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IEB practice · Paper II

IEB Statistics practice

10 newly generated questions across all 13 statistics topics the IEB examines in Paper II, 120 marks in all. Work each one, then open its answer underneath to check yourself. Nothing here is taken from a past paper.

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Cumulative frequency curves

1 questions · 15 marks

Paper II · Section A · 15 marks

QUESTION 1

The ogive (cumulative frequency curve) below shows the mass in kg of 200 parcels.

0510152025050100150200
  1. (a)
    How many data values does the ogive represent?
    (1)
  2. (b)
    Estimate the median.
    (2)
  3. (c)
    Estimate the interquartile range.
    (3)
  4. (d)
    How many of the data values lie between 20 and 25?
    (2)
  5. (e)
    The lowest value is 1 and the highest is 24. Draw a box-and-whisker plot of the data.
    (3)
  6. (f)
    What percentage of the data values are 10 or more?
    (2)
  7. (g)
    Is the data skewed? Use the quartiles to explain.
    (2)

[15]

Show the answers

(a) Answer: 200

  1. the ogive ends at a cumulative frequency of 200

(b) Answer: about 14.4

  1. the median is at a cumulative frequency of 200/2 = 100
  2. read across to the ogive and down: about 14.4

(c) Answer: about 13.2

  1. Q1 at 50: about 4.6; Q3 at 150: about 17.8
  2. IQR = 17.8 − 4.6 = 13.2

(d) Answer: 11

  1. read the ogive at 20 and at 25: 189 and 200
  2. 200 − 189 = 11

(e) Answer: See the box-and-whisker plot below.

051015202514.614.417.824
  1. minimum 1, Q1 ≈ 4.6, median ≈ 14.4, Q3 ≈ 17.8, maximum 24

(f) Answer: 57.5%

  1. 85 values lie below 10, so 115 values are 10 or more
  2. 115/200 × 100 = 57.5%

(g) Answer: Skewed to the left (negatively)

  1. Q3 − median ≈ 3.3 and median − Q1 ≈ 9.8
  2. the lower half of the box is longer

The least-squares line of best fit

1 questions · 10 marks

Paper II · Section A · 10 marks

QUESTION 2

The table and scatter plot show the number of hours (x) a learner studied and the learner's test mark out of 50 (y): (1; 3); (3; 10); (4; 15); (5; 21); (9; 23); (10; 30); (11; 27); (12; 40).

02468101214−1001020304050
  1. (a)
    Use your calculator to determine the equation of the least squares regression line in the form y = A + Bx. Give A and B correct to three decimal places.
    (3)
  2. (b)
    Determine the correlation coefficient r, correct to two decimal places.
    (1)
  3. (c)
    Describe the strength and direction of the correlation.
    (2)
  4. (d)
    Use your equation to predict y when x = 8.
    (2)
  5. (e)
    Is this prediction reliable? Explain.
    (2)

[10]

Show the answers

(a) Answer: y = 2.611 + 2.693x

  1. A = 2.611
  2. B = 2.693

(b) Answer: r = 0.95

(c) Answer: a strong positive correlation

  1. r = 0.95: r > 0 and |r| is at least 0.8

(d) Answer: y = 24.16

  1. y = 2.611 + 2.693(8)

(e) Answer: Yes, reasonably: x lies within the data (interpolation)

  1. the data runs from x = 1 to x = 12; x = 8 lies inside it

Box-and-whisker plots

1 questions · 10 marks

Paper II · Section A · 10 marks

QUESTION 3

The box-and-whisker plots below show the marks (as a percentage) of two classes, A and B, in the same test.

1520253035404550556065A2030384752B3138435160
  1. (a)
    Write down the median of each group.
    (2)
  2. (b)
    Calculate the interquartile range of each group.
    (2)
  3. (c)
    Describe the skewness of the data of group B.
    (2)
  4. (d)
    Approximately what percentage of the values of group A are greater than Q1 of group B?
    (2)
  5. (e)
    Which class did better in the test? Give a reason.
    (2)

[10]

Show the answers

(a) Answer: A: 38; B: 43

  1. the line inside each box

(b) Answer: A: 17; B: 13

  1. A: 47 − 30 = 17
  2. B: 51 − 38 = 13

(c) Answer: Skewed to the right (positively)

  1. Q3 − median = 8 and median − Q1 = 5
  2. the median lies nearer Q1

(d) Answer: 50%

  1. Q1 of group B is 38, which is the median of group A
  2. 50% of the values lie above it

(e) Answer: B: its median and both its quartiles are higher

  1. median 43 > 38, Q1 38 > 30, Q3 51 > 47

The estimated mean of grouped data

2 questions · 25 marks

Paper II · Section A · 10 marks

QUESTION 4

The table shows the time in minutes that 54 people waited at a clinic. One frequency is unknown.

  • 0 ≤ x < 10: frequency x
  • 10 ≤ x < 20: frequency 10
  • 20 ≤ x < 30: frequency 16
  • 30 ≤ x < 40: frequency 10
  • 40 ≤ x < 50: frequency 9
  1. (a)
    The estimated mean is 25. Determine the value of x.
    (3)
  2. (b)
    Write down the modal class.
    (1)
  3. (c)
    In which class does the median lie?
    (2)
  4. (d)
    Complete the cumulative frequencies and draw the ogive.
    (4)

[10]

Show the answers

(a) Answer: x = 9

  1. (1305 + 5x)/(45 + x) = 25
  2. 1305 + 5x = 25(45 + x)

(b) Answer: 20 ≤ x < 30

  1. the highest frequency is 16

(c) Answer: 20 ≤ x < 30

  1. n = 54: the median lies between the 27th and 28th values
  2. cumulative frequencies: 9, 19, 35, 45, 54

(d) Answer: See the ogive below.

010203040500102030405060
  1. cumulative frequencies: 9, 19, 35, 45, 54
  2. plot each against the upper limit of its class, starting at (0; 0)

Paper II · Section A · 15 marks

QUESTION 5

The frequency polygon shows the heights in cm of 37 seedlings, grouped in the classes 10 ≤ x < 20; 20 ≤ x < 30; 30 ≤ x < 40; 40 ≤ x < 50; 50 ≤ x < 60.

010203040506070051015
  1. (a)
    Estimate the mean of the data.
    (3)
  2. (b)
    Write down the modal class.
    (1)
  3. (c)
    Draw an ogive (cumulative frequency curve) for the data.
    (4)
  4. (d)
    Use your ogive to estimate the median.
    (2)
  5. (e)
    Use your ogive to estimate the interquartile range.
    (3)
  6. (f)
    Is the data skewed? Explain, using the mean and the median.
    (2)

[15]

Show the answers

(a) Answer: 41.49

  1. midpoints: 15, 25, 35, 45, 55
  2. Σ f·x = 1535 and n = 37
  3. mean = 1535/37 = 41.49

(b) Answer: 50 ≤ x < 60

  1. the highest frequency is 14

(c) Answer: See the ogive below.

1020304050600510152025303540
  1. cumulative frequencies: 4, 9, 14, 23, 37
  2. plot each against the upper limit of its class, starting at (10; 0)

(d) Answer: about 45.0

  1. the median is at a cumulative frequency of 37/2 = 18.5
  2. read across to the ogive and down: about 45.0

(e) Answer: about 22.9

  1. Q1 at 9.25: about 30.5; Q3 at 27.75: about 53.4
  2. IQR = 53.4 − 30.5 = 22.9

(f) Answer: Skewed to the left (negatively)

  1. mean 41.49 and median about 45.0
  2. mean < median

Mean, median and mode

3 questions · 39 marks

Paper II · Section A · 12 marks

QUESTION 6

The 9 values below give the number of goals scored by a netball team in 9 matches, in ascending order: p; 13; 18; 25; 31; 32; 36; 38; q. The mean of the values is 27 and their range is 30.

  1. (a)
    Determine the values of p and q.
    (4)
  2. (b)
    Write down the median.
    (1)
  3. (c)
    Calculate the standard deviation, correct to two decimal places.
    (2)
  4. (d)
    How many of the values lie within one standard deviation of the mean?
    (2)
  5. (e)
    The value q is removed from the data. Determine the mean and the median of the remaining values. Give the mean correct to two decimal places.
    (3)

[12]

Show the answers

(a) Answer: p = 10 and q = 40

  1. mean: (p + 193 + q)/9 = 27, so p + q = 50
  2. range: q − p = 30
  3. adding: 2q = 80, so q = 40
  4. p = 40 − 30 = 10

(b) Answer: 31

  1. n = 9: the median is the middle value, 31

(c) Answer: 10.45

  1. the data: 10; 13; 18; 25; 31; 32; 36; 38; 40
  2. σ = 10.45 (calculator)

(d) Answer: 5

  1. mean − σ = 27 − 10.45 = 16.55; mean + σ = 37.45
  2. 5 values lie between them

(e) Answer: mean 25.38; median 28

  1. mean = (243 − 40)/8 = 25.38
  2. 8 values: the median is halfway between the two middle ones, (25 + 31)/2 = 28

Paper II · Section A · 15 marks

QUESTION 7

The data below gives the marks out of 50 of 10 learners in a quiz: 13; 17; 19; 20; 20; 21; 24; 24; 25; 27

  1. (a)
    Calculate the mean.
    (1)
  2. (b)
    Calculate the standard deviation, correct to two decimal places.
    (2)
  3. (c)
    How many of the values lie within one standard deviation of the mean?
    (2)
  4. (d)
    Write down the five-number summary.
    (3)
  5. (e)
    Draw a box-and-whisker plot of the data.
    (2)
  6. (f)
    An outlier is a value more than 1.5 × IQR above Q3 or below Q1. Is 27 an outlier? Show your working.
    (3)
  7. (g)
    Every value is increased by 3. What happens to the mean and to the standard deviation?
    (2)

[15]

Show the answers

(a) Answer: 21.00

  1. Σx = 210 and n = 10
  2. mean = 210/10 = 21.00

(b) Answer: σ = 3.95

  1. σ² = Σ(x − 21.00)²/10 = 15.60

(c) Answer: 6

  1. between 17.05 and 24.95
  2. 6 values

(d) Answer: 13; 19; 20.5; 24; 27

  1. minimum 13; Q1 19; median 20.5; Q3 24; maximum 27

(e) Answer: See the plot below.

121416182022242628131920.52427
  1. the box from Q1 to Q3 with the median marked; whiskers to the minimum and maximum

(f) Answer: No: 27 ≤ 31.5

  1. IQR = 24 − 19 = 5
  2. Q3 + 1.5 × IQR = 24 + 7.5 = 31.5
  3. 27 ≤ 31.5

(g) Answer: The mean increases by 3; the standard deviation stays the same.

  1. every value, and so their mean, moves by the same amount
  2. the distances from the mean do not change

Paper II · Section A · 12 marks

QUESTION 8

A data set gives the times in minutes that 20 runners took to finish a race. The mean is 48 and the standard deviation is 8.

  1. (a)
    Calculate the sum of all the values.
    (1)
  2. (b)
    One value, 72, turns out to be a recording error and is removed. Determine the mean of the remaining values, correct to two decimal places.
    (2)
  3. (c)
    Instead, every value in the original data is multiplied by 1.5 and then 10 is added. Determine the new mean and the new standard deviation.
    (3)
  4. (d)
    Determine Σx^2, the sum of the squares of the original values.
    (3)
  5. (e)
    Go back to the original data. A second data set of 8 values of the same kind has a mean of 52. Determine the mean of all 28 values together, correct to two decimal places.
    (3)

[12]

Show the answers

(a) Answer: 960

  1. Σx = n × mean = 20 × 48 = 960

(b) Answer: 46.74

  1. the sum without it: 960 − 72 = 888
  2. new mean = 888/19 = 46.74

(c) Answer: mean 82; standard deviation 12

  1. mean: 1.5 × 48 + 10 = 82
  2. the spread is scaled by 1.5; adding a constant moves every value equally and leaves it: 1.5 × 8 = 12

(d) Answer: Σx^2 = 47 360

  1. σ^2 = Σx^2/n − (mean)^2
  2. 8^2 = Σx^2/20 − 48^2
  3. Σx^2 = 20(64 + 2304) = 47 360

(e) Answer: 49.14

  1. the two sums: 20 × 48 = 960 and 8 × 52 = 416
  2. mean = (960 + 416)/28 = 49.14
  3. not the mean of the two means: the groups are not the same size

Scatter plots

2 questions · 21 marks

Paper II · Section A · 12 marks

QUESTION 9

The scatter plot shows the number of hours (x) a learner practises the piano each week and the learner's music exam mark out of 50 (y): (2; 12); (3; 12); (4; 13); (5; 19); (6; 22); (7; 28); (8; 34); (9; 33); (10; 32); (11; 20); (12; 29). The least squares regression line of all the data is y = 9.473 + 1.945x. Two of the points are labelled P(11; 20) and Q(12; 29).

02468101214510152025303540PQ
  1. (a)
    Without calculating it, which ONE of the following could be the correlation coefficient of all the data? Give a reason. −0.98; −0.75; −0.08; 0.75; 0.98
    (2)
  2. (b)
    P and Q are removed from the data. Does the correlation coefficient increase or decrease? Explain.
    (2)
  3. (c)
    Does the gradient of the least squares regression line increase or decrease when P and Q are removed? Explain.
    (2)
  4. (d)
    Use the regression line to predict y when x = 23. Comment on your answer.
    (3)
  5. (e)
    Instead, go back to all the data. The marks are converted to percentages, so every y-value is doubled. What happens to the correlation coefficient, and what is the gradient of the new least squares regression line?
    (3)

[12]

Show the answers

(a) Answer: r = 0.75: the points rise from left to right, but are spread widely about a line

  1. the trend is upward: r is positive
  2. the points lie widely about a line: the correlation is moderate

(b) Answer: It increases, from 0.75 to 0.96: without P and Q the points lie closer to a line, so the correlation is stronger

  1. with P and Q: r = 0.75
  2. without them: r = 0.96

(c) Answer: It increases, from 1.945 to 3.233: P and Q lie below the trend at the right and pulled that end of the line down

  1. with P and Q: B = 1.945
  2. without them: B = 3.233

(d) Answer: y = 54.21. Not reliable: x = 23 lies far outside the data, which runs from x = 2 to x = 12 (extrapolation), and a mark above 50 is impossible

  1. y = 9.473 + 1.945(23)
  2. y = 54.21

(e) Answer: r stays the same; the new gradient is 3.890

  1. every y-value is multiplied by 2: the pattern of the points does not change, so r stays 0.75
  2. the gradient is multiplied by 2: 1.945 × 2 = 3.890

Paper II · Section A · 9 marks

QUESTION 10

The scatter plot shows the number of absences in a term (x) and a learner's mark out of 100 (y): (1; 94); (2; 90); (3; 83); (5; 77); (6; 98); (7; 67); (9; 58); (11; 51); (13; 46); (14; 38).

024681012141630405060708090100110
  1. (a)
    Identify the outlier in the data.
    (1)
  2. (b)
    The outlier is removed. Determine the equation of the least squares regression line for the remaining data, in the form y = A + Bx, with A and B correct to three decimal places.
    (3)
  3. (c)
    Determine the correlation coefficient r of the remaining data, correct to two decimal places.
    (1)
  4. (d)
    By how much does y change, on average, for each increase of 1 in x? Use your equation.
    (2)
  5. (e)
    Did the outlier make the correlation stronger or weaker? Explain.
    (2)

[9]

Show the answers

(a) Answer: (6; 98)

  1. (6; 98) lies far from the trend of the other points

(b) Answer: y = 97.168 − 4.162x

  1. A = 97.168
  2. B = −4.162

(c) Answer: r = −1.00

(d) Answer: y decreases by about 4.162

  1. B = −4.162 is the change in y for each 1 added to x

(e) Answer: Weaker

  1. with the outlier r = −0.92; without it r = −1.00
  2. |r| is smaller with the outlier: it weakened the correlation

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