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A-Level · Statistics

S2 Statistics: Data Presentation and Interpretation

EDEXCEL 9MA0 · Calculator allowed · about 130 minutes

Total Marks

Name: _______________________________ Date: ____ / ____ / ______

Answer ALL questions. Show all your working.

A company has 850 employees split across three departments: Admin (170 employees), Production (510 employees) and Sales (170 employees). The company wants to survey employees about a proposed new flexible working policy.

(a)State what is meant, in the context of sampling, by a 'sampling frame'.(1)

(b)The company decides to take a stratified sample of 40 employees, proportional to department size. Calculate the number of employees that should be sampled from each department.(3)

(c)Describe how systematic sampling could instead be used to select 40 employees from a numbered list of all 850 employees, and state one advantage this method has over simple random sampling in this context.(3)

(Total for Question 1 is 7 marks)

A cafe owner records, for 8 different days last winter, the outdoor temperature and the number of hot chocolates sold.

(a)State the type of correlation you would expect between outdoor temperature and the number of hot chocolates sold.(1)

(b)Describe the general shape of the scatter diagram you would expect to see for this data, referring to the direction of the trend.(1)

(c)The owner also wants to investigate the relationship between outdoor temperature and the number of cold drinks sold. State the type of correlation expected and explain your reasoning.(2)

(d)On one unusually cold day in December, a local event caused hot chocolate sales to be much higher than the temperature that day would otherwise suggest. Explain the effect this data point is likely to have on the product moment correlation coefficient (PMCC) calculated for temperature against hot chocolate sales across the whole data set.(2)

(Total for Question 2 is 6 marks)

The times, in minutes, taken by 9 runners to complete a 5 km training route are: 34, 36, 29, 41, 38, 33, 30, 45, 37.

(a)Calculate the mean time.(2)

(b)Calculate the standard deviation of these times.(3)

(c)A tenth runner joins the group with a time of 60 minutes, a clear outlier. Without recalculating, state and justify the effect that including this runner's time would have on the mean and on the standard deviation.(2)

(Total for Question 3 is 7 marks)

The typing speeds, in words per minute, of 11 trainees on a data entry course are: 42, 45, 47, 48, 50, 51, 53, 55, 58, 60, 90.

(a)Calculate the mean and the standard deviation of these 11 typing speeds.(4)

(b)Show that the value 90 is an outlier, using the rule that a value is an outlier if it lies more than 2 standard deviations from the mean.(3)

(c)Given that, without the outlier, the sum of the remaining 10 typing speeds is 509 and the sum of their squares is 26201, calculate the new mean and new standard deviation for the remaining 10 trainees.(4)

(d)Comment on the effect that removing the outlier has had on the mean and on the standard deviation of the typing speeds.(2)

(Total for Question 4 is 13 marks)

In a mathematics exam, marks x are coded using y = (x - 30)/5. For 15 students, the coded marks y satisfy: sum of y = 24, sum of y² = 140.

(a)Calculate the mean of the coded marks, y.(1)

(b)Hence calculate the mean exam mark, x.(2)

(c)Calculate the standard deviation of the coded marks, y, and hence calculate the standard deviation of the original exam marks, x.(4)

(Total for Question 5 is 7 marks)

The heights, in cm, of 50 sunflower plants grown in a trial were recorded in the grouped frequency table below.

Height (cm)	150-160	160-170	170-180	180-190	190-200
Frequency	5	12	20	9	4

(a)Using the midpoint of each class, calculate an estimate of the mean height.(3)

(b)Using linear interpolation, estimate the median height.(3)

(c)Using your answer to part (a), calculate an estimate of the variance of the heights.(3)

(Total for Question 6 is 9 marks)

The time, t minutes, taken by 100 children to complete a jigsaw puzzle was recorded in the grouped frequency table below.

Time t (minutes)	0≤t<10	10≤t<20	20≤t<35	35≤t<50	50≤t<70
Frequency	12	28	f	15	10

(a)Given that a total of 100 children completed the puzzle, calculate the value of f.(2)

(b)On a histogram of this data, calculate the frequency density that should be used for the bar representing 20≤t<35.(2)

(c)Using linear interpolation, estimate the median time taken to complete the puzzle.(3)

(d)Assuming the times are evenly spread within the class 50≤t<70, estimate the number of children who took more than 60 minutes to complete the puzzle.(3)

(Total for Question 7 is 10 marks)

The weekly hours worked by 13 part-time staff at a supermarket, listed in ascending order, are: 12, 15, 15, 18, 20, 21, 22, 24, 25, 27, 29, 32, 55.

(a)Find the median, the lower quartile (Q1) and the upper quartile (Q3) of these weekly hours.(3)

(b)Show that 55 hours is an outlier, using the rule that a value is an outlier if it lies more than 1.5 x IQR beyond the nearer quartile.(3)

(c)State the five figures needed to draw a box plot for this data, and explain how the outlier at 55 should be represented on the diagram.(3)

(d)Using your values for the median, Q1 and Q3, comment on the skewness of the distribution of weekly hours.(2)

(Total for Question 8 is 11 marks)

Two classes, A and B, sat the same maths test, out of 50 marks. Box plots of their results gave the following five-figure summaries.
Class A: minimum = 10, Q1 = 18, median = 26, Q3 = 40, maximum = 48
Class B: minimum = 15, Q1 = 27, median = 30, Q3 = 33, maximum = 36

(a)Compare the medians and the interquartile ranges of the two classes, in the context of the test.(3)

(b)Using the quartiles, comment on the skewness of each class's distribution of marks.(2)

(c)A teacher claims 'Class B had less variation in test marks than Class A.' By considering both the range and the interquartile range, comment on whether this claim is fair.(3)

(Total for Question 9 is 8 marks)

For 8 days last spring, a shopkeeper recorded the number of hours of sunshine, x, and the shop's ice cream sales, y, in hundreds of pounds.
Hours of sunshine, x: 2, 3, 5, 6, 7, 8, 9, 10
Ice cream sales, y (GBP hundreds): 1.2, 1.8, 2.6, 3.0, 3.4, 4.0, 4.5, 4.9
You may use: sum(x) = 50, sum(y) = 25.4, sum(xy) = 184.1, sum(x²) = 368, sum(y²) = 92.26

(a)Calculate the value of the product moment correlation coefficient (PMCC), r, for these data.(3)

(b)Interpret the value of r found in part (a), in context.(1)

(c)Find the equation of the regression line of y on x in the form y = a + bx, giving a and b to 3 significant figures.(4)

(d)Use your regression line to estimate the ice cream sales on a day with 6.5 hours of sunshine, and comment on the reliability of this estimate.(3)

(Total for Question 10 is 11 marks)

House prices, in thousands of pounds, were recorded for two streets. For 20 houses in Elm Street: sum(x) = 4200, sum(x²) = 920000. For 15 houses in Birch Street: sum(x) = 3300, sum(x²) = 746000.

(a)Calculate the mean and the standard deviation of house prices in Elm Street.(4)

(b)Calculate the mean and the standard deviation of house prices in Birch Street.(4)

(c)Using your answers to parts (a) and (b), compare house prices in Elm Street and Birch Street.(2)

(Total for Question 11 is 10 marks)

A distribution of delivery times has mean = 45, median = 50, mode = 58 and standard deviation = 12. It is also known that Q1 = 38, Q2 = 50 and Q3 = 54.

(a)Calculate Pearson's coefficient of skewness, given by (mean - mode)/standard deviation.(2)

(b)Calculate the quartile coefficient of skewness, given by (Q3 + Q1 - 2Q2)/(Q3 - Q1).(2)

(c)Interpret the values found in parts (a) and (b) in terms of the skewness of the distribution.(2)

(d)State one advantage of using the quartile coefficient of skewness, found in part (b), rather than Pearson's coefficient of skewness, found in part (a).(1)

(Total for Question 12 is 7 marks)

A researcher recorded, for 10 UK towns, the number of ice cream shops, x, and the number of sunburn cases reported at local hospitals over one summer, y. The PMCC between x and y was found to be r = 0.91, based on values of x ranging from 2 to 15.

(a)The researcher concludes: 'Opening more ice cream shops causes more cases of sunburn.' Comment on this conclusion, suggesting a variable that could explain the high correlation observed.(3)

(b)State one condition that should be checked before using the regression line from this data to predict y for a town with x = 25 ice cream shops, and explain why this matters here.(2)

(c)A large data set recorded daily mean temperature and daily rainfall for a UK location across a whole year. Explain why it would not be appropriate to calculate a single PMCC using all of this data if you wanted to investigate the relationship between temperature and rainfall during summer months only.(2)

(Total for Question 13 is 7 marks)

Mark scheme · S2 Statistics: Data Presentation and Interpretation

Question 1

(a) B1 a list of all the members of the population, from which the sample is taken, oe
(a) Answer: A list of all the members of the population, used to select the sample.
(b) M1 correct method shown for at least one department, e.g. 40 x (170/850)
(b) A1 two of the three values correct
(b) A1 cao all three correct: Admin = 8, Production = 24, Sales = 8
(b) Answer: Admin: 8, Production: 24, Sales: 8.
(c) B1 sampling interval k = 850/40 = 21.25, so k = 21 is used (rounding down so at least 40 employees can be chosen)
(c) B1 choose a random starting point between 1 and 21, then select every 21st employee after that from the numbered list
(c) B1 valid advantage, e.g. quicker/simpler to carry out than generating 40 distinct random numbers, or spreads the sample evenly across the whole ordered list, oe
(c) Answer: k = 850/40 = 21.25, so k = 21; pick a random start between 1 and 21, then take every 21st employee after that. This is quicker to carry out than generating 40 separate random numbers.

Question 2

(a) B1 negative correlation
(a) Answer: Negative correlation.
(b) B1 points generally slope downwards from left to right, i.e. as temperature increases, hot chocolate sales decrease, oe
(b) Answer: The points would generally slope downwards from left to right: as temperature increases, the number of hot chocolates sold decreases.
(c) B1 positive correlation
(c) B1 valid reasoning in context, e.g. customers are more likely to buy cold drinks when it is warmer, so sales rise as temperature rises, oe
(c) Answer: Positive correlation, since customers are more likely to want a cold drink as the outdoor temperature increases.
(d) B1 identifies that this point does not follow the general negative trend of the rest of the data
(d) B1 explains that this weakens the correlation, pulling the PMCC value closer to zero (less strongly negative) than it would otherwise be, oe
(d) Answer: This point does not fit the general negative trend, so it will weaken the correlation, pulling the PMCC value closer to zero than it would be without this day included.

Question 3

(a) M1 attempt sum of the 9 values / 9
(a) A1 awrt 35.9 (minutes)
(a) Answer: awrt 35.9 minutes
(b) M1 attempt sum of squares of the 9 values (or use calculator statistical mode)
(b) M1 correct method: √sum x² / 9 - mean²
(b) A1 awrt 4.82 (minutes)
(b) Answer: awrt 4.82 minutes
(c) B1 mean would increase, since 60 is greater than the current mean (awrt 35.9), pulling the average up, oe
(c) B1 standard deviation would increase, since 60 is a long way from the mean, adding a value with a large deviation increases the spread, oe
(c) Answer: The mean would increase (60 is well above the current mean, pulling it up); the standard deviation would increase (60 is far from the mean, so it adds a large deviation and increases the spread).

Question 4

(a) M1 attempt sum of the 11 values / 11
(a) A1 mean = awrt 54.5
(a) M1 attempt √sum x² / 11 - mean², using sum of squares of the 11 values
(a) A1 SD = awrt 12.4
(a) Answer: mean = awrt 54.5, SD = awrt 12.4 (words per minute)
(b) M1 attempt mean + 2 x (their standard deviation)
(b) M1 correct value: 54.4545 + 2(12.4017) = awrt 79.3
(b) A1 cso: since 90 > 79.3 (their boundary), 90 is an outlier; conclusion must follow correctly from their boundary
(b) Answer: 90 is an outlier (given)
(c) M1 attempt new mean = 509/10
(c) A1 new mean = 50.9 (cao)
(c) dM1 attempt √26201/10 - (their new mean)², dependent on a correct new mean
(c) A1 new SD = awrt 5.41
(c) Answer: new mean = 50.9, new SD = awrt 5.41 (words per minute)
(d) B1 the mean has decreased (from awrt 54.5 to 50.9), since the outlier of 90 was well above the rest of the data and was pulling the mean up, oe
(d) B1 the standard deviation has decreased substantially (from awrt 12.4 to awrt 5.41), since the outlier was a long way from the mean and contributed a large amount to the spread, oe
(d) Answer: Both the mean and the standard deviation have decreased: the mean fell from awrt 54.5 to 50.9 because the high outlier was no longer pulling it up, and the standard deviation fell sharply from awrt 12.4 to awrt 5.41 because the outlier had been contributing a large deviation to the spread.

Question 5

(a) B1 cao mean of y = 1.6
(a) Answer: mean of y = 1.6
(b) M1 correct rearrangement x = 5y + 30 applied to their mean of y
(b) A1 cao mean of x = 38
(b) Answer: mean of x = 38
(c) M1 attempt √sum y² / 15 - (mean of y)²
(c) A1 SD of y = awrt 2.60
(c) M1 correct scaling: SD of x = 5 x (their SD of y), since x = 5y + 30
(c) A1 SD of x = awrt 13.0
(c) Answer: SD of y = awrt 2.60, SD of x = awrt 13.0

Question 6

(a) M1 correct midpoints identified: 155, 165, 175, 185, 195
(a) M1 attempt sum of (frequency x midpoint) / 50
(a) A1 cao mean = 174 (cm)
(a) Answer: estimated mean = 174 cm
(b) M1 correct cumulative frequencies identified (5, 17, 37, 46, 50) and median class 170-180 identified
(b) M1 correct interpolation: 170 + ((25 - 17)/20) x 10
(b) A1 cao median = 174 (cm)
(b) Answer: estimated median = 174 cm
(c) M1 attempt sum of (frequency x midpoint²)
(c) dM1 correct method: (sum f x midpoint²)/50 - (their mean)², dependent on a correct mean from part (a)
(c) A1 cao variance = 113 (cm²)
(c) Answer: estimated variance = 113 cm²

Question 7

(a) M1 100 - 12 - 28 - 15 - 10
(a) A1 cao f = 35
(a) Answer: f = 35
(b) M1 attempt frequency / class width, i.e. their f / 15
(b) A1 awrt 2.33
(b) Answer: frequency density = awrt 2.33
(c) M1 correct cumulative frequencies identified (12, 40, 75, 90, 100) and median class 20≤t<35 identified
(c) M1 correct interpolation: 20 + ((50 - 40)/35) x 15
(c) A1 awrt 24.3 (minutes)
(c) Answer: estimated median = awrt 24.3 minutes
(d) M1 identifies that this requires the proportion of the 50≤t<70 class above t = 60
(d) M1 correct method: 10 x (70 - 60)/(70 - 50)
(d) A1 cao 5 children
(d) Answer: 5 children

Question 8

(a) B1 median = 22 (the 7th of 13 ordered values)
(a) M1 correct method: Q1 as the median of the lower 6 values, Q3 as the median of the upper 6 values
(a) A1 cao Q1 = 16.5 and Q3 = 28
(a) Answer: median = 22, Q1 = 16.5, Q3 = 28 (hours)
(b) M1 IQR = 28 - 16.5 = 11.5
(b) M1 upper boundary = Q3 + 1.5 x IQR = 28 + 1.5(11.5)
(b) A1 cso: boundary = 45.25, and since 55 > 45.25, 55 is an outlier
(b) Answer: 55 is an outlier (given)
(c) B1 minimum = 12, Q1 = 16.5, median = 22, Q3 = 28
(c) B1 the upper whisker is drawn to 32, the highest value that is not an outlier (not to 55)
(c) B1 55 is plotted separately, e.g. as a cross or a dot beyond the whisker, to show it is an outlier
(c) Answer: Five-figure summary: min = 12, Q1 = 16.5, median = 22, Q3 = 28, with the upper whisker drawn to 32 (the largest non-outlier value); 55 is marked separately as a cross beyond the whisker to indicate it is an outlier.
(d) B1ft compares Q3 - median (= 6) with median - Q1 (= 5.5): these are close in size, using their values from part (a)
(d) B1 concludes the quartiles suggest a roughly symmetrical distribution, but the presence of the high outlier (55) suggests some positive skew overall, oe
(d) Answer: Q3 - median = 6 and median - Q1 = 5.5, which are close in size, so the quartiles alone suggest a fairly symmetrical distribution; however, the presence of the high outlier at 55 suggests the overall distribution has some positive skew (a long tail towards higher values).

Question 9

(a) B1 Class B has the higher median (30 compared with 26), so Class B typically scored higher on the test
(a) B1 Class A has the larger IQR (22 compared with 6), so marks in Class A were far more spread out (variable)
(a) B1 valid overall conclusion in context, e.g. Class B performed more consistently well, while Class A's results ranged from much weaker to much stronger performance, oe
(a) Answer: Class B has a higher median (30 vs 26), so its typical mark was higher; Class A has a much larger IQR (22 vs 6), so its marks were far more spread out. Overall, Class B performed more consistently, while Class A had a wider range of ability.
(b) B1 Class A: Q3 - median (14) > median - Q1 (8), so Class A's marks are positively skewed
(b) B1 Class B: Q3 - median (3) = median - Q1 (3), so Class B's marks are symmetrical (based on the quartiles)
(b) Answer: Class A: 40 - 26 = 14 and 26 - 18 = 8, so 14 > 8, meaning Class A's marks are positively skewed. Class B: 33 - 30 = 3 and 30 - 27 = 3, which are equal, so Class B's marks are symmetrical based on the quartiles.
(c) B1 range of Class A = 38, range of Class B = 21, so the range is smaller for Class B
(c) B1 IQR of Class A = 22, IQR of Class B = 6, so the IQR is also smaller for Class B
(c) B1 since both measures of spread agree (both smaller for Class B), the claim is fair/well supported, oe
(c) Answer: Range: Class A = 48 - 10 = 38, Class B = 36 - 15 = 21. IQR: Class A = 22, Class B = 6. Both the range and the IQR are smaller for Class B, so the claim is a fair conclusion, supported by both measures of spread, not just one.

Question 10

(a) M1 attempt Sxy = sum(xy) - sum(x)sum(y)/8, and Sxx = sum(x²) - sum(x)²/8, and Syy = sum(y²) - sum(y)²/8
(a) M1 correct formula: r = Sxy / √Sxx x Syy
(a) A1 awrt 0.998
(a) Answer: r = awrt 0.998
(b) B1 a very strong positive (linear) correlation: more hours of sunshine are associated with higher ice cream sales, oe
(b) Answer: There is a very strong positive linear correlation: days with more hours of sunshine are strongly associated with higher ice cream sales.
(c) M1 attempt b = Sxy / Sxx
(c) A1 b = awrt 0.457
(c) dM1 attempt a = (mean of y) - b(mean of x), dependent on their value of b
(c) A1 a = awrt 0.320
(c) Answer: y = 0.320 + 0.457x
(d) M1 substitute x = 6.5 into their regression equation
(d) A1 awrt 3.29 (i.e. sales of about GBP 329)
(d) B1 x = 6.5 lies within the range of the given data (2 to 10 hours), so this is interpolation and the estimate is reasonably reliable, oe
(d) Answer: y = awrt 3.29 (hundreds of pounds), i.e. sales of about GBP 329; reasonably reliable since 6.5 lies within the given data range.

Question 11

(a) M1 attempt mean = sum(x)/20
(a) A1 cao mean = 210 (GBP thousand)
(a) M1 attempt √sum(x²)/20 - mean²
(a) A1 awrt 43.6 (GBP thousand)
(a) Answer: mean = GBP 210,000, SD = awrt GBP 43,600
(b) M1 attempt mean = sum(x)/15
(b) A1 cao mean = 220 (GBP thousand)
(b) M1 attempt √sum(x²)/15 - mean²
(b) A1 awrt 36.5 (GBP thousand)
(b) Answer: mean = GBP 220,000, SD = awrt GBP 36,500
(c) B1 Birch Street has the higher mean price (GBP 220,000 vs GBP 210,000), so houses there tend to be more expensive, oe
(c) B1 Elm Street has the higher standard deviation (GBP 43,600 vs GBP 36,500), so house prices there are more variable/spread out, oe
(c) Answer: Houses in Birch Street are, on average, more expensive (higher mean), but house prices in Elm Street are more variable (higher standard deviation).

Question 12

(a) M1 substitute values into (mean - mode)/SD, i.e. (45 - 58)/12
(a) A1 awrt -1.08
(a) Answer: awrt -1.08
(b) M1 substitute values into (Q3 + Q1 - 2Q2)/(Q3 - Q1), i.e. (54 + 38 - 100)/(54 - 38)
(b) A1 cao -0.5
(b) Answer: -0.5
(c) B1 both coefficients are negative, indicating the distribution is negatively skewed (a longer tail towards lower values)
(c) B1 this is consistent with mean < median < mode, the typical pattern for a negatively skewed distribution, oe
(c) Answer: Both coefficients are negative, so the distribution of delivery times is negatively skewed, with a longer tail towards lower values. This agrees with the fact that mean (45) < median (50) < mode (58), the pattern typical of negative skew.
(d) B1 the quartile coefficient is not affected by extreme values/outliers (or does not require the mode, which may not be well defined for continuous or multi-modal data), oe
(d) Answer: The quartile coefficient is not affected by extreme values (outliers) and does not rely on the mode, which is not always well defined for continuous data.

Question 13

(a) B1 a strong correlation does not imply causation
(a) B1 identifies a plausible confounding (third) variable, e.g. hours of sunshine or temperature
(a) B1 explains that this variable is likely to affect both x and y directly, e.g. more sunshine leads to both more ice cream shop custom and more sunburn cases, rather than one causing the other, oe
(a) Answer: A strong correlation does not mean one variable causes the other. Here, the amount of sunshine (or temperature) is a likely confounding variable: warmer, sunnier weather leads to both more customers for ice cream shops and more sunburn, rather than the ice cream shops themselves causing sunburn.
(b) B1 identifies that x = 25 lies outside the range of the given data (2 to 15), so this would be extrapolation
(b) B1 explains that the linear relationship is not known to hold outside the observed range, so a prediction here would be unreliable, oe
(b) Answer: x = 25 is outside the observed range of x (2 to 15), so using the regression line here would involve extrapolation; there is no evidence that the linear relationship continues to hold beyond the data collected, so the prediction would be unreliable.
(c) B1 the whole year's data includes other seasons (e.g. winter), where the relationship between temperature and rainfall may be different from summer, oe
(c) B1 combining all seasons could obscure, distort or average out the pattern that actually exists specifically in summer, so the whole-year PMCC would not answer the question about summer months, oe
(c) Answer: The whole year's data includes other seasons, such as winter, where the relationship between temperature and rainfall may behave quite differently to summer. Combining all months together could mask or distort the true summer relationship, so a PMCC calculated across the whole year would not correctly represent the relationship in summer alone; only the summer months' data should be used.

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