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.
Non-calculator
Answer key: Statistics worksheet — GCSE Foundation
MathsUKwww.geekhero.co.uk
- (d) About 510 of the 600 bulbs are likely to have flowered — Method: the proportion found in a random sample is used as an estimate of the proportion in the whole population, and the conclusion is stated as an estimate, never as a fact about every member. Working: 17 of the 20 bulbs dug up had flowered, so the sample proportion is 17 ÷ 20 = 0.85, and applying that proportion to the whole planting gives 0.85 × 600 = 510 bulbs. A different random sample of 20 would very probably give a slightly different figure, so 510 is an estimate. Answer: about 510 of the 600 bulbs are likely to have flowered. The distractors: saying exactly 510 have flowered takes an estimate from a sample of 20 as a count of all 600, which no sample can deliver; saying exactly 17 of the 600 have flowered reports the sample count as though it were the population count, leaving the other 580 bulbs out of the answer altogether; saying about 20 have flowered uses the size of the sample as the estimate, when 20 is the number of bulbs she dug up rather than a number that flowered.
- (d) 1/2 — Method: a percentage is turned into a fraction by writing it over 100 and then cancelling the fraction down to its simplest form. Working: the reading sector is 50% of the pie chart, so as a fraction it is 50/100; dividing the numerator and the denominator by 50 gives 1/2. Answer: 1/2 of the pie chart — a fraction of the chart, not a number of pupils. The distractors: 1/4 comes from reading the share of the sport sector, 25%, instead of the share of the reading sector; 1/3 comes from counting the three sectors and assuming that three sectors must each take a third of the chart, which is only true when the sectors are equal; 3/4 comes from adding the reading and sport sectors together, 50% + 25% = 75%, instead of taking the reading sector on its own.
- (d) 24 — Method: for a sample in proportion to the population, apply the same fraction that each group makes up of the whole population to the size of the sample. Working: women make up 300 out of the 500 members, a fraction of 300 ÷ 500 = 0.6. Applying that fraction to the sample of 40 gives 0.6 × 40 = 24 women. Splitting the sample evenly, 40 ÷ 2 = 20, ignores that the club has more women than men and treats the two groups as equal in size, which they are not. Misreading the sample size as 50 instead of 40, then applying the 3:2 ratio of women to men, 3 ÷ 5 × 50 = 30, uses the right ratio but the wrong sample total. Working out the number of MEN instead of women, 200 ÷ 500 × 40 = 16, answers a different question — how many men, not how many women, belong in the sample. Always apply each group's own share of the population to the sample size, and check which group the question is actually asking about.
- (b) Positive correlation — Method: the type of correlation is named from the direction the points take as the scatter graph is read from left to right. Working: the points rise from left to right, so as the arm span read on the horizontal axis increases, the height read on the vertical axis increases as well; two quantities that increase together show positive correlation. Answer: positive correlation. The distractors: negative correlation comes from naming the direction the wrong way round, since a negative correlation needs the points to fall as the graph is read from left to right; no correlation comes from treating points that are spread out rather than sitting exactly on a line as though they showed no relationship; direct proportion comes from confusing a rising trend with proportion, which would additionally need the line through the points to pass through the origin and would mean doubling one quantity doubles the other.
- (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.
- (b) Interpolation — The salary is being estimated for a value of x between 1 and 15, which is inside the range of x-values that were actually plotted, so this is interpolation. Extrapolation would apply if the estimate used a value of x below 1 or above 15, outside the plotted range. Correlation describes the relationship between the two variables, not the reliability of an estimate, and causation describes one variable actually causing a change in the other, which is a different idea altogether — neither is the word being asked for here.
- (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.
- (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) 72 — Method: when a pie chart is divided into equal sectors, each sector stands for the same share of the people asked, so the fraction of the sectors that are shaded is also the fraction of the people. Working: 6 sectors out of 10 are purple, which is the fraction 6/10 of the whole pie chart; one tenth of the 120 people is 120 ÷ 10 = 12 people, so six tenths is 6 × 12 = 72 people. Answer: 72 people, a count of people rather than a number of sectors. The distractors: 48 comes from working with the 4 sectors that are not purple, 4 × 12, and so answering for the wrong part of the chart; 60 comes from turning the fraction 6/10 into 60% and then writing the 60 down as though it were a number of people; 6 comes from writing down the number of purple sectors instead of the number of people those sectors stand for.
- (d) 19 kg — Method: estimate the mean of grouped data by multiplying each class's midpoint by its frequency, adding the four totals, then dividing by the total frequency. Working: the midpoints are 5, 15, 25 and 35 kg. The weighted totals are 11 × 5 = 55, 5 × 15 = 75, 5 × 25 = 125 and 9 × 35 = 315, which add to 570. Dividing by the 30 dogs gives an estimate of 570 ÷ 30 = 19 kg. Giving 5 kg reads off the midpoint of the modal class, 0 < m ≤ 10, the class with the most dogs — but the class with the most dogs is not where the mean falls, and neither is a substitute for actually calculating it. Giving 20 kg averages the four midpoints, (5 + 15 + 25 + 35) ÷ 4, treating every class as equally likely and ignoring that far more dogs are in the lightest and heaviest classes than in the middle two. Giving 570 kg stops after finding the correct weighted total and forgets the final division by the 30 dogs. Always weight each midpoint by its own frequency, and always finish by dividing by the total frequency, not the number of classes.
- (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) 90° — Method: the sectors of a pie chart share the 360° at the centre of the chart in the same proportion as the data, so a sector's angle is its share of the data multiplied by 360°. Working: a share of 25% is the fraction 25/100, which is one quarter of the whole chart, and one quarter of 360° is 360 ÷ 4 = 90°. Answer: 90°, an angle in degrees rather than a percentage. The distractors: 25° comes from sharing out 100 instead of 360, so the percentage is written straight down as a number of degrees; 14.4° comes from dividing 360 by 25 instead of multiplying 360 by the fraction 25/100, which is the division done the wrong way round; 45° comes from taking a quarter of 180° instead of a quarter of 360°, treating the pie chart as a semicircle.
- (a) Equal means; Class A is more consistent, smaller range. — Method: when two data sets share a measure of location, compare a measure of spread to say more about consistency. Working: both classes have the same mean mark, 14, so on average they performed equally well. Class A has the smaller range, 6, so its marks are more tightly grouped around 14 than Class B's marks, which vary by as much as 14. So Class A's marks were more consistent, even though neither class did better on average. Saying Class B did better because it has the bigger range confuses a wide spread with a high score — a big range describes variability, not performance. Saying Class A did better because it has the smaller range makes the same mistake in the other direction: the two classes are tied on the mean, so neither one 'did better'. Saying the classes cannot be compared because their means are equal misses the whole point of also comparing the range. Always compare both an average AND a spread before describing two data sets — either one alone tells only half the story.
- (c) 20 kg ≤ mass < 30 kg — The modal class is the class with the highest frequency. Reading the plotted points, the frequencies are 6, 10, 16, 6 and 2, so the highest frequency is 16, plotted at the midpoint 25. A class of width 10 centred on 25 runs from 25 − 5 = 20 to 25 + 5 = 30, so the modal class is 20 kg ≤ mass < 30 kg. Writing '25 kg' gives only the midpoint, not the class — the modal class is an interval, not a single value. '10 kg ≤ mass < 20 kg' is the class before the peak, centred on 15, which has frequency 10, not the highest. '30 kg ≤ mass < 40 kg' is the class after the peak, centred on 35, which has frequency 6, not the highest.
- (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.
Build your own mix at the worksheet builder.