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Question
In a school with 125 girls, each student is tested to see how many sit-up exercises (sit-ups) she can do in one minute. The results are given in the table below.
(i) Write down the value of p.
(ii) Find the value of q.
Find the median number of sit-ups.
Find the mean number of sit-ups.
Answer/Explanation
Markscheme
(i) \(p = 65\) A1 N1
(ii) for evidence of using sum is 125 (or \(99 – p\) ) (M1)
\(q = 34\) A1 N2
[3 marks]
evidence of median position (M1)
e.g. 63rd student, \(\frac{{125}}{2}\)
median is 17 (sit-ups) A1 N2
[2 marks]
evidence of substituting into \(\frac{{\sum {fx} }}{{125}}\) (M1)
e.g. \(\frac{{15(11) + 16(21) + 17(33) + 18(34) + 19(18) + 20(8)}}{{125}}\) , \(\frac{{2176}}{{125}}\)
mean \(= 17.4\) A1 N2
[2 marks]
Question
The following is a cumulative frequency diagram for the time t, in minutes, taken by 80 students to complete a task.
Write down the median.
Find the interquartile range.
Complete the frequency table below.
Answer/Explanation
Markscheme
median \(m = 32\) A1 N1
[1 mark]
lower quartile \({Q_1} = 22\) , upper quartile \({Q_3} = 40\) (A1)(A1)
\({\text{interquartile range}} = 18\) A1 N3
[3 marks]
A1A1 N2
[2 marks]
Question
A fisherman catches 200 fish to sell. He measures the lengths, l cm of these fish, and the results are shown in the frequency table below.
Calculate an estimate for the standard deviation of the lengths of the fish.
A cumulative frequency diagram is given below for the lengths of the fish.
Use the graph to answer the following.
(i) Estimate the interquartile range.
(ii) Given that \(40\% \) of the fish have a length more than \(k{\text{ cm}}\), find the value of k.
In order to sell the fish, the fisherman classifies them as small, medium or large.
Small fish have a length less than \(20{\text{ cm}}\).
Medium fish have a length greater than or equal to \(20{\text{ cm}}\) but less than \(60{\text{ cm}}\).
Large fish have a length greater than or equal to \(60{\text{ cm}}\).
Write down the probability that a fish is small.
The cost of a small fish is \(\$ 4\), a medium fish \(\$ 10\), and a large fish \(\$ 12\).
Copy and complete the following table, which gives a probability distribution for the cost \(\$ X\) .
Find \({\text{E}}(X)\) .
Answer/Explanation
Markscheme
evidence of using mid-interval values (5, 15, 25, 35, 50, 67.5, 87.5) (M1)
\(\sigma = 19.8\) (cm) A2 N3
[3 marks]
(i) \({Q_1} = 15\) , \({Q_3} = 40\) (A1)(A1)
\(IQR = 25\) (accept any notation that suggests the interval 15 to 40) A1 N3
(ii) METHOD 1
\(60\% \) have a length less than k (A1)
\(0.6 \times 200 = 120\) (A1)
\(k = 30\) (cm) A1 N2
METHOD 2
\(0.4 \times 200 = 80\) (A1)
\(200 – 80 = 120\) (A1)
\(k = 30\) (cm) A1 N2
[6 marks]
\(l < 20{\text{ cm}} \Rightarrow 70{\text{ fish}}\) (M1)
\({\rm{P(small)}} = \frac{{70}}{{200}}( = 0.35)\) A1 N2
[2 marks]
A1A1 N2
[2 marks]
correct substitution (of their p values) into formula for \({\text{E}}(X)\) (A1)
e.g. \(4 \times 0.35 + 10 \times 0.565 + 12 \times 0.085\)
\({\text{E}}(X) = 8.07\) (accept \(\$ 8.07\)) A1 N2
[2 marks]
Question
The following table gives the examination grades for 120 students.
Find the value of
(i) p ;
(ii) q .
Find the mean grade.
Write down the standard deviation.
Answer/Explanation
Markscheme
(a) (i) evidence of appropriate approach (M1)
e.g. \(9 + 25 + 35\) , \(34 + 35\)
\(p = 69\) A1 N2
(ii) evidence of valid approach (M1)
e.g. \(109 – \) their value of p, \(120 – (9 + 25 + 35 + 11)\)
\(q = 40\) A1 N2
[4 marks]
evidence of appropriate approach (M1)
e.g. substituting into \(\frac{{\sum {fx} }}{n}\), division by 120
mean \(= 3.16\) A1 N2
[2 marks]
1.09 A1 N1
[1 mark]
Question
The cumulative frequency curve below represents the heights of 200 sixteen-year-old boys.
Use the graph to answer the following.
Write down the median value.
A boy is chosen at random. Find the probability that he is shorter than \(161{\text{ cm}}\).
Given that \(82\% \) of the boys are taller than \(h{\text{ cm}}\), find h .
Answer/Explanation
Markscheme
\({\text{median}} = 174 {\text{(cm)}}\) A1 N1
[1 mark]
attempt to find number shorter than 161 (M1)
e.g. line on graph, 12 boys
\(p = \frac{{12}}{{200}}( = 0.06)\) A1 N2
[2 marks]
METHOD 1
\(18\% \) have a height less than h (A1)
\(0.18 \times 200 = 36\) (36 may be seen as a line on the graph) (A1)
\(h = 166\) (cm) A1 N2
METHOD 2
\(0.82 \times 200 = 164\) (164 may be seen as a line on the graph) (A1)
\(200 – 164 = 36\) (A1)
\(h = 166\) (cm) A1 N2
[3 marks]
Question
Consider the following cumulative frequency table.
Find the value of \(p\) .
Find
(i) the mean;
(ii) the variance.
Answer/Explanation
Markscheme
valid approach (M1)
eg \(35 – 26\) , \(26 + p = 36\)
\(p = 9\) A1 N2
[2 marks]
(i) mean \( = 26.7\) A2 N2
(ii) recognizing that variance is (sd)2 (M1)
eg \(11.021{ \ldots ^2}\) , \(\sigma = \sqrt {{\mathop{\rm var}} } \) , \(11.158{ \ldots ^2}\)
\({\sigma ^2} = 121\) A1 N2
[4 marks]
Question
The weights in grams of 80 rats are shown in the following cumulative frequency diagram.
Do NOT write solutions on this page.
Write down the median weight of the rats.
Find the percentage of rats that weigh 70 grams or less.
The same data is presented in the following table.
| Weights \(w\) grams | \(0 \leqslant w \leqslant 30\) | \(30 < w \leqslant 60\) | \(60 < w \leqslant 90\) | \(90 < w \leqslant 120\) |
| Frequency | \(p\) | \(45\) | \(q\) | \(5\) |
Write down the value of \(p\).
The same data is presented in the following table.
| Weights \(w\) grams | \(0 \leqslant w \leqslant 30\) | \(30 < w \leqslant 60\) | \(60 < w \leqslant 90\) | \(90 < w \leqslant 120\) |
| Frequency | \(p\) | \(45\) | \(q\) | \(5\) |
Find the value of \(q\).
The same data is presented in the following table.
| Weights \(w\) grams | \(0 \leqslant w \leqslant 30\) | \(30 < w \leqslant 60\) | \(60 < w \leqslant 90\) | \(90 < w \leqslant 120\) |
| Frequency | \(p\) | \(45\) | \(q\) | \(5\) |
Use the values from the table to estimate the mean and standard deviation of the weights.
Assume that the weights of these rats are normally distributed with the mean and standard deviation estimated in part (c).
Find the percentage of rats that weigh 70 grams or less.
Assume that the weights of these rats are normally distributed with the mean and standard deviation estimated in part (c).
A sample of five rats is chosen at random. Find the probability that at most three rats weigh 70 grams or less.
Answer/Explanation
Markscheme
50 (g) A1 N1
[2 marks]
65 rats weigh less than 70 grams (A1)
attempt to find a percentage (M1)
eg \(\frac{{65}}{{80}},{\text{ }}\frac{{65}}{{80}} \times 100\)
81.25 (%) (exact), 81.3 A1 N3
[2 marks]
\(p = 10\) A2 N2
[2 marks]
subtracting to find \(q\) (M1)
eg \(75 – 45 – 10\)
\(q = 20\) A1 N2
[2 marks]
evidence of mid-interval values (M1)
eg \(15, 45, 75, 105\)
\(\overline x = 52.5\) (exact), \(\sigma = 22.5\) (exact) A1A1 N3
[3 marks]
0.781650
78.2 (%) A2 N2
[2 marks]
recognize binomial probability (M1)
eg \(X \sim {\text{B}}(n,{\text{ }}p)\), \(\left( \begin{array}{c}5\\r\end{array} \right)\) \( \times {0.782^r} \times {0.218^{5 – r}}\)
valid approach (M1)
eg \({\text{P}}(X \leqslant 3)\)
\(0.30067\)
\(0.301\) A1 N2
[3 marks]
Question
The following cumulative frequency graph shows the monthly income, \(I\) dollars, of \(2000\) families.
Find the median monthly income.
(i) Write down the number of families who have a monthly income of \(2000\) dollars or less.
(ii) Find the number of families who have a monthly income of more than \(4000\) dollars.
The \(2000\) families live in two different types of housing. The following table gives information about the number of families living in each type of housing and their monthly income \(I\).
Find the value of \(p\).
A family is chosen at random.
(i) Find the probability that this family lives in an apartment.
(ii) Find the probability that this family lives in an apartment, given that its monthly income is greater than \(4000\) dollars.
Estimate the mean monthly income for families living in a villa.
Answer/Explanation
Markscheme
recognizing that the median is at half the total frequency (M1)
eg\(\;\;\;\)\(\frac{{2000}}{2}\)
\(m = 2500{\text{ (dollars)}}\) A1 N2
[2 marks]
(i) \(500\) families have a monthly income less than \(2000\) A1 N1
(ii) correct cumulative frequency, \(1850\) (A1)
subtracting their cumulative frequency from \(2000\) (M1)
eg\(\;\;\;\)\(2000 – 1850\)
\(150\) families have a monthly income of more than \(4000\) dollars A1 N2
Note: If working shown, award M1A1A1 for \(128{\rm{ }} + {\rm{ }}22{\rm{ }} = {\rm{ }}150\), using the table.
[4 marks]
correct calculation (A1)
eg\(\;\;\;\)\(2000 – (436 + 64 + 765 + 28 + 122),{\text{ }}1850 – 500 – 765\) (A1)
\(p = 585\) A1 N2
[2 marks]
(i) correct working (A1)
eg\(\;\;\;\)\(436 + 765 + 28\)
\(0.6145\;\;\;\)(exact) A1 N2
\(\frac{{1229}}{{2000}},{\text{ }}0.615{\text{ }}[0.614,{\text{ }}0.615]\)
(ii) correct working/probability for number of families (A1)
eg\(\;\;\;\)\(122 + 28,{\text{ }}\frac{{150}}{{2000}},{\text{ 0.075}}\)
\(0.186666\)
\(\frac{{28}}{{150}}\;\;\;\left( { = \frac{{14}}{{75}}} \right),{\text{ }}0.187{\text{ }}[0.186,{\text{ }}0.187]\) A1 N2
[4 marks]
evidence of using correct mid-interval values (\(1500,{\rm{ }}3000,{\rm{ }}4500\)) (A1)
attempt to substitute into \(\frac{{\sum {fx} }}{{\sum f }}\) (M1)
eg\(\;\;\;\)\(\frac{{1500 \times 64 + 3000 \times p + 4500 \times 122}}{{64 + 585 + 122}}\)
\(3112.84\)
\(3110{\text{ }}[3110,{\text{ }}3120]{\text{ (dollars)}}\) A1 N2
[3 marks]
Total [15 marks]
Question
Let \(f(x) = {{\text{e}}^{0.5x}} – 2\).
For the graph of f:
(i) write down the \(y\)-intercept;
(ii) find the \(x\)-intercept;
(iii) write down the equation of the horizontal asymptote.
On the following grid, sketch the graph of \(f\), for \( – 4 \leqslant x \leqslant 4\).
Answer/Explanation
Markscheme
(i) \(y = – 1\) A1 N1
(ii) valid attempt to find \(x\)-intercept (M1)
eg\(\,\,\,\,\,\)\(f(x) = 0\)
1.38629 A1 N2
\(x = 2\ln 2{\text{ (exact), }}1.39\)
(iii) \(y = – 2\) (must be equation) A1 N1
A1A1A1 N3
[3 marks]
Question
Ten students were surveyed about the number of hours, \(x\), they spent browsing the Internet during week 1 of the school year. The results of the survey are given below.
\[\sum\limits_{i = 1}^{10} {{x_i} = 252,{\text{ }}\sigma = 5{\text{ and median}} = 27.} \]
During week 4, the survey was extended to all 200 students in the school. The results are shown in the cumulative frequency graph:
Find the mean number of hours spent browsing the Internet.
During week 2, the students worked on a major project and they each spent an additional five hours browsing the Internet. For week 2, write down
(i) the mean;
(ii) the standard deviation.
During week 3 each student spent 5% less time browsing the Internet than during week 1. For week 3, find
(i) the median;
(ii) the variance.
(i) Find the number of students who spent between 25 and 30 hours browsing the Internet.
(ii) Given that 10% of the students spent more than k hours browsing the Internet, find the maximum value of \(k\).
Answer/Explanation
Markscheme
attempt to substitute into formula for mean (M1)
eg\(\,\,\,\,\,\)\(\frac{{\Sigma x}}{{10}},{\text{ }}\frac{{252}}{n},{\text{ }}\frac{{252}}{{10}}\)
mean \( = 25.2{\text{ (hours)}}\) A1 N2
[2 marks]
(i) mean \( = 30.2{\text{ (hours)}}\) A1 N1
(ii) \(\sigma = 5{\text{ (hours)}}\) A1 N1
[2 marks]
(i) valid approach (M1)
eg\(\,\,\,\,\,\)95%, 5% of 27
correct working (A1)
eg\(\,\,\,\,\,\)\(0.95 \times 27,{\text{ }}27 – (5\% {\text{ of }}27)\)
median \( = 25.65{\text{ (exact), }}25.7{\text{ (hours)}}\) A1 N2
(ii) METHOD 1
variance \( = {({\text{standard deviation}})^2}\) (seen anywhere) (A1)
valid attempt to find new standard deviation (M1)
eg\(\,\,\,\,\,\)\({\sigma _{new}} = 0.95 \times 5,{\text{ }}4.75\)
variance \( = 22.5625{\text{ }}({\text{exact}}),{\text{ }}22.6\) A1 N2
METHOD 2
variance \( = {({\text{standard deviation}})^2}\) (seen anywhere) (A1)
valid attempt to find new variance (M1)
eg\(\,\,\,\,\,\)\({0.95^2}{\text{ }},{\text{ }}0.9025 \times {\sigma ^2}\)
new variance \( = 22.5625{\text{ }}({\text{exact}}),{\text{ }}22.6\) A1 N2
[6 marks]
(i) both correct frequencies (A1)
eg\(\,\,\,\,\,\)80, 150
subtracting their frequencies in either order (M1)
eg\(\,\,\,\,\,\)\(150 – 80,{\text{ }}80 – 150\)
70 (students) A1 N2
(ii) evidence of a valid approach (M1)
eg\(\,\,\,\,\,\)10% of 200, 90%
correct working (A1)
eg\(\,\,\,\,\,\)\(0.90 \times 200,{\text{ }}200 – 20\), 180 students
\(k = 35\) A1 N3
[6 marks]
Question
Adam is a beekeeper who collected data about monthly honey production in his bee hives. The data for six of his hives is shown in the following table.
The relationship between the variables is modelled by the regression line with equation \(P = aN + b\).
Adam has 200 hives in total. He collects data on the monthly honey production of all the hives. This data is shown in the following cumulative frequency graph.
Adam’s hives are labelled as low, regular or high production, as defined in the following table.
Adam knows that 128 of his hives have a regular production.
Write down the value of \(a\) and of \(b\).
Use this regression line to estimate the monthly honey production from a hive that has 270 bees.
Write down the number of low production hives.
Find the value of \(k\);
Find the number of hives that have a high production.
Adam decides to increase the number of bees in each low production hive. Research suggests that there is a probability of 0.75 that a low production hive becomes a regular production hive. Calculate the probability that 30 low production hives become regular production hives.
Answer/Explanation
Markscheme
evidence of setup (M1)
eg\(\,\,\,\,\,\)correct value for \(a\) or \(b\)
\(a = 6.96103,{\text{ }}b = – 454.805\)
\(a = 6.96,{\text{ }}b = – 455{\text{ (accept }}6.96x – 455)\) A1A1 N3
[3 marks]
substituting \(N = 270\) into their equation (M1)
eg\(\,\,\,\,\,\)\(6.96(270) – 455\)
1424.67
\(P = 1420{\text{ (g)}}\) A1 N2
[2 marks]
40 (hives) A1 N1
[1 mark]
valid approach (M1)
eg\(\,\,\,\,\,\)\(128 + 40\)
168 hives have a production less than \(k\) (A1)
\(k = 1640\) A1 N3
[3 marks]
valid approach (M1)
eg\(\,\,\,\,\,\)\(200 – 168\)
32 (hives) A1 N2
[2 marks]
recognize binomial distribution (seen anywhere) (M1)
eg\(\,\,\,\,\,\)\(X \sim {\text{B}}(n,{\text{ }}p),{\text{ }}\left( {\begin{array}{*{20}{c}} n \\ r \end{array}} \right){p^r}{(1 – p)^{n – r}}\)
correct values (A1)
eg\(\,\,\,\,\,\)\(n = 40\) (check FT) and \(p = 0.75\) and \(r = 30,{\text{ }}\left( {\begin{array}{*{20}{c}} {40} \\ {30} \end{array}} \right){0.75^{30}}{(1 – 0.75)^{10}}\)
0.144364
0.144 A1 N2
[3 marks]