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Either paper 2 marks
Isla records the number of visitors entering a museum each hour for 24 hours. Here are the results.
39, 46, 5, 30, 32, 7, 23, 17, 48, 57, 19, 14, 18, 46, 35, 7, 38, 6, 42, 9, 46, 41, 45, 5
Isla makes this tally and frequency chart to put the data into groups.
Number of visitors Tally Frequency 0 to 10 10 to 20 20 to 30 30 to 40 40 to 50 Make two criticisms of Isla's tally and frequency chart. You do not need to complete the chart.
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Either order. (1) There is no group for values above 50, but 57 is in the data. (2) the groups overlap: 30 could go in either "20 to 30" or "30 to 40".Test values from the list against the groups.
The largest value in the list is \(57\). The last group stops at \(50\), so \(57\) belongs to no group.
The list also contains \(30\). The groups on either side of it are \(20\) to \(30\) and \(30\) to \(40\), and \(30\) could go in either of them, so the groups overlap.
Two criticisms, in either order: there is no group for values above \(50\), but \(57\) is in the data; and the groups overlap, because \(30\) could go in either "20 to 30" or "30 to 40".
- B1no group for the largest value: nothing above 50, but 57 is in the data
- B1the groups overlap, shown with 30, which could go in either "20 to 30" or "30 to 40"
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Either paper 3 marks
The table shows information about the weights of the sacks of potatoes on a trailer.
Weight Number of sacks 20 kg 3 30 kg 2 40 kg 5 50 kg 4 60 kg 4 70 kg 2 Show that the total weight of the sacks of potatoes on the trailer is less than 950 kg.
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Total weight \(900\) kg, which is less than \(950\) kgMultiply each weight by the number of sacks of potatoes with that weight:
\(20 \times 3 = 60\)
\(30 \times 2 = 60\)
\(40 \times 5 = 200\)
\(50 \times 4 = 200\)
\(60 \times 4 = 240\)
\(70 \times 2 = 140\)
Add the six products:
\(60 + 60 + 200 + 200 + 240 + 140 = 900\)
The total weight is \(900\) kg, and \(900 < 950\), so the total weight is less than \(950\) kg.
- M1at least one weight multiplied by its frequency, e.g. \(40 \times 5 = 200\) or \(50 \times 4 = 200\)
- M1all six products added to give \(900\) kg
- A1conclusion that \(900\) is less than \(950\)
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Either paper 3 marks
The table shows the number of goals scored by a football team in each of 25 matches.
Number of goals Frequency 0 2 1 7 2 6 3 2 4 8 Total 25 Work out the mean number of goals.
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\(2.28\)Multiply each value by its frequency to find how many goals there were altogether:
\(0 \times 2 = 0\)
\(1 \times 7 = 7\)
\(2 \times 6 = 12\)
\(3 \times 2 = 6\)
\(4 \times 8 = 32\)
That is \(57\) goals in total, shared between the 25 matches:
\(57 \div 25 = 2.28\)
- M1each value multiplied by its frequency, \(7, 12, 6, 32\)
- M1total \(57\) divided by 25
- A1mean \(2.28\)
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Either paper 4 marks
The table shows the number of cars owned by each household on a street.
Number of cars Frequency 0 3 1 5 2 5 3 2 4 8 (a)How many households are there altogether?
(b)Write down the modal number of cars.
(c)Work out the median number of cars.
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(a) \(23\)
(b) \(4\)
(c) \(2\)Add the frequencies:
\(3 + 5 + 5 + 2 + 8 = 23\)
The mode is the value with the highest frequency. The highest frequency in the table is 8, which is against 4.
There are 23 values, so the median is the \(12\)th one when they are put in order.
Running totals of the frequencies: \(3,\ 8,\ 13,\ 15,\ 23\)
The \(12\)th value falls in the row for 2, so the median is 2.
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Either paper 3 marks
The grouped frequency table shows how long each person waited at a clinic.
Time in minutes (\(t\)) Frequency \(0 < t \leq 10\) 5 \(10 < t \leq 20\) 4 \(20 < t \leq 30\) 4 \(30 < t \leq 40\) 11 \(40 < t \leq 50\) 3 (a)State the modal class.
(b)Write down the class that contains the median.
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(a) \(30 < t \leq 40\)
(b) \(30 < t \leq 40\)The modal class is the one with the highest frequency. That is 11, so the modal class is \(30 < t \leq 40\).
There are 27 values altogether, so the median is the \(14\)th value.
Running totals of the frequencies: \(5,\ 9,\ 13,\ 24,\ 27\)
The \(14\)th value has not been reached until the running total passes it, which happens in the class \(30 < t \leq 40\).
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Either paper 5 marks
16 matches were played. Here are the numbers of goals scored in them.
\(2,\ 2,\ 0,\ 0,\ 2,\ 0,\ 3,\ 1,\ 0,\ 0,\ 0,\ 0,\ 0,\ 3,\ 0,\ 1\)
(a)Draw a frequency table for this data, using the values 0, 1, 2 and 3.
(b)Write down the mode.
(c)Calculate the mean number of goals. Give your answer correct to 1 decimal place.
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(a) frequencies \(9,\ 2,\ 3,\ 2\) for 0, 1, 2, 3
(b) \(0\)
(c) \(0.9\)Work through the list once, putting a tally mark against each value, then count the tallies.
The frequencies for 0, 1, 2 and 3 are \(9,\ 2,\ 3,\ 2\).
They add to \(16\), which checks the count against the number of matches.
The mode is the value with the largest frequency. That is \(0\), with a frequency of \(9\).
Multiply each value by its frequency and add:
\(0 \times 9 = 0\), \(1 \times 2 = 2\), \(2 \times 3 = 6\), \(3 \times 2 = 6\)
That is \(14\) goals altogether, shared between the \(16\) matches:
\(14 \div 16 = 0.9\)
- (a) M1values tallied from the list, at least two of the frequencies \(9,\ 2,\ 3,\ 2\) correct
- (a) A1all four frequencies \(9,\ 2,\ 3,\ 2\), adding to \(16\)
- (c) M1values multiplied by frequencies and added: \(14\)
- (c) A1\(14 \div 16 = 0.9\)
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Either paper 5 marks
Each student in a year group was asked how many times they visited a gym last week. The bar chart shows the results.
(a)How many students made more than 1 visit to the gym?
(b)Work out the mean number of visits to the gym. Give your answer as a decimal.
(c)One of the students is chosen at random.
Work out the probability that the student made fewer than 4 visits to the gym. Give your answer as a fraction in its simplest form.
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(a) \(29\)
(b) \(2.725\)
(c) \(\frac{5}{8}\)More than 1 means the bars for 2, 3, 4 and 5. Their heights read off the axis are \(5,\ 9,\ 8,\ 7\) students. Adding them:
\(29\) students
Read each height as a frequency and multiply it by its value:
\(0 \times 6 = 0\), \(1 \times 5 = 5\), \(2 \times 5 = 10\), \(3 \times 9 = 27\), \(4 \times 8 = 32\), \(5 \times 7 = 35\)
The total is \(109\). The number of students is the heights added up:
\(6 + 5 + 5 + 9 + 8 + 7 = 40\)
\(109 \div 40 = 2.725\)
Fewer than 4 means the bars for 0, 1, 2 and 3. Their heights read off the axis are \(6,\ 5,\ 5,\ 9\) students. Adding them:
\(25\) students
The total number of students is \(40\), found by adding all six heights.
\(P = \frac{25}{40} = \frac{5}{8}\)
- (b) M1values multiplied by heights and added: \(109\), with \(40\) students in all
- (b) A1\(109 \div 40 = 2.725\)
- (c) M1\(25\) students in the bars counted, out of \(40\)
- (c) A1probability \(\frac{5}{8}\) in its simplest form
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Either paper 3 marks
The bar chart shows the number of books borrowed from a library by each of 40 members last month.
Work out the mean number of books. Give your answer as a decimal.
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\(2.05\)Read the height of each bar as a frequency: \(12,\ 4,\ 4,\ 10,\ 10\). These add up to \(40\), the number in the sample.
Multiply each value (0, 1, 2, and so on) by its frequency, in order:
\(0,\ 4,\ 8,\ 30,\ 40\)
Adding these gives \(82\) books altogether.
\(82 \div 40 = 2.05\)
- M1each value multiplied by its bar height: \(0,\ 4,\ 8,\ 30,\ 40\)
- M1total \(82\) books
- A1\(82 \div 40 = 2.05\)
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Either paper 3 marks
The table shows the height of each of 50 sunflower seedlings in a greenhouse.
Height, \(h\) (cm) Frequency Midpoint \(4.5 \leq h < 5.0\) 18 \(5.0 \leq h < 5.5\) 17 \(5.5 \leq h < 6.0\) 15 Total 50 Work out an estimate of the mean height of a seedling.
Give your answer as a decimal.
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\(5.22\ \text{cm}\)Each value is only known to be somewhere in its class, so use the midpoint of the class as an estimate.
The midpoints are \(4.75\), \(5.25\) and \(5.75\).
Multiply each midpoint by its frequency:
First class: \(4.75 \times 18 = 85.5\)
Second class: \(5.25 \times 17 = 89.25\)
Third class: \(5.75 \times 15 = 86.25\)
Add the three results:
\(85.5 + 89.25 + 86.25 = 261\)
Divide by the total frequency:
\(261 \div 50 = 5.22\) cm
- B1midpoints \(4.75\), \(5.25\) and \(5.75\)
- M1each midpoint times its frequency, added: \(261\)
- A1dividing by the total frequency: \(5.22\)
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Either paper 4 marks
The table shows the distance cycled by each of 25 members of a cycling club last week.
Distance (\(d\) km) Frequency \(0 < d \leq 10\) 6 \(10 < d \leq 20\) 7 \(20 < d \leq 30\) 4 \(30 < d \leq 40\) 8 Total 25 Work out an estimate for the mean.
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\(20.6\ \text{km}\)The table does not say what the individual values were, only which class each one fell in, so use the midpoint of each class as an estimate of every value in it.
The midpoints are \(5\), \(15\), \(25\) and \(35\).
Multiply each midpoint by its frequency:
\(5 \times 6 = 30\)
\(15 \times 7 = 105\)
\(25 \times 4 = 100\)
\(35 \times 8 = 280\)
These add up to \(515\). Divide by the 25 values:
\(515 \div 25 = 20.6\)
- M1midpoints 5, 15, 25 and 35 used
- M1midpoint times frequency \(30\), \(105\), \(100\), \(280\)
- M1sum of the products \(515\) divided by 25
- A1estimated mean \(20.6\)