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Question
The figure below shows the lengths in centimetres of fish found in the net of a small trawler.
Find the total number of fish in the net.[2]
Find (i) the modal length interval,
(ii) the interval containing the median length,
(iii) an estimate of the mean length.[5]
(i) Write down an estimate for the standard deviation of the lengths.
(ii) How many fish (if any) have length greater than three standard deviations above the mean?[3]
The fishing company must pay a fine if more than 10% of the catch have lengths less than 40cm.
Do a calculation to decide whether the company is fined.[2]
A sample of 15 of the fish was weighed. The weight, W was plotted against length, L as shown below.
Exactly two of the following statements about the plot could be correct. Identify the two correct statements.
Note: You do not need to enter data in a GDC or to calculate r exactly.
(i) The value of r, the correlation coefficient, is approximately 0.871.
(ii) There is an exact linear relation between W and L.
(iii) The line of regression of W on L has equation W = 0.012L + 0.008 .
(iv) There is negative correlation between the length and weight.
(v) The value of r, the correlation coefficient, is approximately 0.998.
(vi) The line of regression of W on L has equation W = 63.5L + 16.5.[2]
Answer/Explanation
Markscheme
Total = 2 + 3 + 5 + 7 + 11 + 5 + 6 + 9 + 2 + 1 (M1)
(M1) is for a sum of frequencies.
= 51 (A1)(G2)[2 marks]
Unit penalty (UP) is applicable where indicated in the left hand column.
(i) modal interval is 60 – 70
Award (A0) for 65 (A1)
(ii) median is length of fish no. 26, (M1)(A1)
also 60 – 70 (G2)
Can award (A1)(ft) or (G2)(ft) for 65 if (A0) was awarded for 65 in part (i).
(iii) mean is \(\frac{{2 \times 25 + 3 \times 35 + 5 \times 45 + 7 \times 55 + …}}{{51}}\) (M1)
(UP) = 69.5 cm (3sf) (A1)(ft)(G1)
Note: (M1) is for a sum of (frequencies multiplied by midpoint values) divided by candidate’s answer from part (a). Accept mid-points 25.5, 35.5 etc or 24.5, 34.5 etc, leading to answers 70.0 or 69.0 (3sf) respectively. Answers of 69.0, 69.5 or 70.0 (3sf) with no working can be awarded (G1).[5 marks]
Unit penalty (UP) is applicable where indicated in the left hand column.
(UP) (i) standard deviation is 21.8 cm (G1)
For any other answer without working, award (G0). If working is present then (G0)(AP) is possible.
(ii) \(69.5 + 3 \times 21.8 = 134.9 > 120\) (M1)
no fish (A1)(ft)(G1)
For ‘no fish’ without working, award (G1) regardless of answer to (c)(i). Follow through from (c)(i) only if method is shown.[3 marks]
5 fish are less than 40 cm in length, (M1)
Award (M1) for any of \(\frac{5}{51}\), \(\frac{46}{51}\), 0.098 or 9.8%, 0.902, 90.2% or 5.1 seen.
hence no fine. (A1)(ft)
Note: There is no G mark here and (M0)(A1) is never allowed. The follow-through is from answer in part (a).[2 marks]
(i) and (iii) are correct. (A1)(A1)[2 marks]
Question
A random sample of 167 people who own mobile phones was used to collect data on the amount of time they spent per day using their phones. The results are displayed in the table below.
Manuel conducts a survey on a random sample of 751 people to see which television programme type they watch most from the following: Drama, Comedy, Film, News. The results are as follows.
Manuel decides to ignore the ages and to test at the 5 % level of significance whether the most watched programme type is independent of gender.
State the modal group.[1]
Use your graphic display calculator to calculate approximate values of the mean and standard deviation of the time spent per day on these mobile phones.[3]
On graph paper, draw a fully labelled histogram to represent the data.[4]
Draw a table with 2 rows and 4 columns of data so that Manuel can perform a chi-squared test.[3]
State Manuel’s null hypothesis and alternative hypothesis.[1]
Find the expected frequency for the number of females who had ‘Comedy’ as their most-watched programme type. Give your answer to the nearest whole number.[2]
Using your graphic display calculator, or otherwise, find the chi-squared statistic for Manuel’s data.[3]
(i) State the number of degrees of freedom available for this calculation.
(ii) State his conclusion.[3]
Answer/Explanation
Markscheme
\(45 \leqslant t < 60\) (A1)[1 mark]
Unit penalty (UP) is applicable in question part (i)(b) only.
(UP) 42.4 minutes (G2)
21.6 minutes (G1)[3 marks]
(A4)[4 marks]
(M1)(M1)(A1)[3 marks]
H0: favourite TV programme is independent of gender or no association between favourite TV programme and gender
H1: favourite TV programme is dependent on gender (must have both) (A1)[1 mark]
\(\frac{{365 \times 217}}{{751}}\) (M1)
\(= 105\) (A1)(ft)(G2)[2 marks]
12.6 (accept 12.558) (G3)[3 marks]
(i) 3 (A1)
(ii) reject H0 or equivalent statement (e.g. accept H1) (A1)(ft)[3 marks]
Question
The diagram shows the cumulative frequency graph for the time t taken to perform a certain task by 2000 men.
Use the diagram to estimate the median time.[1]
Use the diagram to estimate the upper quartile and the lower quartile.[2]
Use the diagram to estimate the interquartile range.[1]
Find the number of men who take more than 11 seconds to perform the task.[3]
55 % of the men took less than p seconds to perform the task. Find p.[2]
The times taken for the 2000 men were grouped as shown in the table below.
Write down the value of a.[1]
Write down the value of b.[1]
Use your graphic display calculator to find an estimate of the mean time.[2]
Use your graphic display calculator to find an estimate of the standard deviation of the time.[1]
Everyone who performs the task in less than one standard deviation below the mean will receive a bonus. Pedro takes 9.5 seconds to perform the task.
Does Pedro receive the bonus? Justify your answer.[3]
Answer/Explanation
Markscheme
Unit penalty (UP) applies in this part
(UP) median = 13 seconds (A1)[1 mark]
Unit penalty (UP) applies in this part
(UP) 16 seconds and 10 seconds (A1)(A1)
Note: Accept 16.1 or 16.2 for the upper quartile value.[2 marks]
IQR = 6 seconds (A1)(ft)
Note: (ft) from reasonable answers to (ii).[1 mark]
value seen 650 (A1)
2000 – value = 2000 – 650 (M1)
= 1350 (A1)(G2)[3 marks]
55 % of 2000 = 1100 (A1)
p = 13.5 (A1)(G2)[2 marks]
a = 500 (A1)[1 mark]
b = 150 (A1)[1 mark]
Unit penalty (UP) applies in this part
(UP) \( \bar t = 13.25 {\text{ seconds}} \) (13.3 seconds) (G2)
OR
\(\bar t = \frac{{7.5 \times 500 + 12.5 \times 850 + 17.5 \times {\text{their }}a + 22.5 \times {\text{their }}b}}{{2000}}\) (M1)
(UP) \(\bar t = 13.25 {\text{ seconds}}\) (13.3 seconds) (A1)(ft)
Note: Award (ft) from their a and their b only if working is seen.[2 marks]
\(\sigma = 4.41{\text{ seconds}}\) (G1)[1 mark]
\(\bar t – \sigma = 8.84\) (A1)(ft)
Their \(\bar t – \sigma\) compared to 9.5 (R1)
Pedro does not receive the bonus (A1)(ft)
Note: Do not award (R0)(A1).[3 marks]
Question
The speed, \(s\) , in \({\text{km }}{{\text{h}}^{ – 1}}\), of \(120\) vehicles passing a point on the road was measured. The results are given below.
Write down the midpoint of the \(60 < s \leqslant 70\) interval.[1]
Use your graphic display calculator to find an estimate for
(i) the mean speed of the vehicles;
(ii) the standard deviation of the speeds of the vehicles.[3]
Write down the number of vehicles whose speed is less than or equal to \({\text{60 km }}{{\text{h}}^{ – 1}}\).[1]
Consider the cumulative frequency table below.
Write down the value of \(a\) , of \(b\) and of \(c\) .[2]
Consider the cumulative frequency table below.
Draw a cumulative frequency graph for the information from the table. Use \(1\) cm to represent \({\text{10 km }}{{\text{h}}^{ – 1}}\) on the horizontal axis and \(1\) cm to represent \(10\) vehicles on the vertical axis.[4]
Use your cumulative frequency graph to estimate
(i) the median speed of the vehicles;
(ii) the number of vehicles that are travelling at a speed less than or equal to \({\text{65 km }}{{\text{h}}^{ – 1}}\).[4]
All drivers whose vehicle’s speed is greater than one standard deviation above the speed limit of \({\text{50 km }}{{\text{h}}^{ – 1}}\) will be fined.
Use your graph to estimate the number of drivers who will be fined.[3]
Answer/Explanation
Markscheme
\(65\) (A1)[1 mark]
(i) \(54{\text{ (km }}{{\text{h}}^{ – 1}})\) (G2)
Note: If the answer to part (b)(i) is consistent with the answer to part (a) then award (G2)(ft) even if no working seen.
(ii) \(19.2\) (\(19.2093 \ldots \)) (G1)
Note: Accept \(19\), do not accept \(20\).[3 marks]
\(76\) (A1)[1 mark]
\(a = 76\), \(b = 98\) (A1)(ft)
Note: Follow through from their answer to part (c) for \(a\) and \(b = \) their \(a + 22\) .
\(c = 118\) (A1)[2 marks]
(A1)(A1)(ft)(A1)(ft)(A1)
Notes: Award (A1) for axes labelled and correct scales. If the axes are reversed do not award this mark but follow through. Award (A2)(ft) for their 6 points correct, (A1)(ft) for at least 3 of these points correct. Award (A1) for smooth curve drawn through all points including (\(0\), \(0\)). If either the \(x\) or the \(y\) axis has a break in it to zero, do not award this final mark.[4 marks]
(i) \(57\) \({\text{(km }}{{\text{h}}^{ – 1}})\) \(( \pm 2)\) (M1)(A1)(ft)(G2)
Note: Award (M1) for clear indication of median on their graph. Follow through from their graph. If their answer is consistent with their incorrect graph but there is no working present on graph then no marks are awarded.
(ii) \(90\) vehicles \(( \pm 2)\) (M1)(A1)(ft)(G2)
Note: Award (M1) for clear indication of method on their graph. Follow through from their graph. If their answer is consistent with their incorrect graph but there is no working present on graph then no marks are awarded.[4 marks]
\(50 + 19.2 = 69.2\) (A1)(ft)
\(24\) \(( \pm 2)\) drivers will be fined (M1)(A1)(ft)(G2)
Notes: Follow through from their graph and from their part (b)(ii). Award (M1) for indication of method on their graph. If their answer is consistent with their incorrect graph but there is no working present on graph then no marks are awarded.[3 marks]
Question
200 people were asked the amount of time T (minutes) they had spent in the supermarket. The results are represented in the table below.
State if the data is discrete or continuous.[1]
State the modal group.[1]
Write down the midpoint of the interval 10 < T ≤ 20 .[1]
Use your graphic display calculator to find an estimate for
(i) the mean;
(ii) the standard deviation.[3]
The results are represented in the cumulative frequency table below, with upper class boundaries of 10, 20, 30, 40, 50.
Write down the value of
(i) q;
(ii) r.[2]
The results are represented in the cumulative frequency table below, with upper class boundaries of 10, 20, 30, 40, 50.
On graph paper, draw a cumulative frequency graph, using a scale of 2 cm to represent 10 minutes (T) on the horizontal axis and 1 cm to represent 10 people on the vertical axis.[4]
Use your graph from part (f) to estimate
(i) the median;
(ii) the 90th percentile of the results;
(iii) the number of people who shopped at the supermarket for more than 15 minutes.[6]
Answer/Explanation
Markscheme
continuous (A1)[1 mark]
20 < T ≤ 30 (A1)[1 mark]
15 (A1)[1 mark]
(i) 21.5 (G2)
(ii) 9.21 (9.20597…) (G1) [3 marks]
(i) q = 194 (A1)
(ii) r = 200 (A1) [2 marks]
(A1)(A2)(ft)(A1)
Notes: Award (A1) for scale and axis labels, (A2)(ft) for 5 correct points, (A1)(ft) for 4 or 3 correct points, (A0) for less than 3 correct points, (A1) for smooth curve through their points, starting at (0, 0). Follow through from their answers to parts (e)(i) and (e)(ii).[4 marks]
(i) 22.5 ± 2 (A1)
(ii) 32 ± 2 (M1)(A1)(ft)(G2)
Note: Award (M1) for lines drawn on graph or some indication of method, follow through from their graph if working is shown.(iii) 44 ± 2 (A1)(ft)
Note: Follow through from their graph if working is shown.
200 − 44 = 156 (M1)(A1)(ft)(G2)
Note: Award (M1) for subtraction from 200, follow through from their graph if working is shown.[6 marks]
Question
A survey was conducted to determine the length of time, \(t\), in minutes, people took to drink their coffee in a café. The information is shown in the following grouped frequency table.
Write down the total number of people who were surveyed.[1]
Write down the mid-interval value for the \(10 < t \leqslant 15\) group.[1]
Find an estimate of the mean time people took to drink their coffee.[2]
The information above has been rewritten as a cumulative frequency table.
Write down the value of \(a\) and the value of \(b\).[2]
This information is shown in the following cumulative frequency graph.
For the people who were surveyed, use the graph to estimate
(i) the time taken for the first \(40\) people to drink their coffee;
(ii) the number of people who take less than \(8\) minutes to drink their coffee;
(iii) the number of people who take more than \(23\) minutes to drink their coffee.[4]
Answer/Explanation
Markscheme
\(60\) (A1)[1 mark]
\(12.5\) (A1)[1 mark]
\(\frac{{3 \times 2.5 + 5 \times 7.5 + \ldots + 10 \times 27.5}}{{60}}\) (M1)
Note: Award (M1) for an attempt to substitute their mid-interval values (consistent with their answer to part (b)) into the formula for the mean.
Award (M1) where a table is constructed with their (consistent) mid-interval values listed along with the frequencies.
\( = \frac{{1075}}{{60}}{\text{ }}\left( {\frac{{215}}{{12}},{\text{ 17.9, 17.9166}} \ldots } \right)\) (A1)(ft)(G2)
Note: Follow through from their answer to part (b).[2 marks]
\(a = 34,{\text{ }}b = 60\) (A1)(A1)[2 marks]
(i) \( \leqslant {\text{21.25 minutes}}\) (A1)
Note: Accept \(21.25\).
Accept any answer between \(21\) and \(21.5\).
(Accept 21.5, but do not accept 21.)
(ii) \(5\) (A1)
Note: Accept \( < 6\). Do not accept \(6\).
Answer must be an integer.
(iii) \(60 – 45\) (M1)
\( = 15\) (A1)(G2)
Notes: Award (M1) for subtraction from \(60\). Accept \(15 \pm 1\).
Answer must be an integer.[4 marks]
Question
The cumulative frequency graph shows the speed, \(s\), in \({\text{km}}\,{{\text{h}}^{ – 1}}\), of \(120\) vehicles passing a hospital gate.
Estimate the minimum possible speed of one of these vehicles passing the hospital gate.[1]
Find the median speed of the vehicles.[2]
Write down the \({75^{{\text{th}}}}\) percentile.[1]
Calculate the interquartile range.[2]
The speed limit past the hospital gate is \(50{\text{ km}}\,{{\text{h}}^{ – 1}}\).
Find the number of these vehicles that exceed the speed limit.[2]
The table shows the speeds of these vehicles travelling past the hospital gate.
Find the value of \(p\) and of \(q\).[2]
The table shows the speeds of these vehicles travelling past the hospital gate.
(i) Write down the modal class.
(ii) Write down the mid-interval value for this class.[2]
The table shows the speeds of these vehicles travelling past the hospital gate.
Use your graphic display calculator to calculate an estimate of
(i) the mean speed of these vehicles;
(ii) the standard deviation.[3]
It is proposed that the speed limit past the hospital gate is reduced to \(40{\text{ km}}\,{{\text{h}}^{ – 1}}\) from the current \(50{\text{ km}}\,{{\text{h}}^{ – 1}}\).
Find the percentage of these vehicles passing the hospital gate that do not exceed the current speed limit but would exceed the new speed limit.[2]
Answer/Explanation
Markscheme
\(10{\text{ (km}}\,{{\text{h}}^{ – 1}})\) (A1)
\(36\) (G2)
\(41.5\) (G1)
\(41.5 – 32.5\) (M1)
\( = 9{\text{ (}} \pm {\text{1)}}\) (A1)(ft)(G2)
Notes: Award (M1) for quartiles seen. Follow through from part (c).
\(120 – 110\) (M1)
\( = 10\) (A1)(G2)
Note: Award (M1) for \(110\) seen.
\(p = 4\;\;\;q = 10\) (A1)(ft)(A1)(ft)
Note: Follow through from part (e).
(i) \(30 < s \leqslant 40\) (A1)
(ii) \(35\) (A1)(ft)
Note: Follow through from part (g)(i).
(i) \(36.8{\text{ (km}}\,{{\text{h}}^{ – 1}})\;\;\;(36.8333)\) (G2)(ft)
Notes: Follow through from part (f).
(ii) \(8.85\;\;\;(8.84904 \ldots )\) (G1)(ft)
Note: Follow through from part (f), irrespective of working seen.
\(\frac{{26}}{{120}} \times 100\) (M1)
Note: Award (M1) for \(\frac{{26}}{{120}} \times 100\) seen.
\( = 21.7{\text{ (}}\% )\;\;\;\left( {21.6666 \ldots ,{\text{ }}21\frac{2}{3},{\text{ }}\frac{{65}}{3}} \right)\) (A1)(G2)
Question
A group of 800 students answered 40 questions on a category of their choice out of History, Science and Literature.
For each student the category and the number of correct answers, \(N\), was recorded. The results obtained are represented in the following table.
A \({\chi ^2}\) test at the 5% significance level is carried out on the results. The critical value for this test is 12.592.
State whether \(N\) is a discrete or a continuous variable.[1]
Write down, for \(N\), the modal class;[1]
Write down, for \(N\), the mid-interval value of the modal class.[1]
Use your graphic display calculator to estimate the mean of \(N\);[2]
Use your graphic display calculator to estimate the standard deviation of \(N\).[1]
Find the expected frequency of students choosing the Science category and obtaining 31 to 40 correct answers.[2]
Write down the null hypothesis for this test;[1]
Write down the number of degrees of freedom.[1]
Write down the \(p\)-value for the test;[1]
Write down the \({\chi ^2}\) statistic.[2]
State the result of the test. Give a reason for your answer.[2]
Answer/Explanation
Markscheme
discrete (A1)[1 mark]
\(11 \leqslant N \leqslant 20\) (A1)[1 mark]
15.5 (A1)(ft)
Note: Follow through from part (b)(i).[1 mark]
\(21.2{\text{ }}(21.2125)\) (G2)[2 marks]
\(9.60{\text{ }}(9.60428 \ldots )\) (G1)[1 marks]
\(\frac{{260}}{{800}} \times \frac{{157}}{{800}} \times 800\)\(\,\,\,\)OR\(\,\,\,\)\(\frac{{260 \times 157}}{{800}}\) (M1)
Note: Award (M1) for correct substitution into expected frequency formula.
\( = 51.0{\text{ }}(51.025)\) (A1)(G2)[2 marks]
choice of category and number of correct answers are independent (A1)
Notes: Accept “no association” between (choice of) category and number of correct answers. Do not accept “not related” or “not correlated” or “influenced”.[1 mark]
6 (A1)[1 mark]
\(0.0644{\text{ }}(0.0644123 \ldots )\) (G1)[1 mark]
\(11.9{\text{ }}(11.8924 \ldots )\) (G2)[2 marks]
the null hypothesis is not rejected (the null hypothesis is accepted) (A1)(ft)
OR
(choice of) category and number of correct answers are independent (A1)(ft)
as \(11.9 < 12.592\)\(\,\,\,\)OR\(\,\,\,\)\(0.0644 > 0.05\) (R1)
Notes: Award (R1) for a correct comparison of either their \({\chi ^2}\) statistic to the \({\chi ^2}\) critical value or their \(p\)-value to the significance level. Award (A1)(ft) from that comparison.
Follow through from part (f). Do not award (A1)(ft)(R0).[2 marks]