Printable · GCSE Higher · ages 14-16
Statistics worksheet — GCSE Higher
Fifteen questions across the statistics statements at Higher 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 Higher
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- 1.In a histogram of the distances, d metres, thrown by some athletes, the bar covering 20 ≤ d < 60 has a constant frequency density of 1.8 per metre. Estimate the number of throws of at least 20 metres but less than 35 metres.
- 2.The number of pets owned by each of 19 pupils in a class is recorded: 0 pets — 7 pupils, 1 pet — 3 pupils, 2 pets — 4 pupils, 3 pets — 5 pupils. Work out the median number of pets.
- 3.The times, t minutes, of 70 visits to a website are grouped into two classes: 0 ≤ t < 4, which contains 30 visits, and 4 ≤ t < 20, which contains 40 visits. A histogram is drawn. Work out how many times taller the bar for 0 ≤ t < 4 is than the bar for 4 ≤ t < 20.
- 4.A school has 1200 pupils. A teacher wants to take a random sample of 60 of them. Write down which of these methods gives a random sample.
- 5.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.
- 6.Work out the median of these six numbers: 13, 21, 22, 36, 37, 47
- 7.A random sample of 50 pupils at a school were asked whether they prefer sport to music. 30 of the 50 pupils said they prefer sport. The school has 500 pupils altogether. Work out an estimate for the number of the 500 pupils who prefer sport.
- 8.The mean of three numbers is 50. A fourth number, 100, is added to the set. Work out the mean of the four numbers.
- 9.Forty pupils in class P and forty pupils in class Q each solved a puzzle. The times, in seconds, were summarised using cumulative frequency. For class P the lower quartile is 24, the median is 38 and the upper quartile is 46. For class Q the lower quartile is 30, the median is 35 and the upper quartile is 44. Write down the statement that correctly compares the two classes.
- 10.A scientist has grouped the lifetimes, in hours, of 300 batteries into classes of unequal width. She wants a diagram in which the number of batteries in a class is given by the area of its bar. Write down the type of diagram she should draw.
- 11.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.
- 12.Four pairs of variables are listed below. Write down the pair that you would expect to show negative correlation.
- 13.The masses, m kg, of 60 parcels are grouped like this: 0 ≤ m < 5, 22 parcels; 5 ≤ m < 10, 20 parcels; 10 ≤ m < 20, 9 parcels; 20 ≤ m < 30, 5 parcels; 30 ≤ m < 50, 4 parcels. Write down the class interval that contains the median mass.
- 14.In a histogram of the lengths, x cm, of some rods, the bar for 10 ≤ x < 30 has a frequency density of 3 per cm. The bar for 30 ≤ x < 45 is twice as tall as the bar for 10 ≤ x < 30. Work out the number of rods with a length in the class 30 ≤ x < 45.
- 15.An estate agent in Leeds lists the prices of the five houses sold on one street last year: £170,000, £150,000, £580,000, £140,000 and £160,000. A newspaper wants to print one figure for a typical price on this street. Work out the mean and the median, and write down which figure the newspaper should print.
Answer key
- (c) 27 — Method: a frequency is the area of the part of the bar being asked about, so frequency = frequency density × the width of that part. Working: the part asked about runs from 20 to 35, so its width is 35 − 20 = 15 metres; the frequency density there is 1.8 per metre, so the estimate is 1.8 × 15 = 27. Answer: about 27 throws. The distractors: 72 comes from taking the whole bar, 1.8 × 40, and so counting every throw from 20 up to 60; 1.8 comes from reading the height of the bar as a frequency, when a height is a density and only an area is a count; 63 comes from using the upper value 35 as the width, 1.8 × 35, instead of the width 35 − 20.
- (b) 1 — Method: for data in a frequency table, find the position of the median using (n + 1) ÷ 2, then read off the value at that position from the cumulative frequencies. Working: there are 19 pupils, so the median is the 10th value. The cumulative frequencies are 7 (up to 0 pets), 10 (up to 1 pet), 14 (up to 2 pets) and 19 (up to 3 pets). The 10th value falls at the end of the '1 pet' group, so the median is 1 pet. Giving 0 pets is the mode — the category with the highest frequency, 7 — not the median. Giving 3, the highest number of pets minus the lowest, finds the range, a different statistic entirely. Giving 19 states the total number of pupils, not a number of pets at all. Find the middle POSITION first, then read off the value it belongs to — do not confuse it with the mode, the range or the total.
- (d) 3 — Method: the height of a bar is its frequency density, frequency ÷ class width, so work out both heights and divide one by the other. Working: the class 0 ≤ t < 4 is 4 minutes wide and holds 30 visits, so its frequency density is 30 ÷ 4 = 7.5 per minute; the class 4 ≤ t < 20 is 16 minutes wide and holds 40 visits, so its frequency density is 40 ÷ 16 = 2.5 per minute; dividing the heights, 7.5 ÷ 2.5 = 3. Answer: the first bar is 3 times as tall. The distractors: 0.75 comes from comparing the frequencies, 30 ÷ 40, as though the frequencies were the heights, which is the mistake the unequal widths are there to expose; 4 comes from comparing the class widths, 16 ÷ 4, instead of the heights; 5 comes from subtracting the two frequency densities, 7.5 − 2.5, which answers how much taller rather than how many times taller.
- (a) Drawing 60 names at random from a list of all 1200 pupils — Method: a sample is random when every member of the population has the same chance of being chosen and nobody, including the pupils themselves, can influence who ends up in it; test each method against that. Working: drawing names from a list of all 1200 pupils gives each pupil the same chance, 60 out of 1200, whatever their year group, class or opinion, so the method is random. Answer: drawing 60 names at random from a list of all 1200 pupils. The distractors: asking the pupils who volunteer is self-selection, and the pupils with the strongest views volunteer first, so they decide the sample; asking the pupils nearest the door is convenience sampling, which reaches only those who happen to be in one place at one time; asking two Year 10 classes samples a cluster, so every pupil in the other year groups has no chance of being chosen at all.
- (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) 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) 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.
- (a) 62.5 — Method: a mean cannot be averaged with a new value — rebuild the total, add the new value to it, then divide by the new count. Working: three numbers with a mean of 50 have a total of 50 × 3 = 150; adding 100 makes the total 150 + 100 = 250; there are now 4 numbers, so the new mean is 250 ÷ 4 = 62.5. Answer: 62.5. The distractors: 75 comes from averaging the old mean with the new value, (50 + 100) ÷ 2, which ignores that three numbers pull against one; 50 comes from assuming an extra value leaves the mean unchanged; 37.5 comes from dividing the old total of 150 by the new count of 4, adding the new value to the count but not to the total.
- (b) Q was faster on average and more consistent — Method: compare the medians for the average and the interquartile ranges for the spread, remembering that a shorter time is faster and a smaller interquartile range means more consistent. Working: the median for class Q is 35 seconds against 38 seconds for class P, so class Q was faster on average; the interquartile range for class P is 46 − 24 = 22 seconds and for class Q it is 44 − 30 = 14 seconds, so class Q's times are more tightly grouped. Answer: class Q was faster on average and more consistent. The distractors: calling Q slower comes from comparing the lower quartiles, 30 against 24, as though a quartile were the average; calling Q less consistent comes from using the gap between the median and the upper quartile as the spread, 44 − 35 = 9 against 46 − 38 = 8, instead of the full interquartile range; the statement that Q was both slower and less consistent comes from making both of those mistakes together.
- (c) A histogram, with frequency density up the vertical axis — Method: decide which diagram makes area stand for frequency, which is the property the question asks for. Working: on a histogram the vertical axis is frequency density, so the area of a bar is frequency density × class width, and that product is the frequency; this is exactly what is wanted, and it is what allows classes of unequal width to be shown fairly. Answer: a histogram, with frequency density up the vertical axis. The distractors: a bar chart plots frequency as the height, so with unequal widths a wide class would cover far more area than a narrow class holding the same number of batteries, and area would measure nothing; a cumulative frequency diagram plots running totals against upper class boundaries, so a point on it gives how many lie below a value rather than how many lie in a class; a pie chart shows each class as a share of the whole 300 and loses the class widths entirely, so no area on it is tied to a scale of hours.
- (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.
- (a) Minutes a candle has burned and length remaining — As a candle burns for longer, less of it remains, so these two variables move in opposite directions as one increases — that is negative correlation. A pupil's shoe size generally increases as they get older, so age and shoe size show positive correlation, not negative, since both rise together. A football team's shirt colour is not a numerical quantity linked to how many matches it wins, so shirt colour and number of wins show no correlation at all. The number of letters in a pupil's name has no real connection to their ability in maths, so that pair also shows no correlation.
- (b) 5 ≤ m < 10 — Method: with 60 values the median is the 60 ÷ 2 = 30th value in order, so build a running total until it first reaches 30. Working: the running totals are 22 after the first class, 22 + 20 = 42 after the second, 51 after the third, 56 after the fourth and 60 after the fifth; the 30th parcel is past 22 but not past 42, so it lies in the second class. Answer: the median lies in the class 5 ≤ m < 10. The distractors: 0 ≤ m < 5 comes from giving the class with the greatest frequency, 22, which is the modal class and not the median class; 10 ≤ m < 20 comes from choosing the middle class in the list of five instead of counting to the middle value; 20 ≤ m < 30 comes from halving the range of the data, 50 ÷ 2 = 25, and giving the class that contains 25 kg rather than the class that contains the 30th parcel.
- (c) 90 — Method: the height of a bar is its frequency density, so twice as tall means twice the frequency density — not twice the frequency, because the two classes have different widths. Then frequency = frequency density × class width. Working: the first bar has frequency density 3 per cm, so the second has frequency density 2 × 3 = 6 per cm; the class 30 ≤ x < 45 is 45 − 30 = 15 cm wide, so its frequency is 6 × 15 = 90. Answer: 90 rods. The distractors: 120 comes from doubling the first bar's frequency instead of its height — the first class holds 3 × 20 = 60 rods, and doubling that ignores the fact that the second class is narrower; 45 comes from using the first bar's frequency density, 3, for the second bar, 3 × 15, and so never using the information that it is twice as tall; 6 comes from stopping at the frequency density of the taller bar and quoting a height as though it were a count.
- (c) The median, £160,000, as one very high price lifts the mean — Method: find both averages, then choose the one that sits closer to the bulk of the data. Working: in order the prices are 140,000, 150,000, 160,000, 170,000 and 580,000, so the median is the third of the five, £160,000. For the mean, 140,000 + 150,000 + 160,000 + 170,000 + 580,000 = 1,200,000 and 1,200,000 ÷ 5 = 240,000, so the mean is £240,000. Four of the five houses sold for £170,000 or less, so a reader told that a typical price is £240,000 would expect to pay at least £70,000 more than any of those four cost. Answer: the median, £160,000, as one very high price lifts the mean. The distractors: £580,000 is the middle value of the list as it is printed, which is the median only when the values have first been put in order; £240,000 is the mean, chosen on the ground that a median ignores three of the five prices, but a median uses all five to find which one is central and is then untroubled by how extreme the outer values are; £155,000 comes from deleting the £580,000 house and taking the mean of what is left, since 140,000 + 150,000 + 160,000 + 170,000 = 620,000 and 620,000 ÷ 4 = 155,000, but a real sale may not be thrown away merely for being large.
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