AQA Further AS Paper 2 Statistics (Further AS Paper 2 Statistics) Specimen

Question 1 1 marks
View details
1 The random variable \(T\) has probability density defined by $$\mathrm { f } ( t ) = \left\{ \begin{array} { c c } \frac { t } { 8 } & 0 \leq t \leq k
0 & \text { otherwise } \end{array} \right.$$ Find the value of \(k\)
[0pt] [1 mark] $$\begin{array} { l l l l } \frac { 1 } { 16 } & \frac { 1 } { 4 } & 4 & 16 \end{array}$$
Question 2
View details
2 The discrete random variable \(X\) has probability distribution defined by $$\mathrm { P } ( X = x ) = \begin{cases} 0.1 & x = 0,1,2,3,4,5,6,7,8,9
0 & \text { otherwise } \end{cases}$$ Find the value of \(\mathrm { P } ( 4 \leq X \leq 7 )\)
Circle your answer.
0.20.30.40.5
Question 3 4 marks
View details
3 The discrete random variable \(R\) has the following probability distribution.
\(\boldsymbol { r }\)- 20\(a\)4
\(\mathbf { P } ( \boldsymbol { R } = \boldsymbol { r } )\)0.3\(b\)\(c\)0.1
It is known that \(\mathrm { E } ( R ) = 0.2\) and \(\operatorname { Var } ( R ) = 3.56\)
Find the values of \(a , b\) and \(c\).
[0pt] [4 marks]
Question 4 3 marks
View details
4 The number of printers, \(V\), bought during one day from the Verigood store can be modelled by a Poisson distribution with mean 4.5 The number of printers, \(W\), bought during one day from the Winnerprint store can be modelled by a Poisson distribution with mean 5.5 4
  1. Find the probability that the total number of printers bought during one day from Verigood and Winnerprint stores is greater than 10.
    [0pt] [2 marks] 4
  2. State the circumstance under which the distributional model you used in part (a) would not be valid.
    [0pt] [1 mark]
Question 5
View details
5 Participants in a school jumping competition gain a total score for each jump based on the length, \(L\) metres, jumped beyond a fixed point and a mark, \(S\), for style.
\(L\) may be regarded as a continuous random variable with probability density function $$\mathrm { f } ( l ) = \left\{ \begin{array} { c c } w l & 0 \leq l \leq 15
0 & \text { otherwise } \end{array} \right.$$ where \(w\) is a constant.
\(S\) may be regarded as a discrete random variable with probability function $$\mathrm { P } ( S = s ) = \left\{ \begin{array} { c l } \frac { 1 } { 15 } s & s = 1,2,3,4,5
0 & \text { otherwise } \end{array} \right.$$ Assume that \(L\) and \(S\) are independent. The total score for a participant in this competition, \(T\), is given by \(T = L ^ { 2 } + \frac { 1 } { 2 } S\) Show that the expected total score for a participant is \(114 \frac { 1 } { 3 }\)
Question 6 5 marks
View details
6 The continuous random variable \(T\) has probability density function defined by $$\mathrm { f } ( t ) = \left\{ \begin{array} { c c } \frac { 1 } { 3 } & 0 \leq t \leq \frac { 3 } { 2 }
\frac { 9 - 2 t } { 18 } & \frac { 3 } { 2 } \leq t \leq \frac { 9 } { 2 }
0 & \text { otherwise } \end{array} \right.$$ 6
    1. Sketch this probability density function below.
      \includegraphics[max width=\textwidth, alt={}, center]{6ccf7d1d-5a7b-47d1-b38e-c7e762204746-07_1009_1041_1073_520} 6
  1. (ii) State the median of \(T\). 6
    1. Find \(\mathrm { E } ( T )\)
      [0pt] [2 marks]
      6
  2. (ii) Given that \(\mathrm { E } \left( T ^ { 2 } \right) = \frac { 15 } { 4 }\), find \(\operatorname { Var } ( 4 T - 5 )\) [3 marks]
Question 7
View details
7 A dairy industry researcher, Robyn, decided to investigate the milk yield, classified as low, medium or high, obtained from four different breeds of cow, \(\mathrm { A } , \mathrm { B } , \mathrm { C }\) and D . The milk yield of a sample of 105 cows was monitored and the results are summarised in contingency Table 1.
\multirow{2}{*}{Table 1}Yield
LowMediumHighTotal
\multirow{4}{*}{Breed}A451221
B106420
C817732
D520732
Total274830105
The sample of cows may be regarded as random.
Robyn decides to carry out a \(\chi ^ { 2 }\)-test for association between milk yield and breed using the information given in Table 1. 7
  1. Contingency Table 2 gives some of the expected frequencies for this test.
    Complete Table 2 with the missing expected values.
    \multirow[t]{2}{*}{Table 2}Yield
    LowMediumHigh
    \multirow{4}{*}{Breed}A6
    B5.149.145.71
    C
    D8.2314.639.14
    7
    1. For Robyn's test, the test statistic \(\sum \frac { ( O - E ) ^ { 2 } } { E } = 19.4\) correct to three significant figures.
      Use this information to carry out Robyn's test, using the \(1 \%\) level of significance.
      7
  2. (ii) By considering the observed frequencies given in Table 1 with the expected frequencies in Table 2, interpret, in context, the association, if any, between milk yield and breed.
Question 8 9 marks
View details
8 In a small town, the number of properties sold during a week in spring by a local estate agent, Keith, can be regarded as occurring independently and with constant mean \(\mu\). Data from several years have shown the value of \(\mu\) to be 3.5 . A new housing development was built on the outskirts of the town and the properties on this development were offered for sale by the builder of the development, not by the local estate agents. During the first four weeks in spring, when properties on the new development were offered for sale by the builder, Keith sold a total of 8 properties. Keith claims that the sale of new properties by the builder reduced his mean number of properties sold during a week in spring. 8
  1. Investigate Keith's claim, using the \(5 \%\) level of significance.
    [0pt] [6 marks]
    8
  2. For your test carried out in part (a) state, in context, the meaning of a Type II error.
    [0pt] [1 mark]
    8
  3. State one advantage and one disadvantage of using a 1\% significance level rather than a 5\% level of significance in a hypothesis test.
    [0pt] [2 marks]