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H20 Skewness

EDEXCEL 1ST0 · Calculator allowed · about 90 minutes

Total Marks

Name: _______________________________ Date: ____ / ____ / ______

Answer ALL questions. Show all your working.

For each statement about skewness, write down whether it is True or False.

(i)In a symmetric distribution, the mean, median and mode are approximately equal.(1)

(ii)A distribution is positively skewed when it has a long tail towards the higher (right-hand) values.(1)

(iii)If a distribution is negatively skewed, then the mean is greater than the median.(1)

(iv)The quartile coefficient of skewness can take negative values as well as positive values.(1)

(Total for Question 1 is 4 marks)

The histogram shows the annual bonus payments, in £, received by 80 employees at a company.

(a)State whether the distribution of bonus payments is symmetric, positively skewed or negatively skewed.(1)

(b)Give two features of the histogram's shape that support your answer to part (a).(2)

(c)State whether the mean bonus payment would be greater than, less than, or about equal to the median bonus payment. Give a reason for your answer.(1)

(Total for Question 2 is 4 marks)

The ages at which employees retire from a large company are known to be negatively skewed: most employees retire close to the official retirement age, with a small number retiring much earlier. On the blank axes provided, sketch a frequency curve to show this negatively skewed distribution. Label the position of the peak and the direction of the tail.

(Total for Question 3 is 3 marks)

In a long jump competition, the distances jumped by 30 athletes had a mean of 4.35 m, a median of 4.60 m and a mode of 4.70 m.

(a)State whether the distribution of jump distances is symmetric, positively skewed or negatively skewed.(1)

(b)Justify your answer to part (a) by referring to the mean, median and mode.(2)

(Total for Question 4 is 3 marks)

The times, in minutes, that 11 customers waited in a queue at a bank are: 2, 3, 3, 4, 4, 4, 5, 6, 7, 9, 25.

(a)Work out the mean waiting time, giving your answer correct to 2 decimal places.(2)

(b)Write down the median waiting time.(1)

(c)Write down the modal waiting time.(1)

(d)State, with a reason based on your answers to parts (a) to (c), whether the distribution of waiting times is symmetric, positively skewed or negatively skewed.(2)

(Total for Question 5 is 6 marks)

The box plot shows the delivery times, in minutes, of 50 parcels handled by a courier company.

(a)Write down the median delivery time.(1)

(b)Work out the interquartile range.(1)

(c)By comparing median - lower quartile with upper quartile - median, state, giving a reason, whether the distribution of delivery times is symmetric, positively skewed or negatively skewed.(2)

(Total for Question 6 is 4 marks)

The box plots show the amount of time, in hours, that Group X and Group Y spent revising for an exam, drawn on the same scale.

(a)Write down the median revision time for Group X.(1)

(b)Write down the median revision time for Group Y.(1)

(c)Work out the interquartile range for Group Y.(1)

(d)Compare the skewness of the two distributions, referring to the shape of each box plot.(2)

(Total for Question 7 is 5 marks)

The histogram shows the rainfall, in mm, recorded on each of 60 days during a rainy season.

(a)State whether the distribution of rainfall amounts is symmetric, positively skewed or negatively skewed.(1)

(b)Give two features of the histogram that support your answer to part (a).(2)

(c)State whether the mean rainfall would be greater than, less than, or about equal to the median rainfall. Give a reason for your answer.(2)

(Total for Question 8 is 5 marks)

The frequency polygon shows the resting heart rate, in beats per minute (bpm), of 40 gym members.

(a)State whether the distribution of resting heart rates is symmetric, positively skewed or negatively skewed.(1)

(b)Give a reason for your answer, based on the shape of the frequency polygon.(2)

(c)State whether you would expect the mean resting heart rate to be greater than, less than, or about equal to the median. Give a brief reason.(1)

(Total for Question 9 is 4 marks)

The annual salaries of the 9 employees at a small company are: £24,000, £25,000, £26,000, £27,000, £28,000, £29,000, £30,000, £31,000 and £104,000 (the company owner).

(a)Work out the mean salary.(2)

(b)Write down the median salary.(1)

(c)Explain, with reference to your answers, why the median is considered a more resistant (robust) measure of average than the mean for this data set.(2)

(Total for Question 10 is 5 marks)

One measure of skewness is the quartile coefficient of skewness, SK, defined as SK = (Q1 + Q3 - 2Q2) / (Q3 - Q1), where Q1 is the lower quartile, Q2 is the median and Q3 is the upper quartile. A value of SK greater than 0 indicates positive skew, a value less than 0 indicates negative skew, and SK = 0 indicates a symmetric quartile spacing. The times, in minutes, taken by 60 runners to complete a park run have lower quartile Q1 = 24, median Q2 = 30 and upper quartile Q3 = 34.

(a)Calculate the value of SK for the park run times.(3)

(b)State whether the distribution of park run times is symmetric, positively skewed or negatively skewed.(1)

(c)The times for a second, larger park run have a quartile coefficient of skewness SK = 0.05. Compare the skew of the two data sets, referring to both the sign and the relative size of SK.(2)

(Total for Question 11 is 6 marks)

The number of emails received per day by 15 employees, arranged in order, are: 3, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 18, 40.

(a)Find the lower quartile, the median and the upper quartile of this data.(3)

(b)Calculate the quartile coefficient of skewness, SK = (Q1 + Q3 - 2Q2) / (Q3 - Q1), using your answers to part (a).(2)

(c)Calculate the mean number of emails, giving your answer correct to 2 decimal places.(2)

(d)The value of SK suggests the data is symmetric, but the mean is noticeably larger than the median. Explain why these two pieces of evidence appear to disagree.(2)

(Total for Question 12 is 9 marks)

The table shows summary statistics for the exam results (out of 100) of two schools.

School	Lower quartile (Q1)	Median (Q2)	Upper quartile (Q3)
A	45	58	62
B	40	58	90

The quartile coefficient of skewness is given by SK = (Q1 + Q3 - 2Q2) / (Q3 - Q1).

(a)Calculate SK for School A.(2)

(b)Calculate SK for School B.(2)

(c)Compare the two schools' distributions of exam results, referring to both the sign and the size of SK.(2)

(Total for Question 13 is 6 marks)

A student calculates a quartile coefficient of skewness of SK = 0 for a data set and concludes: 'Since SK = 0, the distribution must be perfectly symmetric, so the mean, median and mode are all equal.' Evaluate this claim, explaining any limitations of using SK to judge symmetry.

(Total for Question 14 is 3 marks)

Tom, a Year 11 student, wants to investigate the claim: 'The distribution of house prices in a small village is positively skewed.'

(a)Suggest an appropriate sampling method Tom could use to select a sample of houses from the village, and give one reason for your choice.(2)

(b)Tom collects the following sale prices, in £000s, for 9 houses: 180, 185, 190, 195, 200, 205, 210, 215, 490. Calculate the mean price.(2)

(c)Find the median house price, and use it with the mean to assess whether Tom's data supports the claim that house prices are positively skewed.(3)

(d)Tom only sampled 9 houses from the village. Suggest one way he could improve the reliability of his conclusion about skewness.(1)

(Total for Question 15 is 8 marks)

A clinic recorded the waiting times, in minutes, of 100 patients. The grouped frequency table shows the results.

Waiting time (minutes)	0-10	10-20	20-30	30-40	40-50
Frequency	8	12	20	40	20

(a)Calculate an estimate of the mean waiting time, using the midpoint of each class.(3)

(b)Use linear interpolation to estimate the median waiting time.(2)

(c)Use linear interpolation to estimate the lower quartile of the waiting times.(2)

(d)Use linear interpolation to estimate the upper quartile of the waiting times.(2)

(e)Calculate the quartile coefficient of skewness, SK = (Q1 + Q3 - 2Q2) / (Q3 - Q1), using your estimates from parts (b) to (d).(2)

(f)Compare your estimates of the mean and median, and use this comparison together with the sign of SK to describe the skew of the waiting times, explaining whether the two methods agree.(3)

(Total for Question 16 is 14 marks)

Mark scheme · H20 Skewness

Question 1

(i) B1 True cao
(i) Answer: True
(ii) B1 True cao
(ii) Answer: True
(iii) B1 False cao
(iii) Answer: False
(iv) B1 True cao
(iv) Answer: True

Question 2

(a) B1 Positively skewed cao
(a) Answer: Positively skewed
(b) B1 the tallest bar (the peak, frequency 30) occurs in the lowest bonus class (£0-2000)
(b) B1 frequencies decrease gradually across the remaining four classes (22, 15, 8, 5), giving a long tail stretching towards the higher bonus payments
(b) Answer: The peak (30) is in the lowest class, and frequencies fall away gradually across the higher classes, giving a tail towards the larger bonuses.
(c) B1 Greater than - the small number of employees with much higher bonuses (the long tail) pulls the mean upward, while the median is unaffected by how far these extreme values are from the centre
(c) Answer: Greater than the median

Question 3

B1 a single smooth, unimodal curve drawn (one peak, not multiple humps)
B1 peak positioned towards the higher end of the value axis, with a longer tail stretching towards the lower values (a negatively skewed shape)
B1 the peak and the direction of the tail are clearly labelled on the sketch
Answer: A smooth, single-peaked curve with its peak towards the higher end of the value axis (near the official retirement age) and a longer tail stretching towards the lower values, with the peak and tail direction labelled.

Question 4

(a) B1 Negatively skewed cao
(a) Answer: Negatively skewed
(b) B1 identifies mean < median < mode (4.35 < 4.60 < 4.70), oe
(b) B1 correct interpretation: most athletes jumped distances bunched towards the higher end, with a small tail of shorter jumps pulling the mean down below the median and mode
(b) Answer: Since 4.35 < 4.60 < 4.70, most jump distances are bunched near the top of the range, with a tail of shorter jumps pulling the mean down below the median and mode.

Question 5

(a) M1 2+3+3+4+4+4+5+6+7+9+25 (=72), oe divided by 11
(a) A1 6.55 cao (awrt 6.55)
(a) Answer: 6.55
(b) B1 4 cao
(b) Answer: 4
(c) B1 4 cao
(c) Answer: 4
(d) B1 mean (6.55) is greater than the median and mode (both 4)
(d) B1 Positively skewed - one extreme value (25 minutes) pulls the mean up while barely affecting the median or mode
(d) Answer: Positively skewed

Question 6

(a) B1 15 cao
(a) Answer: 15
(b) B1 20 cao
(b) Answer: 20
(c) M1 15 - 12 (=3) compared with 32 - 15 (=17), oe
(c) A1 Positively skewed - the upper quartile is much further from the median than the lower quartile is, so the longer delivery times are more spread out cao
(c) Answer: Positively skewed

Question 7

(a) B1 14 cao
(a) Answer: 14
(b) B1 6 cao
(b) Answer: 6
(c) B1 12 cao
(c) Answer: 12
(d) B1 Group X is symmetric (median - LQ = 4 = UQ - median, the two box halves are equal)
(d) B1 Group Y is positively skewed (median - LQ = 2 is much smaller than UQ - median = 10, so the higher revision times are more spread out - a long tail towards the higher times)
(d) Answer: Group X is symmetric; Group Y is positively skewed, with a tail towards the longer revision times.

Question 8

(a) B1 Positively skewed cao
(a) Answer: Positively skewed
(b) B1 the tallest bar (peak, frequency 25) is in the lowest class (0-10 mm)
(b) B1 frequencies decrease gradually across the remaining five classes, giving a long tail stretching towards the higher rainfall values
(b) Answer: The peak is at 0-10 mm, near the low end, and frequencies decline gradually across the higher classes, giving a tail towards the wetter days.
(c) B1 Greater than the median
(c) B1 the small number of days with much higher rainfall (the long tail) pulls the mean upward, while the median is unaffected by how far these extreme values are from the centre
(c) Answer: Greater than the median

Question 9

(a) B1 Negatively skewed cao
(a) Answer: Negatively skewed
(b) B1 the peak (highest frequency, 16) is towards the higher end of the heart rates (the 75 bpm class)
(b) B1 frequencies rise gradually across the four lower classes (55 to 70) but fall sharply after the peak, giving a longer tail towards the lower heart rates
(b) Answer: The peak is at 75 bpm, near the top of the range, and frequencies build up gradually from the lower classes, giving a tail stretching towards the lower heart rates.
(c) B1 Less than - the tail of lower heart rates pulls the mean down below the median, oe
(c) Answer: Less than the median

Question 10

(a) M1 24+25+26+27+28+29+30+31+104 (=324, in £000s), oe divided by 9
(a) A1 £36,000 cao
(a) Answer: £36,000
(b) B1 £28,000 cao
(b) Answer: £28,000
(c) B1 the mean is affected by the size of an extreme value (the owner's £104,000 salary pulls it upward), while the median only depends on the middle value's position, not its size
(c) B1 the median (£28,000) better represents a typical employee's salary, since the mean (£36,000) is higher than 8 of the 9 salaries
(c) Answer: The owner's £104,000 salary is an extreme value that raises the mean to £36,000, which is higher than eight of the nine salaries; the median (£28,000) is unaffected by its size and better reflects a typical salary.

Question 11

(a) M1 24 + 34 - 2(30) (=-2), oe
(a) M1 their numerator divided by (34 - 24) (=10), dep on first method mark
(a) A1 -0.2 cao (allow -1/5)
(a) Answer: -0.2
(b) B1 Negatively skewed, since SK < 0 cao
(b) Answer: Negatively skewed
(c) B1 this park run is negatively skewed (SK = -0.2), while the second park run is very slightly positively skewed, almost symmetric (SK = 0.05)
(c) B1 this distribution shows noticeably more skew, since its SK value is much further from 0 (magnitude 0.2 is greater than 0.05)
(c) Answer: This park run is negatively skewed while the second is almost symmetric (very slightly positive); this one shows more skew since its SK is further from 0.

Question 12

(a) B1 lower quartile = 7 (4th of 15 values)
(a) B1 median = 11 (8th of 15 values)
(a) B1 upper quartile = 15 (12th of 15 values)
(a) Answer: Q1 = 7, median = 11, Q3 = 15
(b) M1 7 + 15 - 2(11) (=0), oe divided by (15-7)
(b) A1 SK = 0 cao
(b) Answer: 0
(c) M1 3+5+6+7+8+9+10+11+12+13+14+15+16+18+40 (=187), oe divided by 15
(c) A1 12.47 cao (awrt 12.47)
(c) Answer: 12.47
(d) B1 SK only depends on the positions of Q1, Q2 and Q3, so it is not affected by how extreme any values beyond Q3 are (the value of 40 does not change Q1, Q2 or Q3)
(d) B1 the mean uses every value in its calculation, so the one very high value (40 emails) pulls it above the median, revealing skew that SK does not detect
(d) Answer: SK = 0 only reflects the spread of the middle 50% of the data around the median, which happens to be symmetric here; it ignores the extreme value of 40, which the mean is sensitive to, pulling it well above the median.

Question 13

(a) M1 45 + 62 - 2(58) (=-9), oe divided by (62-45)
(a) A1 -0.53 cao (awrt -0.53)
(a) Answer: -0.53
(b) M1 40 + 90 - 2(58) (=14), oe divided by (90-40)
(b) A1 0.28 cao
(b) Answer: 0.28
(c) B1 School A has negative SK (-0.53), showing negative skew - the lower marks are more spread out than the higher marks; School B has positive SK (0.28), showing positive skew - the higher marks are more spread out
(c) B1 School A's distribution is more skewed than School B's, since the size of SK for A (0.53) is greater than for B (0.28)
(c) Answer: School A is negatively skewed, School B is positively skewed, and School A shows the stronger skew since 0.53 > 0.28.

Question 14

B1 SK = 0 only shows that the middle 50% of the data (between Q1 and Q3) is evenly spread either side of the median - it does not take into account values beyond the quartiles
B1 a data set can therefore have SK = 0 while still containing an outlier or extreme values beyond Q1 or Q3, so the distribution need not be perfectly symmetric overall
B1 the claim that the mean, median and mode must all be exactly equal is not fully justified by SK = 0 alone; other checks (e.g. comparing the mean and median directly, or a histogram) should also be used
Answer: The student's conclusion is not fully justified. SK = 0 only shows the middle 50% of the data is evenly spread either side of the median; it ignores values beyond the quartiles, so a data set could still contain an outlier that makes the mean differ from the median, meaning the mean, median and mode are not guaranteed to be exactly equal.

Question 15

(a) B1 valid sampling method named (e.g. simple random sample using the electoral roll or council tax register)
(a) B1 valid reason linked to representativeness or avoiding bias, matched to the method given
(a) Answer: e.g. A simple random sample, using a numbered list of all houses in the village and a random number generator, so that every house has an equal chance of being chosen and the sample is not biased.
(b) M1 180+185+190+195+200+205+210+215+490 (=2070), oe divided by 9
(b) A1 £230,000 cao
(b) Answer: £230,000
(c) B1 median = £200,000 (5th of 9 ordered values)
(c) B1 mean (£230,000) is greater than the median (£200,000)
(c) B1 yes, this supports the claim of positive skew - one much higher-priced house (£490,000) pulls the mean above the median, creating a tail towards the higher prices
(c) Answer: Yes, the data supports the claim: median = £200,000 < mean = £230,000, consistent with positive skew caused by the one much higher-priced house.
(d) B1 e.g. increase the sample size (or survey all houses in the village) so the estimate of skew is less affected by any one unusual value, oe
(d) Answer: Increase the sample size, so the conclusion is less affected by any single unusually priced house.

Question 16

(a) M1 correct midpoints used: 5, 15, 25, 35, 45
(a) M1 sum of f x midpoint = 8(5)+12(15)+20(25)+40(35)+20(45) (=3020), oe divided by 100
(a) A1 30.2 cao (awrt 30.2)
(a) Answer: 30.2
(b) M1 identifies the median class (30-40, since cumulative frequencies are 8, 20, 40, 80, 100) and forms 30 + [(50-40)/40] x 10, oe
(b) A1 32.5 cao
(b) Answer: 32.5
(c) M1 identifies the lower quartile class (20-30) and forms 20 + [(25-20)/20] x 10, oe
(c) A1 22.5 cao
(c) Answer: 22.5
(d) M1 identifies the upper quartile class (30-40) and forms 30 + [(75-40)/40] x 10, oe
(d) A1 38.75 cao
(d) Answer: 38.75
(e) M1 22.5 + 38.75 - 2(32.5) (=-3.75), oe divided by (38.75-22.5)
(e) A1 -0.23 cao (awrt -0.23, ft their b to d)
(e) Answer: -0.23
(f) B1 mean (30.2) is less than the median (32.5), which is consistent with negative skew
(f) B1 SK is negative (-0.23), also indicating negative skew
(f) B1 the two methods agree: the waiting times are negatively skewed, with a longer tail towards the shorter waiting times (most patients wait close to or above the modal class of 30-40 minutes, with fewer experiencing a much shorter wait)
(f) Answer: Both methods agree: mean (30.2) < median (32.5) and SK is negative (-0.23), so the waiting times are negatively skewed, with a tail towards the shorter waiting times.

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