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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Answer key: Statistics worksheet — GCSE Foundation
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- (d) 75 — Method: to find a total from a bar chart, add together the height of every bar. Working: 12 + 18 + 9 + 15 + 21 = 75 books. Leaving out Wednesday's bar by mistake, 12 + 18 + 15 + 21 = 66, misses one day out of the total. Giving 21 states only Friday's bar, the tallest one, not the total of all five days. Dividing the total by the number of days, 75 ÷ 5 = 15, finds the mean number of books per day, not the total borrowed. Add up every single bar — do not stop at the biggest one, and do not divide once you have added them all.
- (d) 6 — Method: the mode, or modal value, is the value that occurs most often in the data set, and it is a value from the data rather than a count. Working: size 4 occurs twice, size 5 occurs once, size 6 occurs three times and size 9 occurs once, so the highest frequency is three and the size it belongs to is 6. Answer: 6. The distractors: 3 comes from writing down the frequency of the most common size instead of the size itself; 9 comes from picking the largest size in the list, which confuses the mode with the maximum; 4 comes from stopping at the first size that repeats rather than checking which size repeats most often.
- (a) The temperature is rising steadily — Method: the trend of a line graph is the overall direction of the readings as time goes on, found by comparing each reading with the one before it. Working: from 10 °C to 14 °C is a rise of 4 °C, and the same comparison from 14 °C to 18 °C, from 18 °C to 22 °C, from 22 °C to 26 °C and from 26 °C to 30 °C gives a rise of 4 °C every time; every reading is greater than the one before it and none of them falls. Answer: the temperature is rising steadily — steadily because the rise is the same size each hour. The distractors: falling steadily comes from reading the six values from right to left, which reverses the direction of time; stays the same comes from noticing that the step of 4 °C is the same each hour and describing the step as constant instead of the temperature; rises and then falls comes from assuming that a line graph has to turn at some point rather than reading the values that are actually given.
- (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.
- (a) 1,400 and 1,240, so combine the samples for one estimate — Method: scale each sample up to the whole stock, then use the fact that a larger sample gives a more reliable estimate than a smaller one. Working: the first sample gives 35 ÷ 50 = 0.7 and 0.7 × 2,000 = 1,400 paperbacks; the second gives 31 ÷ 50 = 0.62 and 0.62 × 2,000 = 1,240 paperbacks. Two random samples of the same size are expected to differ a little, so neither estimate is wrong. Putting the two together gives 35 + 31 = 66 paperbacks in 100 books, and 66 ÷ 100 = 0.66 with 0.66 × 2,000 = 1,320, an estimate resting on twice as many books as either volunteer checked. Answer: 1,400 and 1,240, so combine the samples for one estimate. The distractors: keeping 1,400 because it is larger picks an estimate by its size, when both samples held 50 books and neither has a stronger claim; saying a volunteer must have miscounted assumes two random samples ought to agree exactly, which is precisely what random sampling does not promise; 1,750 and 1,550 come from 35 × 50 = 1,750 and 31 × 50 = 1,550, multiplying each count by the size of the sample instead of scaling by 2,000 ÷ 50.
- (d) 38, and fairly confident, as 12 °C is inside the range — Method: substitute the forecast temperature into the equation of the line of best fit, then judge the prediction by where that temperature sits among the data the line was drawn from. Working: putting x = 12 into y = −3x + 74 gives −3 × 12 + 74 = 38, so the line predicts 38 hot chocolates. The recorded temperatures run from 4 °C to 18 °C, and 12 °C lies inside that interval, so this is interpolation, the safer kind of prediction. Answer: 38, and fairly confident, as 12 °C is inside the range; the owner should still expect the true figure to differ a little, since the points only lie near the line and not on it. The distractors: being completely certain treats a line of best fit as a rule that fixes each day's sales, when it describes a trend that individual days depart from; saying 12 °C is outside the range misreads the interval 4 °C to 18 °C, and the wrong warning would be attached to a sound prediction; 110 comes from −3 × 12 being taken as +36, giving 36 + 74 = 110, which loses the negative gradient and so predicts that a warm day sells more hot chocolate than a cold one.
- (a) x = 0 gives y = −20: a negative number sold — The y-intercept is the value the line predicts when x = 0: y = 3 × 0 − 20 = −20. A kiosk cannot sell a negative number of ice creams, so this is not a sensible estimate. The 3 in the equation is the gradient, not the intercept, so an option claiming x = 0 gives y = 3 has swapped the two numbers around — substituting x = 0 makes the 3x term equal 0, leaving −20, not 3. The danger of extrapolating to very high temperatures is a real issue with this line, but it is a different issue from the y-intercept, so it does not answer this question. And whether x = 0 could occur on a trading day is beside the point: the model still makes that prediction, and it is the prediction itself, −20, that is impossible.
- (d) Route 1, as its times vary by 6 minutes rather than 20 — Method: work out an average and a measure of spread for each route, then decide which matters to a commuter who must arrive on time every day. Working: for Route 1, 22 + 23 + 24 + 24 + 25 + 25 + 26 + 26 + 27 + 28 = 250 and 250 ÷ 10 = 25, so the mean is 25 minutes, and the range is 28 − 22 = 6 minutes. For Route 2, 18 + 19 + 20 + 20 + 21 + 22 + 26 + 30 + 36 + 38 = 250 and 250 ÷ 10 = 25, so the mean is also 25 minutes, but the range is 38 − 18 = 20 minutes. The means give no reason to prefer either route; the spreads do, because a commuter who must never be late has to allow for the worst day, which is 28 minutes on Route 1 and 38 minutes on Route 2. Answer: Route 1, as its times vary by 6 minutes rather than 20. The distractors: saying Route 2 has the lower mean assumes that its quicker-looking early times must pull the average down, when both routes total 250 minutes over the ten days; choosing Route 2 for its fastest journey of 18 minutes judges a route by its best day, and the commuter has to survive its worst; saying either route will do uses the equal means and ignores the spread altogether, which is the one thing that separates the two routes.
- (b) Treat them as outliers and check them before deciding — Method: a point lying a long way from the pattern the rest of the data make is called an outlier, and an outlier is investigated before anything is done with it, because it may be an error in the data or it may be a genuine but unusual case. Working: 48 of the 50 points lie close to the rising line of best fit, so the trend is set by those 48; the two remaining points do not follow it, so they are identified as outliers and checked — a mistake in measuring or recording would be corrected, while a genuine reading would be kept and reported. Answer: treat them as outliers and check them before deciding what to do with them. The distractors: deleting them at once assumes that every point far from the line must be an error, which throws away real data; moving the line so that it passes through them assumes a line of best fit must touch particular points, when it is drawn to follow all 50; taking them as proof that there is no correlation lets two points overturn the pattern that the other 48 agree on.
- (d) 29 — Method: with an even number of values there is no single middle value, so the median is the mean of the two values either side of the middle. Working: the six numbers are already in order and 6 ÷ 2 = 3, so the middle pair are the third and fourth values, 22 and 36; their mean is (22 + 36) ÷ 2 = 58 ÷ 2 = 29. Answer: 29, which lies between the two middle values as a median of an even data set must. The distractors: 22 comes from taking the lower of the two middle values and stopping there instead of averaging the pair; 36 comes from taking the larger value of that pair because it sits just past the halfway point of the list; 34 comes from working out the range, 47 − 13, instead of a measure of centre.
- (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.
- (d) How much the temperatures varied over the seven days — Method: the range of a set of values is the largest value take away the smallest, so it is built from two values only and it measures the gap they leave between them. Working: the largest of the seven readings is 7 °C and the smallest is 2 °C, so the range is 7 − 2 = 5 °C. That figure says the week's readings covered a band 5 °C wide; it names no particular day and no particular reading. Answer: the range describes how much the temperatures varied over the seven days. The distractors: the temperature that occurred most often is the mode, which here is 3 °C, and a mode counts repeats instead of measuring a gap; the temperature typical of the week is an average, and the range is not an average, since it throws away every value lying between the two extremes; the number of different temperatures recorded is 6, a count of how many distinct values appear, while the range is a difference between two of them.
- (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.
- (d) 300 pupils — Method: an estimate for a whole population is made by finding the proportion in the sample and applying that same proportion to the population. Working: in the sample 30 of the 50 pupils prefer sport, a proportion of 30 ÷ 50 = 0.6, and applying that proportion to the school gives 0.6 × 500 = 300 pupils. Answer: 300 pupils, and it is only an estimate, because a different random sample of 50 would give a slightly different figure. The distractors: 200 pupils comes from scaling up the 20 pupils in the sample who did not prefer sport, 20 × 10, which answers the opposite question; 150 pupils comes from reading 30 out of 50 as 30% and taking 30% of 500; 60 pupils comes from working out the proportion correctly as 60% and then writing the 60 down as a number of pupils instead of applying it to the 500.
- (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.
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