Question
The table below shows the number of words in the extended essays of an IB class.
Draw a histogram on the grid below for the data in this table.
[3]
Write down the modal group.[1]
The maximum word count is \(4000\) words.
Write down the probability that a student chosen at random is on or over the word count.[2]
Answer/Explanation
Markscheme
(A3) (C3)
Notes: (A3) for correct histogram, (A2) for one error, (A1) for two errors, (A0) for more than two errors.
Award maximum (A2) if lines do not appear to be drawn with a ruler.
Award maximum (A2) if a frequency polygon is drawn.[3 marks]
\({\text{Modal group}} = 3800 \leqslant w < 4000\) (A1) (C1)[1 mark]
\({\text{Probability}} = \frac{3}{{35}}{\text{ }}(0.0857{\text{, }}8.57\% )\) (A1)(A1) (C2)
Note: (A1) for correct numerator (A1) for correct denominator.[2 marks]
Question
The following histogram shows the weights of a number of frozen chickens in a supermarket. The weights are grouped such that \(1 \leqslant {\text{weight}} < 2\), \(2 \leqslant {\text{weight}} < 3\) and so on.
Find the total number of chickens.[1]
Write down the modal group.[1]
Gabriel chooses a chicken at random.
Find the probability that this chicken weighs less than \(4{\text{ kg}}\).[2]
Answer/Explanation
Markscheme
\(96\) (A1) (C1)[1 mark]
\(3 \leqslant {\text{weight}} < 4{\text{ kg}}\) . Accept \(3 – 4{\text{ kg}}\) (A1) (C1)[1 mark]
For adding three heights or subtracting \(14\) from \(96\) (M1)
\(\frac{{82}}{{96}}{\text{ }}(0.854{\text{ or }}\frac{{41}}{{48}}{\text{, }}85.4\% )\) (ft) from (b). (A1)(ft) (C2)[2 marks]
Question
The distribution of the weights, correct to the nearest kilogram, of the members of a football club is shown in the following table.
On the grid below draw a histogram to show the above weight distribution.
[2]
Write down the mid-interval value for the \(40 – 49\) interval.[1]
Find an estimate of the mean weight of the members of the club.[2]
Write down an estimate of the standard deviation of their weights.[1]
Answer/Explanation
Markscheme
(A1)(A1) (C2)
Notes: (A1) for all correct heights, (A1) for all correct end points (\(39.5\), \(49.5\) etc.).
Histogram must be drawn with a ruler (straight edge) and endpoints must be clear.
Award (A1) only if both correct histogram and correct frequency polygon drawn. [2 marks]
\(44.5\) (A1) (C1)
Note: If (b) is given as \(45\) then award
(b) \(45\) (A0)
(c) \(58.8{\text{ kg}}\) (M1)(A1)(ft) or (C2)(ft) if no working seen.
(d) \(8.44\) (C1)[1 mark]
Unit penalty (UP) applies in this question.
\({\text{Mean}} = \frac{{44.5 \times 6 + 54.5 \times 18 + \ldots }}{{42}}\) (M1)
Note: (M1) for a sum of frequencies multiplied by midpoint values divided by \(42\).
\( = 58.3{\text{ kg}}\) (A1)(ft) (C2)
Note: Award (A1)(A0)(AP) for \(58\).
Note: If (b) is given as \(45\) then award
(b) \(45\) (A0)
(c) \(58.8{\text{ kg}}\) (M1)(A1)(ft) or (C2)(ft) if no working seen.
(d) \(8.44\) (C1)[2 marks]
\({\text{Standard deviation}} = 8.44\) (A1) (C1)
Note: If (b) is given as \(45\) then award
(b) \(45\) (A0)
(c) \(58.8{\text{ kg}}\) (M1)(A1)(ft) or (C2)(ft) if no working seen.
(d) \(8.44\) (C1)[1 mark]
Question
120 Mathematics students in a school sat an examination. Their scores (given as a percentage) were summarized on a cumulative frequency diagram. This diagram is given below.
Complete the grouped frequency table for the students.
[3]
Write down the mid-interval value of the \(40 < x \leqslant 60\) interval.[1]
Calculate an estimate of the mean examination score of the students.[2]
Answer/Explanation
Markscheme
(A1)(A1)(A1) (C3)[3 marks]
50 (A1) (C1)[1 mark]
\({\text{Mean}} = \frac{{10 \times 14 + ……. + 90 \times 6}}{{120}}\) (M1)
Note: Award (M1) for correct substitution of their values from (a) in mean formula.
\( = 45\frac{2}{3}(45.7)\) (A1)(ft) (C2)[2 marks]
Question
The weights of 90 students in a school were recorded. The information is displayed in the following table.
Write down the mid interval value for the interval \(50 \leqslant w \leqslant 60\).[1]
Use your graphic display calculator to find an estimate for the mean weight.[2]
Use your graphic display calculator to find an estimate for the standard deviation.[1]
Find the weight that is 3 standard deviations below the mean.[2]
Answer/Explanation
Markscheme
55 (A1) (C1)[1 mark]
\(62.\bar 5{\text{ }}\left( {62.6} \right)\) (A2)(ft) (C2)[2 marks]
8.86 (A1) (C1)
Note: Follow through from their answer to part (a).[1 mark]
62.6 – 3 × 8.86 = 36.0 (M1)(A1)(ft) (C2)
Note: Accept 36.
Follow through from their values in part (b) only if working is seen.[2 marks]
Question
The weights in kg, of 80 adult males, were collected and are summarized in the box and whisker plot shown below.
Write down the median weight of the males.[1]
Calculate the interquartile range.[2]
Estimate the number of males who weigh between \(61\) kg and \(66\) kg.[1]
Estimate the mean weight of the lightest \(40\) males.[2]
Answer/Explanation
Markscheme
\(61\) kg (A1) (C1)[1 mark]
\(66 – 52\) (A1)
\( = 14\) (A1)(ft) (C2)
Note: Award (A1) for identifying quartiles, (A1)(ft) for correct subtraction of their quartiles.[2 marks]
\(20\) (A1) (C1)[1 marks]
\(\frac{{49.5 \times 20 + 56.5 \times 20}}{{40}}\) (M1)
Note: Award (M1) for multiplication of midpoints by frequencies.
\(= 53\) kg (A1) (C2)[2 marks]
Question
Toronto’s annual snowfall, x, in cm, has been recorded for the past 176 years. The results are shown in the table.
Write down the modal class.[1]
Write down the mid interval value for the class 6 ≤ x < 10 .[1]
Calculate an estimate of the mean annual snowfall.[2]
Find the number of years for which the annual snowfall was at least 18 cm.[2]
Answer/Explanation
Markscheme
14 ≤ x < 18 (A1) (C1)[1 mark]
8 (A1) (C1)[1 mark]
\(\frac{{{\text{4}} \times {\text{30 + 8}} \times {\text{26 + 12}} \times {\text{29 + 16}} \times {\text{32 + 20}} \times {\text{18 + 24}} \times {\text{27 + 28}} \times {\text{14}}}}{{176}}\) (M1)
Notes: 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.
= 14.7 (cm) (14.7045…) (A1)(ft) (C2)
Notes: Follow through from their answer to part (b). If a final incorrect answer that is consistent with their (b) is given award (M1)(A1)(ft) even if no working is seen.[2 marks]
18 + 27 + 14 (M1)
Note: Award (M1) for adding 18, 27 and 14.
= 59 (A1) (C2)[2 marks]
Question
The cumulative frequency graph represents the speed, s, in \({\text{km }}{{\text{h}}^{ – 1}}\), of 80 cars passing a speed camera.
Write down the number of cars passing the camera with speed of less than or equal to 50 \({\text{km}}\,{{\text{h}}^{ – 1}}\).[1]
Complete the following grouped frequency table for \(s\), the speed of the cars passing the camera.
[1]
Write down the mid-interval value of the \(50 < s \leqslant 70\) interval.[1]
Use your graphic display calculator to find an estimate of
(i) the mean speed of the cars passing the camera;
(ii) the standard deviation of the speed of the cars passing the camera.[3]
Answer/Explanation
Markscheme
10 (A1) (C1)[1 mark]
(A1)(ft) (C1)
Note: Follow through from their answer to part (a).[1 mark]
60 (A1) (C1)[1 mark]
(i) \(67.5 \left( {{\text{km}}\,{{\text{h}}^{ – 1}}} \right)\) (A2)(ft)
Notes: Award (M1) for an attempt to use the formula for the mean with at least two midpoint values consistent with their answer to part (c). Follow through from their table in part (b).
(ii) \(18.6 (18.6413…)\) (A1)(ft) (C3)
Note: Follow through from their table in part (b).[3 marks]
Question
A tetrahedral (four-sided) die has written on it the numbers 1, 2, 3 and 4. The die is rolled many times and the scores are noted. The table below shows the resulting frequency distribution.
The die was rolled a total of 100 times.
The mean score is 2.71.
Write down an equation, in terms of \(x\) and \(y\), for the total number of times the die was rolled.[1]
Using the mean score, write down a second equation in terms of \(x\) and \(y\).[2]
Find the value of \(x\) and of \(y\).[3]
Answer/Explanation
Markscheme
\(18 + x + y + 22 = 100\) or equivalent (A1) (C1)[1 mark]
\(\frac{{18 + 2x + 3y + 88}}{{100}} = 2.71\) or equivalent (M1)(A1) (C2)
Note: Award (M1) for a sum including \(x\) and \(y\), divided by 100 and equated to 2.71, (A1) for a correct equation.[2 marks]
\(x + y = 60\) and \(2x + 3y = 165\) (M1)
Note: Award (M1) for obtaining a correct linear equation in one variable from their (a) and their (b).
This may be implied if seen in part (a) or part (b).
\(x = 15;{\text{ }}y = 45\) (A1)(ft)(A1)(ft) (C3)
Notes: Follow through from parts (a) and (b), irrespective of working seen provided the answers are positive integers.[3 marks]
Question
The histogram shows the time, t, in minutes, that it takes the customers of a restaurant to eat their lunch on one particular day. Each customer took less than 25 minutes.
The histogram is incomplete, and only shows data for 0 ≤ t < 20.
The mean time it took all customers to eat their lunch was estimated to be 12 minutes.
It was found that k customers took between 20 and 25 minutes to eat their lunch.
Write down the mid-interval value for 10 ≤ t < 15.[1]
Write down the total number of customers in terms of k.[1]
Calculate the value of k.[3]
Hence, complete the histogram.[1]
Answer/Explanation
Markscheme
12.5 (A1) (C1)[1 mark]
33 + k OR 10 + 8 + 5 + 10 + k (A1)
Note: Award (A1) for “number of customers = 33 + k”.[1 mark]
\(\frac{{2.5 \times 10 + 7.5 \times 8 + \ldots + 22.5 \times k}}{{33 + k}} = 12\) (M1)(A1)(ft)
Note: Award (M1) for substitution into the mean formula and equating to 12, (A1)(ft) for their correct substitutions.
(k =) 7 (A1)(ft) (C4)
Note: Follow through from part (b)(i) and their mid-interval values, consistent with part (a). Do not award final (A1) if answer is not an integer.[3 marks]
(A1)(ft) (C1)
Note: Follow through from their part (b)(ii) but only if the value is between 1 and 10, inclusive.[1 mark]
Question
Consider the frequency histogram for the distribution of the time, \(t\) , in minutes of telephone calls that Helen made last week.
Complete the frequency table for this distribution.
[2]
Write down the modal class.[1]
Write down the mid interval value of the \(10 < t \leqslant 15\) class.[1]
Use your graphic display calculator to find an estimate for the mean time.[2]
Answer/Explanation
Markscheme
(A2) (C2)
Note: Award (A2) for four correct entries, (A1) for three correct entries, (A0) otherwise.
\(0 < t \leqslant 5\) (A1) (C1)
\(12.5\) (A1) (C1)
\(\frac{{275}}{{30}}\) (M1)
Note: Award (M1) for division with \(275\) seen.
\(= 9.17\) (\(9.16666 \ldots \)) (A1)(ft) (C2)
Note: Follow through from their parts (a) and (c), irrespective of whether working is shown.