left( 1.5 < X _ { 1 } + X _ { 2 } \leqslant 3.5 \right)$\\

\hfill \mbox{\textit{Edexcel S1 2011 Q6 [14]}}
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