GCSE Maths Statistics (Higher): check-up and exam-style questions

Key words. An average is one number that stands for a whole set of data. The mean is the total divided by how many values there are. The median is the middle value when the data is in order. The mode is the value that appears most often. The range is the biggest value minus the smallest value.

Method. The mode is the value that appears most often.

1. Count how many times each value appears. The mode is the value that occurs most often.
   3 appears 3 times
   5 appears 2 times
   8 appears 1 time
2. Find the value that appears most often. 3 appears 3 times, more than any other value.
   Mode = 3

Check. Look back at the list: no other value appears more times. ✓

Watch out. The mode is the value, not how many times it appears.

Key words. An average is one number that stands for a whole set of data. The mean is the total divided by how many values there are. The median is the middle value when the data is in order. The mode is the value that appears most often. The range is the biggest value minus the smallest value.

Method. Range = biggest value − smallest value.

1. Find the biggest value. The range measures how spread out the data is, from one end to the other.
   Biggest = 8
2. Find the smallest value.
   Smallest = 3
3. Subtract: biggest minus smallest. The range is the gap between the two ends.
   8 − 3 = 5

Check. Count on from 3 by 5 and you reach 8. ✓

Watch out. The range is one number (the difference), not the pair “from 3 to 8”.

Method. In a two-way table every row adds up to its row total and every column adds up to its column total. Find a row or column with only one number missing.

1. Find the missing number in the year 10 row, car column. The row total is 35, and the cells in a row add up to the row total, so subtract the numbers we know.
   35 − 18 − 12 = 5
2. Write the answer. This is the number in the cell the question asks about.
   5

Check. Add up the completed row and column again: they agree with the totals. ✓

Watch out. Look for a row or column with exactly one missing number first.

Method. In a two-way table every row adds up to its row total and every column adds up to its column total. Find a row or column with only one number missing.

1. Find the missing number in the year 10 row, car column. The row total is 35, and the cells in a row add up to the row total, so subtract the numbers we know.
   35 − 18 − 12 = 5
2. Find the missing number in the year 11 row, walk column. The row total is 30, and the cells in a row add up to the row total, so subtract the numbers we know.
   30 − 9 − 7 = 14
3. Write the answer. This is the number in the cell the question asks about.
   14

Check. Add up the completed row and column again: they agree with the totals. ✓

Watch out. Look for a row or column with exactly one missing number first.

Method. In a two-way table every row adds up to its row total and every column adds up to its column total. Find a row or column with only one number missing.

1. Find the missing number in the year 10 row, car column. The row total is 35, and the cells in a row add up to the row total, so subtract the numbers we know.
   35 − 18 − 12 = 5
2. Find the missing number in the year 11 row, walk column. The row total is 30, and the cells in a row add up to the row total, so subtract the numbers we know.
   30 − 9 − 7 = 14
3. Find the missing number in the total row, bus column. The row total is 65, and the cells in a row add up to the row total, so subtract the numbers we know.
   65 − 32 − 12 = 21
4. Write the answer. This is the number in the cell the question asks about.
   21

Check. Add up the completed row and column again: they agree with the totals. ✓

Watch out. Look for a row or column with exactly one missing number first.

Key words. The population is everything we want to know about. A sample is a smaller part of it that we actually measure. A good sample is random (everyone has an equal chance) and representative (a fair mix).

Method. Capture–recapture: the fraction of the second sample that is marked is the same as the fraction of the whole population that is marked.

1. Write down what we know. First we catch, mark and release some. Later we catch a second sample and count how many of those are marked.
   Marked and released: 60
   Second sample: 50
   Marked in the second sample: 10
2. Write the proportion marked in the second sample. If the animals have mixed well, this is a fair estimate of the proportion marked in the whole population.
   10 out of 50 = 1050
3. Write the proportion marked in the whole population. Let N be the whole population. The 60 marked ones make up 60 out of N.
   60N
4. Make the two proportions equal. We assume the second sample is representative of the population.
   1050 = 60N
5. Solve for N. Multiply both sides by N and by the second-sample size, then divide by the number marked in the second sample: N = marked × second sample ÷ marked in second sample.
   N = 60 × 50 ÷ 10 = 3000 ÷ 10 = 300

Check. Of the whole population of 300, 60 were marked: 60300 ≈ 0.200, and 1050 = 0.200. The two proportions agree (to within rounding). ✓

Watch out. N = marked × second sample ÷ marked in the second sample. Do not divide by the second sample size or multiply by the number marked in it.

Key words. The population is everything we want to know about. A sample is a smaller part of it that we actually measure. A good sample is random (everyone has an equal chance) and representative (a fair mix).

Method. Think about what must be true for the second sample to be fair.

1. Think about mixing. If the marked fish stayed together, the second sample would not be fair.
   The marked fish mixed evenly with the rest.
2. Think about change. The population must not change between the two catches.
   No fish were born, died or moved in or out.

Check. Either assumption is accepted. ✓

Watch out. Mention the animals, not the arithmetic.

Method. Look at the direction and at how close the points are to the line.

1. Direction. The points slope upwards.
   Positive
2. Strength. The points are close to a line.
   Strong

Check. More hours of revision go with higher marks. ✓

Watch out. Say positive and strong.

Method. Read across or up from the line of best fit. The answer is an estimate, so a small tolerance is allowed.

1. Find 6 hours on the horizontal axis. We know the hours-value, so start on the hours-axis.
   hours = 6 hours
2. Go straight up to the line of best fit. The line of best fit shows the trend, so it gives us the best estimate.
   Draw a vertical line up from the axis until it meets the line.
3. Go across to the vertical axis and read the mark-value. Read the scale carefully.
   mark ≈ 73

Check. The point you read should lie on the line of best fit, not on one of the crosses. ✓

Watch out. Read from the line, not from the nearest cross.

Method. Compare 14 hours with the range of the data.

1. Range of the data. The hours in the graph go from 1 to 9.
   14 > 9: extrapolation
2. Check the answer. The line gives about 123, which is more than 100.
   A mark cannot exceed 100.

Check. Extrapolation is not reliable. ✓

Watch out. Name extrapolation and give a reason.

Key words. Grouped data is data put into classes (groups), such as 10 < x ≤ 20. We lose the exact values, so calculations are estimates. The midpoint of a class is halfway between its two ends.

Method. The modal class is the class with the highest frequency.

1. Look down the frequency column for the biggest number. The class that has the most values is the modal class.
   Frequencies: 6, 14, 15, 5. The highest is 15.
2. Read the class that goes with it. The answer is the class, not the frequency.
   Modal class = 160 < h ≤ 170 cm

Check. No other class has as many as 15 values. ✓

Watch out. Give the class (the group), not the frequency.

Key words. Grouped data is data put into classes (groups), such as 10 < x ≤ 20. We lose the exact values, so calculations are estimates. The midpoint of a class is halfway between its two ends.

Method. Estimate the mean of grouped data by using the midpoint of each class in place of the unknown values.

1. Find the midpoint of each class. We do not know the exact values inside a class, so we pretend every value is at the middle of its class. Midpoint = (lower end + upper end) ÷ 2.
   (140 + 150) ÷ 2 = 145
   (150 + 160) ÷ 2 = 155
   (160 + 170) ÷ 2 = 165
   (170 + 180) ÷ 2 = 175
2. Multiply each frequency by its midpoint. This estimates the total of the values in that class.

   Height	Frequency (f)	Midpoint	f × midpoint
   140 < h ≤ 150 cm	6	145	870
   150 < h ≤ 160 cm	14	155	2170
   160 < h ≤ 170 cm	15	165	2475
   170 < h ≤ 180 cm	5	175	875
   Total	40		6390

3. Add up the frequencies and the (f × midpoint) column. The first total is the number of values. The second is an estimate of the total of all the values.
   Total frequency = 40
   Total of f × midpoint = 6390
4. Divide to estimate the mean. estimated mean = total of (f × midpoint) ÷ total frequency.
   6390 ÷ 40 = 159.75 cm

Check. The estimate should be inside the range of the classes, from 140 to 180, and 159.75 is. It is only an estimate because the real values are not known. ✓

Watch out. Multiply the frequency by the midpoint, not by the class limits. Divide by the total frequency, not by the number of classes.

Method. Find Q2 Y3 on the horizontal axis and read the point.

1. Find the point. Q2 of Year 3 is the 10th point.
   Read up and across
2. Read the value.
   49 hundred

Check. The value is between Q1 (30) and Q3 (68). ✓

Watch out. Read the units: hundreds.

Method. Compare the quarters inside a year.

1. Look at each year. The values rise from Q1 to a peak in Q3 and then fall in Q4.
   Peak in Q3
2. Write it in context.
   Ice cream sales are highest in summer and lowest in winter.

Check. The same shape repeats every year. ✓

Watch out. A seasonal pattern repeats each year.

Method. The Q3 sales rise by 5 each year, so add 5 to Year 3.

1. Find the yearly rise.
   58, 63, 68: up 5 each year
2. Continue.
   68 + 5 = 73 hundred
3. Say why it may be wrong. It is an extrapolation: the trend may not continue.
   Unreliable.

Check. The sequence goes up by 5 each year. ✓

Watch out. State that the estimate assumes the trend continues.

Key words. The population is everything we want to know about. A sample is a smaller part of it that we actually measure. A good sample is random (everyone has an equal chance) and representative (a fair mix).

Method. Ask whether everyone in the town had an equal chance of being asked.

1. Who is in the sample? People who have just been to the cinema.
   They go to the cinema more often than most people.
2. Say what this does. The sample does not represent the whole town.
   The sample is biased.

Check. People who never go to the cinema had no chance of being asked. ✓

Watch out. Explain why the sample is unfair.

Key words. The population is everything we want to know about. A sample is a smaller part of it that we actually measure. A good sample is random (everyone has an equal chance) and representative (a fair mix).

Method. Give everyone in the town an equal chance of being asked.

1. Use a list of the whole town. For example, the electoral register.
   Number everyone on the list.
2. Choose at random. Use random numbers or a computer to pick the sample.
   Everyone has an equal chance.

Check. Everyone in the population could be chosen. ✓

Watch out. A better sample is random and covers the whole population.

Key words. Cumulative frequency is a running total of the frequencies. A cumulative frequency graph plots each running total against the upper end of its class.

Method. On a cumulative frequency graph, the median is at height n ÷ 2, the lower quartile at n ÷ 4 and the upper quartile at 3n ÷ 4. Read across to the curve and down to the axis.

1. Find the total number of values. The last cumulative frequency is the total, n.
   n = 80
2. Work out the heights to read at. Lower quartile at n ÷ 4, median at n ÷ 2, upper quartile at 3n ÷ 4.
   n ÷ 4 = 20
   n ÷ 2 = 40
   3n ÷ 4 = 60
3. Draw a horizontal line across from each height to the curve, then straight down to the axis. The value on the horizontal axis is the quartile.
4. Read the three values. Read the scale carefully. The answers are estimates.
   Q1 ≈ 23.5 min
   Median ≈ 31.5 min
   Q3 ≈ 39.5 min

Check. Going up from each reading to the curve and across should give the heights again. ✓

Watch out. Read heights from the cumulative frequency axis and answers from the data axis. Do not use n + 1 here.

Key words. Cumulative frequency is a running total of the frequencies. A cumulative frequency graph plots each running total against the upper end of its class.

Method. On a cumulative frequency graph, the median is at height n ÷ 2, the lower quartile at n ÷ 4 and the upper quartile at 3n ÷ 4. Read across to the curve and down to the axis.

1. Find the total number of values. The last cumulative frequency is the total, n.
   n = 80
2. Work out the heights to read at. Lower quartile at n ÷ 4, median at n ÷ 2, upper quartile at 3n ÷ 4.
   n ÷ 4 = 20
   n ÷ 2 = 40
   3n ÷ 4 = 60
3. Draw a horizontal line across from each height to the curve, then straight down to the axis. The value on the horizontal axis is the quartile.
4. Read the three values. Read the scale carefully. The answers are estimates.
   Q1 ≈ 23.5 min
   Median ≈ 31.5 min
   Q3 ≈ 39.5 min
5. Subtract to find the interquartile range. IQR = Q3 − Q1.
   39.5 − 23.5 = 16 min

Check. Going up from each reading to the curve and across should give the heights again. ✓

Watch out. Read heights from the cumulative frequency axis and answers from the data axis. Do not use n + 1 here.

Key words. Cumulative frequency is a running total of the frequencies. A cumulative frequency graph plots each running total against the upper end of its class.

Method. Read the cumulative frequency at the value. This is the number of values up to that value; subtract from the total to get the number above it.

1. Find 45 minutes on the horizontal axis and go up to the curve. The curve shows how many values are at or below each value.
2. Read across to the cumulative frequency axis. This gives the number of values up to 45.
   Cumulative frequency = 69
3. Subtract from the total to find how many are more than that. The last cumulative frequency is the total.
   80 − 69 = 11

Check. 69 below plus 11 above makes 80, the total. ✓

Watch out. The reading gives the number below the value. For “more than” you must subtract from the total.

Key words. A histogram has bars of different widths. The height is the frequency density = frequency ÷ class width, so the area of a bar is the frequency.

Method. For a histogram, plot frequency density = frequency ÷ class width on the vertical axis.

1. Find the width of each class. The bars have different widths, so the height cannot be the frequency. Width = upper end − lower end.
   10 − 0 = 10
   20 − 10 = 10
   40 − 20 = 20
   50 − 40 = 10
   70 − 50 = 20
2. Divide each frequency by its class width. frequency density = frequency ÷ class width. This is the height of the bar, and it makes the area of the bar equal to the frequency.
   8 ÷ 10 = 0.8
   14 ÷ 10 = 1.4
   30 ÷ 20 = 1.5
   16 ÷ 10 = 1.6
   12 ÷ 20 = 0.6
3. Write the results in a table.

   Time	Frequency	Class width	Frequency density
   0 < t ≤ 10 min	8	10	0.8
   10 < t ≤ 20 min	14	10	1.4
   20 < t ≤ 40 min	30	20	1.5
   40 < t ≤ 50 min	16	10	1.6
   50 < t ≤ 70 min	12	20	0.6

Check. Multiply back: 0.8 × 10 = 8, 1.4 × 10 = 14, 1.5 × 20 = 30, 1.6 × 10 = 16, 0.6 × 20 = 12. These are the frequencies. ✓

Watch out. The vertical axis is frequency density, not frequency.

Key words. A histogram has bars of different widths. The height is the frequency density = frequency ÷ class width, so the area of a bar is the frequency.

Method. The class 20 to 40 has frequency density 1.5. The part from 30 to 40 is 10 wide.

1. Find the frequency density. 30 ÷ 20.
   1.5
2. Multiply by the width wanted.
   1.5 × 10 = 15

Check. Half of the class frequency (30) is 15. ✓

Watch out. Use only the width of the part.

Key words. A histogram has bars of different widths. The height is the frequency density = frequency ÷ class width, so the area of a bar is the frequency.

Method. Estimate the median by finding which class holds the middle value, then assuming the values are spread evenly across that class.

1. Work out the frequencies and the total. Frequency = frequency density × class width for each bar.
   Frequencies: 8, 14, 30, 16, 12; total n = 80
2. Find the position of the median. It is half way through the data.
   80 ÷ 2 = 40
3. Use a running total to find the median class. The running total first reaches 40 in the class 20 < x ≤ 40 min.

   Class	Frequency	Running total
   0 < x ≤ 10 min	8	8
   10 < x ≤ 20 min	14	22
   20 < x ≤ 40 min	30	52
   40 < x ≤ 50 min	16	68
   50 < x ≤ 70 min	12	80

4. Find how far into the class we need to go. Before this class there are 22 values, so we need 18 more from this class of 30.
   18 ÷ 30 = 0.6 of the class
5. Find the estimate of the median. Start at the lower end and move that fraction of the class width.
   20 + 0.6 × 20 = 32 min

Check. The estimate lies inside the median class. ✓

Watch out. This is an estimate: it assumes the values are spread evenly within the class.

Key words. The quartiles split ordered data into four equal parts. Q1 is the lower quartile, Q2 is the median and Q3 is the upper quartile. The interquartile range (IQR) is Q3 minus Q1: the spread of the middle half.

Method. Quartiles split ordered data into four equal parts. Q1 is the median of the lower half, Q3 is the median of the upper half, and IQR = Q3 − Q1.

1. Write the values in order. Quartiles are found from ordered data.
   12, 14, 15, 17, 18, 20, 22, 60
2. Find the median (Q2). The median splits the data into a lower half and an upper half.
   n = 8, median = 17.5 (halfway between the two middle values)
3. Split into two equal halves. With an even number of values, the lower half is the first half of the list.
   Lower half: 12, 14, 15, 17
   Upper half: 18, 20, 22, 60
4. Q1 is the median of the lower half. It is the value a quarter of the way through the data.
   Q1 = 14.5
5. Q3 is the median of the upper half. It is the value three quarters of the way through the data.
   Q3 = 21
6. Subtract to find the interquartile range. IQR = Q3 − Q1: the spread of the middle half of the data.
   21 − 14.5 = 6.5

Check. Half of the 8 values lie between Q1 and Q3. ✓

Watch out. Do not forget to put the data in order first, and leave out the median when you split an odd number of values.

Key words. The quartiles split ordered data into four equal parts. Q1 is the lower quartile, Q2 is the median and Q3 is the upper quartile. The interquartile range (IQR) is Q3 minus Q1: the spread of the middle half.

Method. An outlier is a value that is far from the rest. One rule says a value is an outlier if it is more than 1.5 × IQR below Q1 or above Q3.

1. Work out the IQR. IQR = Q3 − Q1.
   21 − 14.5 = 6.5
2. Work out 1.5 × IQR. This is the allowed distance from the quartiles.
   1.5 × 6.5 = 9.75
3. Work out the lower limit. Take it away from Q1.
   14.5 − 9.75 = 4.75
4. Work out the upper limit. Add it to Q3.
   21 + 9.75 = 30.75
5. Compare each value with the limits. A value below the lower limit or above the upper limit is an outlier.
   Outliers: 60

Check. Every other value lies between 4.75 and 30.75. ✓

Watch out. The limits are measured from Q1 and Q3, not from the median.