AQA S2 (Statistics 2) 2014 June

Question 1 7 marks
View details
1 Vanya collected five samples of air and measured the carbon dioxide content of each sample, in parts per million by volume (ppmv). The results were as follows. $$\begin{array} { l l l l l } 387 & 375 & 382 & 379 & 381 \end{array}$$
  1. Assuming that these data form a random sample from a normal distribution with mean \(\mu\) ppmv, construct a \(90 \%\) confidence interval for \(\mu\).
    [0pt] [6 marks]
  2. Vanya repeated her sampling procedure on each of 30 days and, for each day's results, a \(90 \%\) confidence interval for \(\mu\) was constructed. On how many of these 30 days would you expect \(\mu\) to lie outside that day's confidence interval?
    [0pt] [1 mark]
Question 2 1 marks
View details
2 A large multinational company recruits employees from all four countries in the UK. For a sample of 250 recruits, the percentages of males and females from each of the countries are shown in Table 1. \begin{table}[h]
\captionsetup{labelformat=empty} \caption{Table 1}
\cline { 2 - 5 } \multicolumn{1}{c|}{}EnglandScotlandWales
Northern
Ireland
Male22.817.610.86.8
Female15.617.27.61.6
\end{table}
  1. Add the frequencies to the contingency table, Table 2, below.
  2. Carry out a \(\chi ^ { 2 }\)-test at the \(10 \%\) significance level to investigate whether there is an association between country and gender of recruits.
  3. By comparing observed and expected values, make one comment about the distribution of female recruits.
    [0pt] [1 mark] \begin{table}[h]
    \captionsetup{labelformat=empty} \caption{Table 2}
    EnglandScotlandWalesNorthern IrelandTotal
    Male145
    Female105
    Total250
    \end{table}
Question 3 3 marks
View details
3 A box contains a large number of pea pods. The number of peas in a pod may be modelled by the random variable \(X\). The probability distribution of \(X\) is tabulated below.
\(\boldsymbol { x }\)2 or fewer345678 or more
\(\mathbf { P } ( \boldsymbol { X } = \boldsymbol { x } )\)00.10.2\(a\)0.3\(b\)0
  1. Two pods are picked randomly from the box. Find the probability that the number of peas in each pod is at most 4.
  2. It is given that \(\mathrm { E } ( X ) = 5.1\).
    1. Determine the values of \(a\) and \(b\).
    2. Hence show that \(\operatorname { Var } ( X ) = 1.29\).
    3. Some children play a game with the pods, randomly picking a pod and scoring points depending on the number of peas in the pod. For each pod picked, the number of points scored, \(N\), is found by doubling the number of peas in the pod and then subtracting 5. Find the mean and the standard deviation of \(N\).
      [0pt] [3 marks]
Question 4
View details
4 A continuous random variable \(X\) has a probability density function defined by $$f ( x ) = \begin{cases} \frac { 1 } { k } & a \leqslant x \leqslant b
0 & \text { otherwise } \end{cases}$$ where \(b > a > 0\).
    1. Prove that \(k = b - a\).
    2. Write down the value of \(\mathrm { E } ( X )\).
    3. Show, by integration, that \(\mathrm { E } \left( X ^ { 2 } \right) = \frac { 1 } { 3 } \left( b ^ { 2 } + a b + a ^ { 2 } \right)\).
    4. Hence derive a simplified formula for \(\operatorname { Var } ( X )\).
  1. Given that \(a = 4\) and \(\operatorname { Var } ( X ) = 3\), find the numerical value of \(\mathrm { E } ( X )\).
    \includegraphics[max width=\textwidth, alt={}]{34517557-011e-4956-be7d-b26fe5e64d0a-08_1347_1707_1356_153}
Question 5 3 marks
View details
5 Peter, a geologist, is studying pebbles on a beach. He uses a frame, called a quadrat, to enclose an area of the beach. He then counts the number of quartz pebbles, \(X\), within the quadrat. He repeats this procedure 40 times, obtaining the following summarised data. $$\sum x = 128 \quad \text { and } \quad \sum ( x - \bar { x } ) ^ { 2 } = 126.4$$ Peter believes that the distribution of \(X\) can be modelled by a Poisson distribution with \(\lambda = 3.2\).
  1. Use the summarised data to support Peter's belief.
  2. Using Peter's model, calculate the probability that:
    1. a single placing of the quadrat contains more than 5 quartz pebbles;
    2. a single placing of the quadrat contains at least 3 quartz pebbles but fewer than 8 quartz pebbles;
    3. when the quadrat is placed twice, at least one placing contains no quartz pebbles.
  3. Peter also models the number of flint pebbles enclosed by the quadrat by a Poisson distribution with mean 5 . He assumes that the number of flint pebbles enclosed by the quadrat is independent of the number of quartz pebbles enclosed by the quadrat. Using Peter's models, calculate the probability that a single placing of the quadrat contains a total of either 9 or 10 pebbles which are quartz or flint.
    [0pt] [3 marks]
Question 6 15 marks
View details
6 South Riding Alarms (SRA) maintains household burglar-alarm systems. The company aims to carry out an annual service of a system in a mean time of 20 minutes.
Technicians who carry out an annual service must record the times at which they start and finish the service.
  1. Gary is employed as a technician by SRA and his manager, Rajul, calculates the times taken for 8 annual services carried out by Gary. The results, in minutes, are as follows. $$\begin{array} { l l l l l l l l } 24 & 25 & 29 & 16 & 18 & 27 & 19 & 23 \end{array}$$ Assume that these times may be regarded as a random sample from a normal distribution. Carry out a hypothesis test, at the \(10 \%\) significance level, to examine whether the mean time for an annual service carried out by Gary is 20 minutes.
    [0pt] [8 marks]
  2. Rajul suspects that Gary may be taking longer than 20 minutes on average to carry out an annual service. Rajul therefore calculates the times taken for 100 annual services carried out by Gary. Assume that these times may also be regarded as a random sample from a normal distribution but with a standard deviation of 4.6 minutes. Find the highest value of the sample mean which would not support Rajul's suspicion at the \(5 \%\) significance level. Give your answer to two decimal places.
    [0pt] [4 marks]
    \(7 \quad\) A continuous random variable \(X\) has the probability density function defined by $$f ( x ) = \begin{cases} \frac { 4 } { 5 } x & 0 \leqslant x \leqslant 1
    \frac { 1 } { 20 } ( x - 3 ) ( 3 x - 11 ) & 1 \leqslant x \leqslant 3
    0 & \text { otherwise } \end{cases}$$
  3. Find \(\mathrm { P } ( X < 1 )\).
    1. Show that, for \(1 \leqslant x \leqslant 3\), the cumulative distribution function, \(\mathrm { F } ( x )\), is given by $$\mathrm { F } ( x ) = \frac { 1 } { 20 } \left( x ^ { 3 } - 10 x ^ { 2 } + 33 x - 16 \right)$$
    2. Hence verify that the median value of \(X\) lies between 1.13 and 1.14 .
      [0pt] [3 marks] QUESTION
      PART Answer space for question 7
      REFERENCE REFERENCE
      \includegraphics[max width=\textwidth, alt={}]{34517557-011e-4956-be7d-b26fe5e64d0a-20_2290_1707_221_153}