Statistical sampling and data – Year 12 A level maths worksheet answers to download as PDF

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1. Answer to Question 1

   1. Stratified: sample proportional to each group. The fraction is \(\tfrac{60}{600}=\tfrac1{10}\), so year 10: 15, year 11: 13, year 12: 19, year 13: 13 (total 60). Select each group randomly (e.g. random numbers).
   2. The interval is \(\dfrac{600}{60}=10\). Choose a random start between 1 and 10 (say 7) and then take every 10th student: 7, 17, 27, \dots
   3. It is an opportunity sample: early arrivals may be unrepresentative (e.g. only those who use buses or who are keen), so the sample is biased and not random.

2. Answer to Question 2

   1. Mean \(=\dfrac{143}7=\) 20.4. Median (4th value) \(=\) 18. Mode \(=\) 15.
   2. New data: 12, 15, 15, 18, 20, 23. Mean \(=\dfrac{103}6=\) 17.2. Median \(=\dfrac{15+18}2=\) 16.5.
   3. The mean changes by 3.2 and the median by 1.5. The mean uses every value so it is pulled towards an extreme value; the median depends only on the middle.

3. Answer to Question 3

   1. Multiply each value \(x\) by its frequency \(f\), and each \(x^2\) by \(f\), then add. \(\sum fx=4+14+30+24+15=\) 87. \(\sum fx^2=4+28+90+96+75=\) 293.
   2. Use \(\bar x=\dfrac{\sum fx}{\sum f}\) and \(\sigma^2=\dfrac{\sum fx^2}{\sum f}-\bar x^2\). Mean \(=\dfrac{87}{30}=\) 2.9. Variance \(=\dfrac{293}{30}-2.9^2=9.767-8.41=1.357\). Standard deviation \(=\sqrt{1.357}=\) 1.16. On average a household has 2.9 pets, with a standard deviation of about 1.16.
   3. With 30 values, the median is the mean of the 15th and 16th values; use cumulative frequencies. The 15th and 16th values lie in the group \(x=3\) (cumulative frequencies 4, 11, 21), so the median is 3.

4. Answer to Question 4

   1. Find the mean first, then use \(\sigma^2=\dfrac{\sum x^2}n-\bar x^2\). Mean \(=9\). \(\sum x^2=16+49+64+121+225=475\). Variance \(=\dfrac{475}5-81=14\). Standard deviation \(=\sqrt{14}=\) 3.74. The values lie on average about 3.74 from the mean of 9.
   2. Coding with \(y=x-50\) shifts every value by 50; find \(\bar y\) and \(\sigma_y\) first, then decode. Let \(y=x-50\). \(\bar y=\dfrac{30}{10}=3\), so \(\bar x=\) 53. Variance of \(y=\dfrac{250}{10}-9=16\), so the standard deviation is \(\sqrt{16}=\) 4 (subtracting a constant does not change the spread).

5. Answer to Question 5

   1. Median \(=\dfrac{9+10}2=\) 9.5. Lower half 3, 5, 6, 8, 9 gives \(Q_1=\) 6; upper half 10, 12, 13, 14, 31 gives \(Q_3=\) 13. IQR \(=13-6=\) 7.
   2. \(Q_3+1.5\times7=23.5\) and \(Q_1-1.5\times7=-4.5\).
   3. 31 is above 23.5, so it is an outlier. On a box plot it is marked with a cross beyond the whisker, and the whisker stops at the largest non-outlier (14).

6. Answer to Question 6

   1. In a histogram the height of each bar is the frequency density. Frequency density \(=\dfrac{\text{frequency}}{\text{class width}}=\dfrac{24}{20}=1.2\). So the bar for 30-50 has height 1.2.
   2. Use linear interpolation: 25-30 is half of the class 20-30 and 30-40 is half of the class 30-50. Assume frequencies are evenly spread within each class. 25-30: half of 20 \(=10\). 30-40: half of 24 \(=12\). Total \(\approx\) 22. About 22 students took between 25 and 40 minutes.
   3. In a histogram the area (not the height) represents frequency; a wide class with few students has a low frequency density.

7. Answer to Question 7

   1. Look at the pattern of the points: as \(x\) increases, \(y\) increases and the points lie close to a straight line. Strong positive linear correlation.
   2. The gradient is the change in \(y\) for each increase of 1 in \(x\). Each extra hour of revision is associated with an increase of about 1.03 marks.
   3. Substitute \(x=4.5\) into the regression line; 4.5 hours is within the range of the data. \(y=2.26+1.026\times4.5=\) 6.88 (interpolation, reliable). The estimated mark is about 6.9 out of 10.
   4. 12 hours is outside the data range (extrapolation): the line gives 14.6, impossible for a test marked out of 10, and the linear relationship may not continue.

8. Answer to Question 8

   1. Mia: opportunity (convenience) sample. Omar: simple random sample.
   2. Advantage: quick and cheap. Disadvantage: biased, because it only includes people who already use a leisure centre on Saturday mornings.
   3. Advantage: every adult on the register has an equal chance of selection, so it is unbiased. Disadvantage: the register excludes some residents (e.g. under 18s), and some people may not respond.
   4. Omar\'s method, because it is representative of the whole population and removes the selection bias.

9. Answer to Question 9

   1. Mean \(=\dfrac{140.6}{10}=\) 14.06. \(\sum x^2=2062.56\), so the standard deviation is 2.93.
   2. Ordered: 11.8, 12.1, 12.5, 12.9, 13.0, 13.3, 13.4, 14.0, 15.2, 22.4. Median \(=\dfrac{13.0+13.3}{2}=\) 13.15. \(Q_1=12.5\) and \(Q_3=14.0\) (medians of the halves).
   3. IQR \(=14.0-12.5=1.5\). Upper limit \(=14.0+1.5\times1.5=16.25\). Since \(22.4>16.25\), it is an outlier.
   4. Because of the outlier the median (13.15) and IQR (1.5) are better than the mean and standard deviation, which are pulled upwards.

10. Answer to Question 10

    1. The modal class has the greatest frequency density: \(\tfrac{12}{2}=6\), \(\tfrac{25}2=12.5\), \(\tfrac{30}4=7.5\), \(\tfrac8{8}=1\). So £10-£12.
    2. Mid-points 9, 11, 14, 20. \(\sum fx=108+275+420+160=963\). Mean \(\approx\dfrac{963}{75}=\) £12.84.
    3. \(n=75\), median at the 37.5th value. Cumulative: 12, 37, 67. It lies in 12-16: \(12+\dfrac{37.5-37}{30}\times4=\) £12.07.
    4. The exact values within each class are not known, so we assume each value is at the mid-point (or evenly spread).

11. Answer to Question 11

    1. Frequency density is frequency divided by class width. \(\dfrac{30}{30}=1\). The value in the table is correct.
    2. Assume the values are spread evenly within each class and use proportions of the class widths. 25-30: half of 75 \(=37.5\). 30-40: \(\dfrac{10}{15}\) of 45 \(=30\). Total \(\approx\) 67.5 (about 68).
    3. Use the mid-point of each class as a representative value. Mid-points 5, 15, 25, 37.5, 60: \(\sum fx=150+900+1875+1687.5+1800=6412.5\). Mean \(\approx\dfrac{6412.5}{240}=\) 26.7 minutes.
    4. Positively skewed: a long tail to the right (a few very long journeys).

12. Answer to Question 12

    1. Class A: \(62-35=27\). Class B: \(58-40=18\).
    2. The medians are similar (48 and 50), so a typical student scored about the same, but Class A has a larger spread (IQR 27 against 18, range 78 against 55).
    3. Class B has the smaller spread, so it was more consistent.
    4. Upper limit \(=62+1.5\times27=102.5\) and lower limit \(=35-40.5=-5.5\). Max 90 and min 12 lie inside the limits.

13. Answer to Question 13

    1. Gradient: each extra hour of revision is associated with about 5 extra marks. Intercept: a student who does no revision would be expected to score about 34 marks (although this is outside the data range).
    2. \(34.4+4.96\times4=\) 54.2 marks (interpolation).
    3. \(34.4+4.96\times12=93.9\), but 12 hours is outside the range of the data (2 to 10 hours): extrapolation, and the relationship may level off.
    4. \(r=0.996\) is very close to 1: very strong positive linear correlation. (It does not prove that revision causes higher scores.)

14. Answer to Question 14

    1. Use \(\bar y=\dfrac{\sum y}n\) and \(\sigma_y^2=\dfrac{\sum y^2}n-\bar y^2\) with \(n=40\). \(\bar y=\dfrac{64}{40}=1.6\). Variance \(=\dfrac{360}{40}-1.6^2=9-2.56=6.44\). \(\sigma_y=\sqrt{6.44}=2.54\). So \(\bar y=1.6\) and \(\sigma_y\approx2.54\).
    2. Rearrange the coding to \(x=10y+100\). Adding 100 changes the mean but not the spread; multiplying by 10 multiplies both the mean and the standard deviation by 10. \(x=10y+100\), so \(\bar x=10\times1.6+100=\) 116 g and \(\sigma_x=10\times2.54=\) 25.4 g. So the potatoes have a mean mass of 116 g and a standard deviation of about 25.4 g.
    3. Find the value two standard deviations above the mean and compare. Mean \(+2\sigma=116+50.8=166.8\) g. 175 g is greater, so yes. So this potato is unusually heavy compared with the rest of the sample.

15. Answer to Question 15

    1. Mean \(=\dfrac{278}8=\) 34.75. Median \(=\dfrac{27+28}2=\) 27.5.
    2. \(Q_1=\dfrac{25+26}2=25.5\), \(Q_3=\dfrac{29+31}2=30\). IQR \(=4.5\). Upper limit \(=30+6.75=36.75\). \(90>36.75\), so it is an outlier.
    3. Without 90: mean \(=\dfrac{188}7=\) 26.9. The outlier raised the mean by nearly 8 minutes but the median only by 0.5, so the median is the better measure of a typical delivery.
    4. A traffic incident or a mistake in recording; investigate before deciding whether to remove it.

16. Answer to Question 16

    1. Frequency = density \(\times\) width: \(30\), \(40\), \(50\), \(40\), \(10\). Total \(=\) 170.
    2. Median at the 85th value. Cumulative: 30, 70, 120. It lies in 30-40: \(30+\dfrac{85-70}{50}\times10=\) 33.0 years.
    3. Mid-points 10, 25, 35, 50, 70: \(\sum fx=300+1000+1750+2000+700=5750\). Mean \(\approx\dfrac{5750}{170}=\) 33.8 years.
    4. 45-60: \(\dfrac{15}{20}\times40=30\); 60-80: 10. Total \(\approx\) 40.

17. Answer to Question 17

    1. Each extra £100 spent on advertising is associated with about £604 extra sales.
    2. £450 means \(x=4.5\): \(y=1.52+0.604\times4.5=4.24\), i.e. £4 240 (interpolation).
    3. \(10=1.52+0.604x\) gives \(x=14.05\), i.e. about £1 400 of advertising. This is far outside the range of data (£100 to £800), so the estimate is unreliable.
    4. \(r\) is very close to 1: very strong positive linear correlation. Correlation does not prove cause: other factors such as the season might affect both.

18. Answer to Question 18

    1. IQR \(=Q_3-Q_1\). May: \(3.2-0=3.2\). August: \(4.6-0=4.6\).
    2. Find the upper limit \(Q_3+1.5\times\text{IQR}\) for each month and compare the largest value. May: upper limit \(=3.2+1.5\times3.2=8.0\); \(14.2>8.0\), outlier. August: upper limit \(=4.6+1.5\times4.6=11.5\); \(12.4>11.5\), outlier.
    3. Mean \(>\) median means positive skew: many dry days and a few very wet days.
    4. August: its IQR (4.6) is larger than May\'s (3.2). A larger IQR means the middle half of the data is more spread out.

19. Answer to Question 19

    1. \(\dfrac{120}{1200}=\dfrac1{10}\).
    2. Boys: 14, 13, 15, 18. Girls: 16, 17, 15, 12. Total \(=60+60=120\).
    3. A simple random sample could by chance miss or over-represent a group; stratified sampling guarantees that each group is in proportion.
    4. Number the 150 Year 12 girls and use random numbers (or a calculator) to select 15 different numbers.

20. Answer to Question 20

    1. Use the formulas \(S_{xx}=\sum x^2-\dfrac{(\sum x)^2}n\), \(S_{xy}=\sum xy-\dfrac{\sum x\sum y}n\) and \(S_{yy}=\sum y^2-\dfrac{(\sum y)^2}n\) with \(n=10\). \(S_{xx}=300-\dfrac{50^2}{10}=50\). \(S_{xy}=640-\dfrac{50\times120}{10}=40\). \(S_{yy}=1480-\dfrac{120^2}{10}=40\). These summary values are used in the next two parts.
    2. The gradient is \(b=\dfrac{S_{xy}}{S_{xx}}\), and the line passes through the mean point \((\bar x,\bar y)\). \(b=\dfrac{S_{xy}}{S_{xx}}=\dfrac{40}{50}=0.8\). \(\bar x=5\), \(\bar y=12\), so \(a=12-0.8\times5=8\). Line: \(y=8+0.8x\). Check: when \(x=5\), \(y=8+4=12=\bar y\) ✓.
    3. Use the formula for the product-moment correlation coefficient. \(r=\dfrac{S_{xy}}{\sqrt{S_{xx}S_{yy}}}=\dfrac{40}{\sqrt{2000}}=\) 0.894. This shows strong positive linear correlation between \(x\) and \(y\).
    4. Substitute \(x=6\) into the regression line. \(y=8+0.8\times6=\) 12.8. Since \(x=6\) is close to the mean of the data, this estimate should be reliable.

21. Answer to Question 21

    1. The combined total is the sum of the two group totals, \(n\bar x\) for each group. Total \(=20\times60+30\times70=3300\); mean \(=\dfrac{3300}{50}=\) 66. The combined mean is 66, closer to 70 because group B is larger.
    2. Rearrange \(\sigma^2=\dfrac{\sum x^2}n-\bar x^2\) to get \(\sum x^2=n(\sigma^2+\bar x^2)\) for each group, then add. Group A: \(20(25+3600)=72\,500\). Group B: \(30(64+4900)=148\,920\). Combined \(\sum x^2=221\,420\). These are the sums of squares of all the scores.
    3. Use \(\sigma^2=\dfrac{\sum x^2}n-\bar x^2\) for the combined group of 50 students. Variance \(=\dfrac{221420}{50}-66^2=4428.4-4356=72.4\). Standard deviation \(=\sqrt{72.4}=\) 8.51. The combined standard deviation (8.51) is larger than either group's, because the two group means differ by 10.