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 diagram is a box and whisker plot for a set of data.
The interquartile range is 20 and the range is 40.
Write down the median value.
Find the value of
(i) \(a\) ;
(ii) \(b\) .
Answer/Explanation
Markscheme
18 A1 N1
[1 mark]
(i) 10 A2 N2
(ii) 44 A2 N2
[4 marks]
Question
The following frequency distribution of marks has mean 4.5.
Find the value of x.
Write down the standard deviation.
Answer/Explanation
Markscheme
\(\sum {fx = 1(2) + 2(4) + \ldots + 7(4)} \) , \(\sum {fx = 146 + 5x} \) (seen anywhere) A1
evidence of substituting into mean \(\frac{{\sum {fx} }}{{\sum f }}\) (M1)
correct equation A1
e.g. \(\frac{{146 + 5x}}{{34 + x}} = 4.5\) , \(146 + 5x = 4.5(34 + x)\)
\(x = 14\) A1 N2
[4 marks]
\(\sigma = 1.54\) A2 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 histogram below shows the time T seconds taken by 93 children to solve a puzzle.
The following is the frequency distribution for T .
(i) Write down the value of p and of q .
(ii) Write down the median class.
A child is selected at random. Find the probability that the child takes less than 95 seconds to solve the puzzle.
Consider the class interval \(45 \le T < 55\) .
(i) Write down the interval width.
(ii) Write down the mid-interval value.
Hence find an estimate for the
(i) mean;
(ii) standard deviation.
John assumes that T is normally distributed and uses this to estimate the probability that a child takes less than 95 seconds to solve the puzzle.
Find John’s estimate.
Answer/Explanation
Answer/Explanation
Markscheme
(i) \(p = 17\) , \(q = 11\) A1A1 N2
(ii) \(75 \le T < 85\) A1 N1
[3 marks]
evidence of valid approach (M1)
e.g. adding frequencies
\(\frac{{76}}{{93}} = 0.8172043 \ldots \)
\({\rm{P}}(T < 95) = \frac{{76}}{{93}} = 0.817\) A1 N2
[2 marks]
(i) 10 A1 N1
(ii) 50 A1 N1
[2 marks]
(i) evidence of approach using mid-interval values (may be seen in part (ii)) (M1)
\(79.1397849\)
\(\overline x = 79.1\) A2 N3
(ii) \(16.4386061\)
\(\sigma = 16.4\) A1 N1
[4 marks]
e.g. standardizing, \(z = 0.9648 \ldots \)
\(0.8326812\)
\({\rm{P}}(T < 95) = 0.833\) A1 N2
[2 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 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]