Pre-U Pre-U 9794/1 2011 June — Question 14 9 marks

Exam BoardPre-U
ModulePre-U 9794/1 (Pre-U Mathematics Paper 1)
Year2011
SessionJune
Marks9
TopicBivariate data
TypeFind missing data values
DifficultyStandard +0.3 This is a straightforward two-part question testing basic statistics concepts. Part (a) requires solving a simple quadratic equation from the sum of products, then recognizing perfect linear correlation. Part (b) is a standard Venn diagram probability problem with conditional probability. Both parts use routine techniques with no novel insight required, making this slightly easier than average.
Spec2.02f Measures of average and spread2.03b Probability diagrams: tree, Venn, sample space2.03c Conditional probability: using diagrams/tables

  1. The table below relates the values of two variables \(x\) and \(y\).
    \(x\)1\(A\)\(A + 3\)10
    \(y\)2\(A - 1\)\(A\)5
    \(A\) is a positive integer and \(\sum xy = 92\).
    1. Calculate the value of \(A\). [3]
    2. Explain how you can tell that the product-moment correlation coefficient is 1. [1]
  2. A music society has 300 members. 240 like Puccini, 100 like Wagner and 50 like neither.
    1. Calculate the probability that a member chosen at random likes Puccini but not Wagner. [3]
    2. Calculate the probability that a member chosen at random likes Puccini given that this member likes Wagner. [2]

AnswerMarks Guidance
(a) (i) \(A(A - 1) + 4(A + 3) + 50 + 2 = 92\)M1 Attempt xy products
\(A^2 + A - 20 = 0\) or equivA1 Obtain \(A^2 + A - 20 = 0\) or equiv 3 termed expression. State \(A = 4\) only
\(A = 4\)A1
(ii) The points exactly lie on a straight lineB1 The line is \(3y - x = 5\)
(b) (i) \(240 - x + x + 100 - x = 250\)M1 Valid method seen
\(X\) or \(P(A \cap B) = 90\)A1 Award if 90 seen in the diagram
\(\frac{150}{300}\)A1 [9]
(ii) \(\frac{90}{100}\)M1 Use conditional probability. \(\frac{\text{their}x}{100}\) or \(\frac{100}{300}\)
B1[9] Obtain 0.9 or equiv
(a) (i) $A(A - 1) + 4(A + 3) + 50 + 2 = 92$ | M1 | Attempt xy products

$A^2 + A - 20 = 0$ or equiv | A1 | Obtain $A^2 + A - 20 = 0$ or equiv 3 termed expression. State $A = 4$ only

$A = 4$ | A1 |

(ii) The points exactly lie on a straight line | B1 | The line is $3y - x = 5$

(b) (i) $240 - x + x + 100 - x = 250$ | M1 | Valid method seen

$X$ or $P(A \cap B) = 90$ | A1 | Award if 90 seen in the diagram

$\frac{150}{300}$ | A1 | [9] | State $\frac{150}{300}$ aef

(ii) $\frac{90}{100}$ | M1 | Use conditional probability. $\frac{\text{their}x}{100}$ or $\frac{100}{300}$

| B1 | [9] | Obtain 0.9 or equiv
\begin{enumerate}[label=(\alph*)]
\item The table below relates the values of two variables $x$ and $y$.

\begin{center}
\begin{tabular}{|c|c|c|c|c|}
\hline
$x$ & 1 & $A$ & $A + 3$ & 10 \\
\hline
$y$ & 2 & $A - 1$ & $A$ & 5 \\
\hline
\end{tabular}
\end{center}

$A$ is a positive integer and $\sum xy = 92$.

\begin{enumerate}[label=(\roman*)]
\item Calculate the value of $A$. [3]
\item Explain how you can tell that the product-moment correlation coefficient is 1. [1]
\end{enumerate}

\item A music society has 300 members. 240 like Puccini, 100 like Wagner and 50 like neither.

\begin{enumerate}[label=(\roman*)]
\item Calculate the probability that a member chosen at random likes Puccini but not Wagner. [3]
\item Calculate the probability that a member chosen at random likes Puccini given that this member likes Wagner. [2]
\end{enumerate}
\end{enumerate}

\hfill \mbox{\textit{Pre-U Pre-U 9794/1 2011 Q14 [9]}}