In a study, samples of soil were collected during the summer. Soil samples of dimensions \(25 \mathrm {~cm} \times 25 \mathrm {~cm} \times 40 \mathrm {~cm}\) were collected for analysis. The study found that there were, on average, 11 earthworms per sample.
a) Explain briefly the conditions under which a Poisson distribution could be used to model the number of earthworms per sample.
b) In July, pupils at a primary school are asked to dig a smaller hole, \(25 \mathrm {~cm} \times 25 \mathrm {~cm} \times 10 \mathrm {~cm}\), and to count the number of earthworms they find. Calculate the probability that the pupils find exactly 5 earthworms.
c) In the autumn, the average number of earthworms per sample is greater than in the summer. The probability that, in the autumn, there are fewer than 13 earthworms in a soil sample of dimensions \(25 \mathrm {~cm} \times 25 \mathrm {~cm} \times 40 \mathrm {~cm}\) is close to \(36 \%\). Find the mean number of earthworms, to the nearest whole number, per \(25 \mathrm {~cm} \times 25 \mathrm {~cm} \times 40 \mathrm {~cm}\) soil sample in the autumn.
Jessica is studying the relationship between hip girth, \(h \mathrm {~cm}\), and thigh girth, \(t \mathrm {~cm}\), for American adults who are physically active. She takes a random sample of 11 people from a very large dataset which she has downloaded into a spreadsheet software package. The results are shown below.
| \(h ( \mathrm {~cm} )\) | \(98 \cdot 6\) | \(112 \cdot 1\) | \(97 \cdot 9\) | \(110 \cdot 2\) | \(89 \cdot 2\) | \(111 \cdot 7\) | \(87 \cdot 0\) | \(94 \cdot 7\) | \(100 \cdot 4\) | \(104 \cdot 0\) | \(88 \cdot 4\) |
| \(t ( \mathrm {~cm} )\) | \(48 \cdot 3\) | \(87 \cdot 2\) | \(55 \cdot 2\) | \(68 \cdot 0\) | \(48 \cdot 5\) | \(63 \cdot 2\) | \(49 \cdot 5\) | \(55 \cdot 7\) | \(59 \cdot 1\) | \(64 \cdot 0\) | \(52 \cdot 4\) |
a) Jessica notes that, for the thigh girth data, the lower quartile is 49.5 and the upper quartile is \(64 \cdot 0\).
i) Show that 87.2 should be classified as an outlier for \(t\).
ii) Give a reason why Jessica might exclude the outlier.
iii) Give a reason why Jessica might include the outlier.
Jessica decides to exclude the outlier and produces the following scatter diagram.
\section*{Thigh girth versus Hip girth}
\includegraphics[max width=\textwidth, alt={}, center]{77c62e6d-58e4-42d3-9982-5a8325e8e826-04_647_1250_1439_404}
b) Interpret, in context, the correlation in the data shown in the diagram.
The equation of the regression line of \(t\) on \(h\) for this sample is
$$t = 0.69 h - 11.26$$
c) Interpret the gradient of the regression line in this context.
d) Use your knowledge of large data sets and spreadsheet software packages to suggest a way in which Jessica could improve her investigation.
A company, Run4Lyfe, sponsors an athletic event. The organisers of the event claim that \(70 \%\) of the participants know the name of the sponsoring company. Run4Lyfe is concerned that the proportion, \(p\), of participants knowing the name of the sponsoring company is less than \(70 \%\). They decide to survey 60 randomly selected participants to carry out a significance test.
a) State suitable hypotheses for carrying out the test.
b) i) Explain what is meant by the critical region for this test.
ii) Determine the critical region if the test is to be carried out at a significance level as close as possible to, but not exceeding, \(5 \%\).
iii) Given that 40 participants out of the 60 in the sample know the name of the company, complete the significance test.
c) State, with a reason, how you would advise Run4Lyfe with regards to sponsoring the event next year.
The fertility rate for a country is the average number of children that are born to a woman over her lifetime. The graphs and table below show some data on the fertility rates for 197 countries in the years 1914 and 2014.
\begin{figure}[h]
\captionsetup{labelformat=empty}
\caption{Fertility rates in 1914}
\includegraphics[alt={},max width=\textwidth]{77c62e6d-58e4-42d3-9982-5a8325e8e826-06_671_1483_593_283}
\end{figure}
\begin{figure}[h]
\captionsetup{labelformat=empty}
\caption{Fertility rates in 2014}
\includegraphics[alt={},max width=\textwidth]{77c62e6d-58e4-42d3-9982-5a8325e8e826-06_616_1219_1434_287}
\end{figure}
\begin{figure}[h]
\captionsetup{labelformat=empty}
\caption{Decreases in fertility rates from 1914 to 2014}
\includegraphics[alt={},max width=\textwidth]{77c62e6d-58e4-42d3-9982-5a8325e8e826-06_476_613_2270_388}
\end{figure}
| Minimum value | - 0.71 | | Lower quartile | 2.08 | | Median | 3.19 | | Upper quartile | 3.94 | | Maximum value | 6.49 |
a) Comment on the shapes of the distributions of fertility rates for 1914 and 2014.
b) Interpret the minimum value, \(- 0 \cdot 71\), in the boxplot.
You are also given the following information:
| Country | | | | France | | 1.98 | | Ethiopia | | 4.4 |
c) i) Find the best possible estimate for the decrease in the fertility rate from 1914 to 2014 for France.
ii) Find the best possible estimate for the decrease in the fertility rate from 1914 to 2014 for Ethiopia.
iii) Give one possible reason why the answers to i) and ii) are so different.
iv) Explain why these estimates may not be very accurate.
\section*{Section B: Mechanics}
The diagram below shows a vehicle of mass 1300 kg towing a trailer of mass 500 kg by means of a light horizontal tow bar. The vehicle is moving forward along a straight horizontal road such that a constant resistance of magnitude 650 N acts on the vehicle and a constant resistance of magnitude 320 N acts on the trailer. The vehicle's engine produces a constant driving force of \(F \mathrm {~N}\).
\includegraphics[max width=\textwidth, alt={}]{77c62e6d-58e4-42d3-9982-5a8325e8e826-08_158_851_781_609}
Given that the acceleration of the vehicle and trailer is \(0.85 \mathrm {~ms} ^ { - 2 }\), show that \(F = 2500\) and determine the tension in the tow bar.
CAIE
FP2
2012
June
Q8
9 marks
Standard +0.3
The number of flaws in a randomly chosen 100 metre length of ribbon is modelled by a Poisson distribution with mean 1.6. The random variable \(X\) metres is the distance between two successive flaws. Show that the distribution function of \(X\) is given by
$$\text{F}(x) = \begin{cases}
1 - e^{-0.016x} & x \geq 0, \\
0 & x < 0,
\end{cases}$$
and deduce that \(X\) has a negative exponential distribution, stating its mean. [4]
Find
- the median distance between two successive flaws, [3]
- the probability that there is a distance of at least 50 metres between two successive flaws. [2]
CAIE
FP2
2012
June
Q8
9 marks
Standard +0.3
The number of flaws in a randomly chosen 100 metre length of ribbon is modelled by a Poisson distribution with mean 1.6. The random variable \(X\) metres is the distance between two successive flaws. Show that the distribution function of \(X\) is given by
$$\text{F}(x) = \begin{cases}
1 - e^{-0.016x} & x \geqslant 0, \\
0 & x < 0,
\end{cases}$$
and deduce that \(X\) has a negative exponential distribution, stating its mean.
[4]
Find
- the median distance between two successive flaws, [3]
- the probability that there is a distance of at least 50 metres between two successive flaws. [2]
CAIE
FP2
2018
November
Q10
12 marks
Standard +0.8
The number of accidents, \(x\), that occur each day on a motorway are recorded over a period of 40 days. The results are shown in the following table.
| Number of accidents | 0 | 1 | 2 | 3 | 4 | 5 | 6 | \(\geqslant 7\) | | Observed frequency | 3 | 5 | 8 | 10 | 5 | 7 | 2 | 0 |
\begin{enumerate}[label=(\roman*)]
\item Show that the mean number of accidents each day is 2.95 and calculate the variance for this sample. Explain why these values suggest that a Poisson distribution might fit the data.
[3]
\item A Poisson distribution with mean 2.95, as found from the data, is used to calculate the expected frequencies, correct to 2 decimal places. The results are shown in the following table.
| Number of accidents | 0 | 1 | 2 | 3 | 4 | 5 | 6 | \(\geqslant 7\) | | Observed frequency | 3 | 5 | 8 | 10 | 5 | 7 | 2 | 0 | | Expected frequency | 2.09 | 6.18 | 9.11 | 8.96 | 6.61 | 3.90 | 1.92 | 1.23 |
Show how the expected frequency of 6.61 for \(x = 4\) is obtained.
[2]
\item Test at the 5% significance level the goodness of fit of this Poisson distribution to the data.
[7]
\end{enumerate]
CAIE
S2
2021
June
Q1
4 marks
Standard +0.3
Accidents at two factories occur randomly and independently. On average, the numbers of accidents per month are 3.1 at factory \(A\) and 1.7 at factory \(B\).
Find the probability that the total number of accidents in the two factories during a 2-month period is more than 3. [4]
CAIE
S2
2021
June
Q5
7 marks
Standard +0.3
On average, 1 in 75000 adults has a certain genetic disorder.
- Use a suitable approximating distribution to find the probability that, in a random sample of 10000 people, at least 1 has the genetic disorder. [3]
- In a random sample of \(n\) people, where \(n\) is large, the probability that no-one has the genetic disorder is more than 0.9.
Find the largest possible value of \(n\). [4]
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