Printable · GCSE Foundation · ages 14-16
Statistics worksheet — GCSE Foundation
Fifteen questions across the statistics statements at Foundation tier. Choose the non-calculator filter to rehearse Paper 1, which counts for a third of the marks.
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Statistics worksheet — GCSE Foundation
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- 1.A quality inspector weighs a random sample of 50 packets of crisps from one day's production and finds their mean mass is 32.4 g. The factory makes 20,000 packets that day. Work out an estimate for the total mass, in kg, of all the packets made that day.
- 2.On a scatter graph the horizontal axis shows height in centimetres and the vertical axis shows mass in kilograms. One point is plotted at (170, 65). Write down the height and the mass of that person.
- 3.Isla wants to find out the favourite television programme of the pupils at her school. She asks only her own close group of friends. Write down what is wrong with her sample.
- 4.A sports shop in Cardiff sold 40 pairs of football boots last month: 4 pairs of size 6, 5 pairs of size 7, 8 pairs of size 8, 13 pairs of size 9 and 10 pairs of size 10. The manager will order 40 pairs for next month and wants as many pairs as possible to be in a size customers will buy. Work out the mean size and the modal size, and write down which of the two he should use.
- 5.Priya wants to find out the favourite sport of the 900 pupils at her school. She asks 5 pupils, chosen at random. Give a reason why her sample may not give a reliable result.
- 6.A bar chart shows the highest temperature recorded on each of five days. The temperatures were Monday 20 °C, Tuesday 22 °C, Wednesday 18 °C, Thursday 25 °C and Friday 23 °C. Write down the day on which the temperature was highest.
- 7.A stem-and-leaf diagram, described in words, shows the ages of 9 people at a family party. The stem is the tens digit: stem 1 has leaves 4 and 8; stem 2 has leaves 0, 3, 5 and 9; stem 3 has leaves 1 and 6; stem 4 has leaf 2. Work out the median age.
- 8.A sports centre in Ipswich has 2,000 members. It wants to know what its members think of its opening hours, so it asks a random sample of 100 of them. Write down what the population is in this survey.
- 9.A café sold 10 sandwiches on each of four days: 10, 10, 10, 10. Work out the range of the numbers sold.
- 10.The mean of the four numbers 10, 15, 20 and x is 18. Work out the value of x.
- 11.Noah wrote down the numbers 9, 3, 7, 1, 5 and said, “The median is 7, because 7 is in the middle of my list.” Is Noah right? Give a reason for your answer.
- 12.Grace asked 12 children how many brothers and sisters they have. Her results, in order, were 0, 0, 1, 1, 1, 1, 2, 2, 3, 3, 4, 6. Work out the median number of brothers and sisters.
- 13.Seven pupils were asked how many books they had read last month. Their answers were 3, 5, 5, 7, 8, 5, 3. Write down the mode.
- 14.Two classes sat the same test. The 30 pupils in Class A had a mean mark of 72. The 20 pupils in Class B had a mean mark of 82. Work out the mean mark of all 50 pupils.
- 15.The mean mass of four parcels is 17 kg. Three of the parcels have masses 12 kg, 16 kg and 18 kg. Work out the mass of the fourth parcel.
Answer key
- (d) 648 kg — Method: to estimate a total from a sample, multiply the sample's mean by the number of items in the whole population, then check the units the question asks for. Working: 32.4 g × 20,000 = 648,000 g. Converting to kilograms, 648,000 ÷ 1,000 = 648 kg. This is only an estimate, not an exact total, because it assumes every one of the 20,000 packets has exactly the sample mean mass, when in reality individual packets vary above and below it. Giving 1.62 kg multiplies the mean by 50, the SAMPLE size, instead of by 20,000, the number of packets actually made that day — this finds the total mass of the 50 sampled packets, not the day's production. Giving 32.4 kg treats the sample mean itself, in grams, as if it already were the day's total mass in kilograms, skipping the scaling up altogether. Giving 648,000 kg correctly scales the mean up to the whole day's production but never converts the answer from grams to kilograms, leaving it 1,000 times too large. Always scale a sample's mean up by the SIZE OF THE WHOLE POPULATION, and always finish by checking the units the question asks for.
- (c) Height 170 cm, mass 65 kg — Method: a point on a scatter graph is written as a pair of coordinates in which the horizontal value is written first and the vertical value second, so each value is matched to the quantity named on its own axis. Working: in (170, 65) the value 170 is the horizontal coordinate and the horizontal axis shows height in centimetres, so the height is 170 cm; the value 65 is the vertical coordinate and the vertical axis shows mass in kilograms, so the mass is 65 kg. Answer: height 170 cm, mass 65 kg, each with the unit named on its own axis. The distractors: height 65 cm and mass 170 kg come from reading the pair the wrong way round, which would describe an impossible person; height 170 cm and mass 170 kg come from reading the horizontal coordinate for both quantities and never using the second number; height 235 cm and mass 105 kg come from combining the two coordinates, 170 + 65 and 170 − 65, instead of reading them separately.
- (c) It is not representative, as she picked her own friends — Method: judge a sample by asking whether the pupils in it were chosen in a way that gives the whole school a fair chance of being heard. Working: Isla's friends are a group she formed herself, and friends tend to share tastes, so their favourite programme is likely to match hers rather than the school's, and pupils in other year groups and other friendship groups had no chance at all of being asked; the fault lies in how the pupils were selected, not in how many of them there were. Answer: it is not representative, as she picked her own friends. The distractors: the reply blaming the size claims the pupils were picked at random, which is false here, and it is the common mistake of thinking a biased sample can be cured by making it bigger; the reply calling the sample too large is false in the other direction, as a survey is never spoilt by collecting more replies; the reply that the sample is fine treats attending the school as enough, which would make any group of pupils in the building a fair sample.
- (a) The modal size, 9, bought by more customers than any other — Method: work out both averages from the frequencies, then choose the one the shop can act on. Working: for the mean, multiply each size by the number of pairs sold at it and add: 6 × 4 + 7 × 5 + 8 × 8 + 9 × 13 + 10 × 10 = 340, and 340 ÷ 40 = 8.5, so the mean size is 8.5. The largest frequency is 13, which belongs to size 9, so the modal size is 9. The mean 8.5 is a size no customer in the record asked for, so 40 pairs of it would sit unsold, while 13 of the 40 customers wanted size 9, more than wanted any other size. Answer: the modal size, 9, bought by more customers than any other. The distractors: the mean size 8.5 does take account of all 40 pairs, but a mean of sizes is a summary figure and not a size the month's customers were buying; the mean size 8 comes from averaging the five sizes on sale, 6 + 7 + 8 + 9 + 10 = 40 and 40 ÷ 5 = 8, which ignores how many pairs were sold at each size and so treats the 4 pairs of size 6 as equal in weight to the 13 pairs of size 9; the range 4 comes from 10 − 6 and measures spread, so it says how wide a set of sizes the shop must stock, not which size to stock most of.
- (a) 5 pupils are far too few to represent 900 pupils — Method: a sample can only support a claim about a population if it is chosen fairly and if it is large enough for the pattern in it to be more than chance. Working: Priya's method of choosing is fair, because the 5 pupils were picked at random, so every pupil had the same chance of being asked. The difficulty is the size: 900 ÷ 5 = 180, so each pupil she asks stands for 180 pupils. If two of the five happen to play in the same netball team, netball takes 40% of her sample on the strength of two answers, and a second sample of 5 could easily give a different favourite sport. Answer: 5 pupils are far too few to represent 900 pupils. The distractors: saying the pupils were not chosen at random contradicts the question, which states that they were; saying the 5 may each name a different sport describes what often happens in a small sample, but disagreement is not the fault, since 5 pupils who all named the same sport would be just as weak a basis for a claim about 900; saying a sample must hold at least half of the population is an invented rule, and a properly chosen sample of a few hundred can describe a population of many thousands.
- (c) Thursday — Method: on a bar chart the tallest bar belongs to the greatest value, so compare the five temperatures and then read off the day that the greatest one belongs to. Working: the temperatures are 20 °C, 22 °C, 18 °C, 25 °C and 23 °C; in order of size these are 18, 20, 22, 23 and 25, so the greatest temperature is 25 °C, and the day recorded with 25 °C is Thursday. Answer: Thursday — the answer is a day, not a temperature. The distractors: Friday comes from stopping at the final temperature listed instead of comparing all five; Wednesday comes from picking out the shortest bar, 18 °C, and so answering for the lowest temperature rather than the highest; Monday comes from writing down the first day in the chart without comparing any of the temperatures at all.
- (d) 25 — In order, the nine ages are 14, 18, 20, 23, 25, 29, 31, 36 and 42, and with 9 values the median is the 5th one, which is 25. Choosing 23 takes the 4th value instead of the 5th. Choosing 29 takes the 6th value instead of the 5th. Choosing 26 comes from averaging the 4th and 6th values, 23 + 29 = 52, and 52 ÷ 2 = 26, a method that is only needed when there is an even number of values.
- (b) All 2,000 members of the sports centre. — Method: in a survey, the population is the whole group the survey is trying to find out about, and the sample is the smaller group actually asked. Working: this survey wants to know what the sports centre's members think, so the population is every one of the 2,000 members — whether or not they were personally asked. Saying the population is the 100 members who were asked names the sample, not the population; the sample is drawn FROM the population, so it is smaller than it, not the same as it. Saying the population is everybody who lives in Ipswich widens the group far beyond who the survey is actually about — plenty of Ipswich residents are not members of the sports centre at all, so they are outside this survey altogether. Saying the population is the members who say they are unhappy confuses the population with a result of the survey: whether a member turns out to be happy or unhappy is something the survey finds out, not part of the definition of who is being studied. The population is always the whole group the question is about, before any sampling or any results come in.
- (d) 0 — Method: the range is the largest value minus the smallest value, whatever those two values turn out to be. Working: every value is 10, so the largest value is 10 and the smallest value is 10 as well, and the range is 10 − 10 = 0. Answer: 0 — a range of nothing says the data do not vary at all. The distractors: 10 comes from writing down the repeated value itself instead of the difference between the extremes; 20 comes from adding the largest and the smallest, 10 + 10, instead of subtracting; 40 comes from adding all four values, which gives the total sold and not a measure of spread.
- (a) 27 — Method: turn the mean into a total using total = mean × number of values, then subtract the numbers that are already known. Working: four numbers with a mean of 18 have a total of 18 × 4 = 72; the three known numbers give 10 + 15 + 20 = 45; so x = 72 − 45 = 27. Answer: 27, and checking, (10 + 15 + 20 + 27) ÷ 4 = 72 ÷ 4 = 18. The distractors: 72 comes from stopping at the total the four numbers must reach and never subtracting the known three; 18 comes from assuming the missing number must equal the mean; 45 comes from stopping at the total of the three known numbers.
- (b) No — in order the numbers are 1, 3, 5, 7, 9, so the median is 5. — Method: the median is the middle value of the data in order of size, so the data must be sorted before any position is read off. Working: Noah's list 9, 3, 7, 1, 5 is not in order; sorted it becomes 1, 3, 5, 7, 9, and with 5 values the middle position is the third, which now holds 5 rather than 7. Noah has read the third value of the unsorted list. Answer: no — in order the numbers are 1, 3, 5, 7, 9, so the median is 5. The distractors: the reply giving 3 as the median sorts the data correctly but then reads the value in the second place instead of the third; the reply that 7 is the third number he wrote accepts a position in the unsorted list, which is exactly the mistake the question is about; the reply using the mean claims a value of 7 for it, but the mean is 25 ÷ 5 = 5, so that reasoning is false as well.
- (a) 1.5 — Method: with an even number of values the median is the mean of the two middle values, which for 12 values are the 6th and the 7th once the data are in order. Working: the results are already in order, and 12 ÷ 2 = 6, so the middle pair are the 6th value, 1, and the 7th value, 2; the median is (1 + 2) ÷ 2 = 1.5. Answer: 1.5 brothers and sisters. The distractors: 1 comes from reading the 6th value and stopping there instead of averaging the middle pair; 2 comes from working out the mean, 24 ÷ 12, instead of the median; 6 comes from working out the range, 6 − 0, which measures spread rather than centre.
- (c) 5 — Method: the mode is the value that occurs most often, so count how many times each different value appears and compare the counts. Working: 3 appears twice, 5 appears three times, 7 appears once and 8 appears once, so the highest frequency is three and the value carrying it is 5. Answer: 5. The distractors: 3 comes from writing down the frequency of the most common answer instead of the answer itself; 7 comes from taking the middle number of the list as it was written, which applies the median without ordering the data and without answering the question asked; 8 comes from picking the largest value, which confuses the mode with the maximum.
- (c) 76 marks — Method: a mean of means only works when the groups are the same size, so rebuild each class's total mark, add the totals and divide by the number of pupils altogether. Working: Class A scored 30 × 72 = 2160 marks and Class B scored 20 × 82 = 1640 marks, giving 2160 + 1640 = 3800 marks between 50 pupils, so the overall mean is 3800 ÷ 50 = 76 marks. Answer: 76 marks. The distractors: 77 marks comes from averaging the two class means, (72 + 82) ÷ 2, which ignores the different class sizes; 78 marks comes from attaching each mean to the other class's size, (30 × 82 + 20 × 72) ÷ 50; 3800 marks comes from stopping at the combined total and never dividing by 50.
- (c) 22 kg — Method: multiply the mean by the number of parcels to rebuild the total mass, then subtract the masses that are known. Working: four parcels with a mean mass of 17 kg have a total mass of 17 × 4 = 68 kg; the three known parcels total 12 + 16 + 18 = 46 kg; so the fourth parcel has mass 68 − 46 = 22 kg. Answer: 22 kg. The distractors: 68 kg comes from stopping at the total mass of all four parcels; 17 kg comes from assuming the missing parcel must have the mean mass; 5 kg comes from multiplying the mean by 3, the number of parcels whose mass is given, leaving 51 − 46 = 5.
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