Exam practice

cf: 4, 14, 34, 44, 50; N = 50.
Q₁: the 12.5th item, in 10–20: $10 + \frac{12.5 - 4}{10} \times 10 = 18.5$.
Q₃: the 37.5th item, in 30–40: $30 + \frac{37.5 - 34}{10} \times 10 = 33.5$.
Q.D. $= \frac{33.5 - 18.5}{2} = 7.5$; coefficient $= \frac{15}{52} \approx 0.29$.

cf: 4, 10, 13, 21, 33, 40; N = 40.
Q₁: the 10th item, in 5–10: $5 + \frac{10 - 4}{6} \times 5 = 10$.
Q₃: the 30th item, in 20–25: $20 + \frac{30 - 21}{12} \times 5 = 23.75$.
Q.D. $= \frac{23.75 - 10}{2} = 6.875 \approx 6.88$ thousand; coefficient $= \frac{13.75}{33.75} \approx 0.41$.

cf: 4, 9, 19, 34, 42, 49, 50; N = 50.
Q₁: the 12.5th item, in 40–50: $40 + \frac{12.5 - 9}{10} \times 10 = 43.5$.
Q₃: the 37.5th item, in 60–70: $60 + \frac{37.5 - 34}{8} \times 10 = 64.375$.
Q.D. $= \frac{64.375 - 43.5}{2} \approx 10.44$ kg; coefficient $= \frac{20.875}{107.875} \approx 0.19$.

10 ≤ x < 20 is the class 10–20 written exactly: 10 is in it, 20 is not. cf: 5, 15, 30, 40, 45; N = 45.
Q₁: the 11.25th item, in 10–20: $10 + \frac{11.25 - 5}{10} \times 10 = 16.25$.
Q₃: the 33.75th item, in 30–40: $30 + \frac{33.75 - 30}{10} \times 10 = 33.75$.
Q.D. $= \frac{33.75 - 16.25}{2} = 8.75$; coefficient $= \frac{17.5}{50} = 0.35$.

Taking the classes 10 wide, the first is 10–20 (5 students), then 20–30 (7), 30–40 (15), 40–50 (13), 50–60 (10). cf: 5, 12, 27, 40, 50; N = 50.
Q₁: the 12.5th item, in 30–40: $30 + \frac{12.5 - 12}{15} \times 10 = 30.33$.
Q₃: the 37.5th item, in 40–50: $40 + \frac{37.5 - 27}{13} \times 10 = 48.08$.
Q.D. $= \frac{48.08 - 30.33}{2} \approx 8.87$ (8.88 from the rounded quartiles); coefficient $= \frac{17.74}{78.41} \approx 0.23$. The first class doesn’t matter here: neither quartile is in it.

A “more than” table counts from the top: 100 earn more than Rs 100, and 80 of them more than Rs 200, so 100 − 80 = 20 earn Rs 100–200. The classes are 100–200, 200–300, 300–400, 400–500 and 500–600, each with 20 workers (the last class taken as 100 wide like the rest). cf: 20, 40, 60, 80, 100.
Q₁: the 25th item, in 200–300: $200 + \frac{25 - 20}{20} \times 100 = 225$.
Q₃: the 75th item, in 400–500: $400 + \frac{75 - 60}{20} \times 100 = 475$.
Q.D. $= \frac{475 - 225}{2} =$ Rs 125; coefficient $= \frac{250}{700} \approx 0.36$.

(10–20: 12, 18, 15, 19, 14; 20–30: 25, 22, 29, 20, 24, 21, 26, 23; 30–40: 36, 33, 38, 30, 37, 35, 31; 40–50: 45, 42, 41, 48, 44, 40, 49; 50–60: 54, 52, 50. Total 30.)

cf: 5, 13, 20, 27, 30. Q₁ is the 7.5th item, in 20–30: $Q_1 = 20 + \frac{7.5 - 5}{8} \times 10 =$ 23.13.

Q₃ is the 22.5th item, in 40–50: $Q_3 = 40 + \frac{22.5 - 20}{7} \times 10 =$ 43.57.

Q.D. $= \frac{43.57 - 23.13}{2} =$ 10.22.
(If the classes are taken as 10–19, 20–29, … with class boundaries 9.5, 19.5, …, both quartiles come out 0.5 lower, 22.63 and 43.07, and the quartile deviation is the same, 10.22.)

$\frac{43.57 - 23.13}{43.57 + 23.13} = \frac{20.44}{66.70} \approx$ 0.31.

m: 2, 6, 10, 14, 18, 22; N = 52, $\Sigma fm = 624$, X = 12.
|m − 12|: 10, 6, 2, 2, 6, 10; f|D|: 70, 42, 20, 30, 42, 60; $\Sigma f|D| = 264$.
MD $= \frac{264}{52} \approx 5.08$ kg; coefficient $= \frac{5.08}{12} \approx 0.42$.

m: 5, 15, …, 55; N = 100, $\Sigma fm = 3100$, X = 31.
|m − 31|: 26, 16, 6, 4, 14, 24; f|D|: 208, 192, 120, 160, 168, 192; $\Sigma f|D| = 1040$.
MD $= \frac{1040}{100} = 10.40$ marks; coefficient $= \frac{10.4}{31} \approx 0.34$.

cf: 7, 19, 37, 65, 81, 95; N = 95. The median is the 47.5th item, in 50–60: $M_d = 50 + \frac{47.5 - 37}{28} \times 10 = 53.75$.
|m − 53.75|: 28.75, 18.75, 8.75, 1.25, 11.25, 21.25; f|D|: 201.25, 225, 157.5, 35, 180, 297.5; $\Sigma f|D| = 1096.25$.
MD $= \frac{1096.25}{95} \approx 11.54$ kg; coefficient $= \frac{11.54}{53.75} \approx 0.21$.

cf: 1, 5, 11, 15, 16; N = 16. The median is the 8th item, in 7–9: $M_d = 7 + \frac{8 - 5}{6} \times 2 = 8$.
|m − 8|: 4, 2, 0, 2, 4; f|D|: 4, 8, 0, 8, 4; $\Sigma f|D| = 24$.
MD $= \frac{24}{16} = 1.5$ kg; coefficient $= \frac{1.5}{8} = 0.1875$.

The classes: 0–10 (1), 10–20 (2), 20–30 (4), 30–40 (2), 40–50 (1). $\Sigma fm = 5 + 30 + 100 + 70 + 45 = 250$, X = 25.
|m − 25|: 20, 10, 0, 10, 20; f|D|: 20, 20, 0, 20, 20; $\Sigma f|D| = 80$.
MD $= \frac{80}{10} = 8$ marks; coefficient $= \frac{8}{25} = 0.32$.

The median is the 5th item, in 20–30 (cf = 3, f = 4): $M_d = 20 + \frac{5 - 3}{4} \times 10 = 25$. It equals the mean (the table is symmetric), so MD = 8 marks and coefficient = 0.32, as in part a.

The classes: 100–200 (4), 200–300 (6), 300–400 (10), 400–500 (20), 500–600 (10), 600–700 (6), 700–800 (4). N = 60, $\Sigma fm = 27000$, X = 450.
|m − 450|: 300, 200, 100, 0, 100, 200, 300; f|D|: 1200, 1200, 1000, 0, 1000, 1200, 1200; $\Sigma f|D| = 6800$.
MD $= \frac{6800}{60} \approx$ Rs 113.33; coefficient $= \frac{113.33}{450} \approx 0.25$.

Tally carefully: 20 goes in 20–30, 50 in 50–60, 70 in 70–80. Check that the frequencies add up to 28.

a.

• b.$\Sigma fm = 1240$, $\overline{X} = \frac{1240}{28} \approx 44.29$. $\Sigma f|m - 44.29| = 442.86$. MD $\approx \frac{442.86}{28} = 15.82$ marks; coefficient $\approx \frac{15.82}{44.29} = 0.36$.
• c.cf: 3, 8, 12, 17, 21, 25, 28. The median is the 14th item, in 40–50: $M_d = 40 + \frac{14 - 12}{5} \times 10 = 44$. |m − 44|: 29, 19, 9, 1, 11, 21, 31; f|D|: 87, 95, 36, 5, 44, 84, 93; $\Sigma f|D| = 444$. MD $= \frac{444}{28} \approx 15.86$ marks; coefficient $= \frac{15.86}{44} \approx 0.36$.

The standard deviation is the square root of the mean of the squared distances of the values from their mean, $\sigma = \sqrt{\frac{\Sigma f(m - \overline{X})^2}{N}}$. It measures how far a typical value is from the mean, in the data’s own unit. Its square is the variance.

The coefficient of standard deviation written as a percentage: $\text{C.V.} = \frac{\sigma}{\overline{X}} \times 100\%$. It compares the spread of data sets with different means or units; the smaller C.V. belongs to the more uniform set.

Both measure the spread of every value about a central value. The mean deviation takes the distances without their signs, |m − X|, and averages them; it can be taken from the mean, the median or the mode. The standard deviation squares the distances, averages the squares and takes the square root; it is always taken from the mean. Squaring follows the rules of algebra, so the standard deviation is used in further statistics, and it gives more weight to values far from the mean. For the same data it is never smaller than the mean deviation from the mean.

They are the same comparison: the coefficient of SD is $\frac{\sigma}{\overline{X}}$, a plain number, and the C.V. is that number times 100, a percentage. A coefficient of SD of 0.25 is a C.V. of 25%.

Mean $= \frac{400}{100} =$ 4. $\sigma = \sqrt{\frac{4000}{100} - 4^2} = \sqrt{40 - 16} = \sqrt{24} \approx$ 4.899.

$\sigma = \sqrt{\frac{5000}{40} - \left(\frac{400}{40}\right)^2} = \sqrt{125 - 100} = \sqrt{25} =$ 5.

N = 40, $\Sigma fm = 1390$, $\Sigma fm^2 = 54400$. Mean = 34.75.
$\sigma = \sqrt{\frac{54400}{40} - 34.75^2} = \sqrt{1360 - 1207.56} = \sqrt{152.44} \approx$ 12.35 kg; coefficient $= \frac{12.35}{34.75} \approx$ 0.355.

N = 75, $\Sigma fm = 3025$, $\Sigma fm^2 = 138275$. Mean $= \frac{3025}{75} \approx 40.33$.
$\sigma = \sqrt{\frac{138275}{75} - \left(\frac{3025}{75}\right)^2} = \sqrt{1843.67 - 1626.78} = \sqrt{216.89} \approx$ 14.73 marks; coefficient $= \frac{14.73}{40.33} \approx$ 0.365. (With the mean rounded to 40.33 before squaring, the root comes out 14.74.)

N = 15, $\Sigma fm = 126$, $\Sigma fm^2 = 1132$. Mean = 8.4.
$\sigma = \sqrt{\frac{1132}{15} - 8.4^2} = \sqrt{75.47 - 70.56} = \sqrt{4.91} \approx$ 2.22 kg; coefficient $= \frac{2.22}{8.4} \approx$ 0.264.

The classes: 0–10 (5), 10–20 (6), 20–30 (14), 30–40 (10), 40–50 (5). $\Sigma fm = 25 + 90 + 350 + 350 + 225 = 1040$. Mean $= \frac{1040}{40} =$ 26.

$\Sigma fm^2 = 125 + 1350 + 8750 + 12250 + 10125 = 32600$.
$\sigma = \sqrt{\frac{32600}{40} - 26^2} = \sqrt{815 - 676} = \sqrt{139} \approx$ 11.79 marks; coefficient $= \frac{11.79}{26} \approx$ 0.453.

The classes: 600–700 (20), 700–800 (25), 800–900 (30), 900–1000 (10), 1000–1100 (7), 1100–1200 (4), and 1200–1300 (4), taking the last class 100 wide like the others. With A = 950 and h = 100, d′ is −3, −2, −1, 0, 1, 2, 3, and $\Sigma fd' = -60 - 50 - 30 + 0 + 7 + 8 + 12 = -113$.
Mean $= 950 + \frac{-113}{100} \times 100 =$ Rs 837.

$\Sigma fd'^2 = 180 + 100 + 30 + 0 + 7 + 16 + 36 = 369$.
$\sigma = \sqrt{\frac{369}{100} - \left(\frac{-113}{100}\right)^2} \times 100 = \sqrt{3.69 - 1.2769} \times 100 = \sqrt{2.4131} \times 100 \approx$ Rs 155.34; coefficient $= \frac{155.34}{837} \approx$ 0.186.

$\sigma^2 = 2.4131 \times 100^2 =$ 24,131 (square rupees).

C.V. $= 0.1856 \times 100\% \approx$ 18.56%.

Total = number of workers × mean salary. A: 40 × 35000 = Rs 14,00,000. B: 50 × 45000 = Rs 22,50,000. Factory B spends more each month.

C.V. of A $= \frac{40}{35000} \times 100\% \approx 0.114\%$; C.V. of B $= \frac{35}{45000} \times 100\% \approx 0.078\%$. Factory B’s C.V. is smaller, so its salaries are more uniform.

$\Sigma fm = 45 + 175 + 210 + 225 + 275 + 325 + 375 = 1630$. Mean $= \frac{1630}{36} \approx$ 45.28.

$\Sigma fm^2 = 675 + 4375 + 7350 + 10125 + 15125 + 21125 + 28125 = 86900$.
$\sigma = \sqrt{\frac{86900}{36} - \left(\frac{1630}{36}\right)^2} = \sqrt{2413.89 - 2050.08} = \sqrt{363.81} \approx$ 19.07; coefficient $= \frac{19.07}{45.28} \approx$ 0.42.

C.V. $= 0.4213 \times 100\% \approx$ 42.1%.

Fertilizer A: its lowest yield is 48 kg, Fertilizer B’s only 30 kg.

Fertilizer A. Its box runs from 50 to 52 kg (an interquartile range of 2 kg) and its whole range is only 6 kg: year after year it gives about 51 kg. Fertilizer B’s interquartile range is 64 − 41 = 23 kg, and in its worst year it gave only 30 kg. The medians are almost the same (51 and 52 kg), so B gives no more in a typical year; it is only less predictable.

Yes. B’s right whisker stretches from its Q₃, 64 kg, all the way to 95 kg, much longer than any other part of the plot: in at least one year B gave far more than in its usual years. (A long whisker shows the stretch, not how many years are in it; here, with ten years, the top quarter is two or three years, and the highest of them was 95 kg.)

79 hours.

88 hours.

A: 90 − 70 = 20 hours. B: 110 − 60 = 50 hours.

It depends on what matters more, and either answer with the right reason is correct.
Battery B if you want the longer life: its median, 88 hours, is 9 hours more than A’s, and three quarters of B batteries (those above its Q₁, 79 hours) last at least as long as a typical A battery.
Battery A if you need to be sure, for a torch on a trek or a medical device: its range is only 20 hours and its shortest life is 70 hours, while some B batteries die after 60 hours.

With 7 values, Q₁ is the $\frac{7+1}{4} =$ 2nd, the median the 4th, Q₃ the 6th.
Shop A, in order 5, 7, 8, 9, 11, 14, 18: 5, 7, 9, 14, 18.
Shop B, in order 2, 3, 6, 8, 10, 12, 16: 2, 3, 8, 12, 16.

Median: A 9, B 8: on a typical day Shop A sells one phone more.

Interquartile range: A 14 − 7 = 7, B 12 − 3 = 9: Shop A’s daily sales are a little more consistent.

So Shop A sells a little more, and more steadily.

• a.Q.D. $= \dfrac{Q_3 - Q_1}{2}$.
• b.N = 60, $\Sigma fm = 3340$, $\Sigma fm^2 = 200300$. $\overline{X} = \frac{3340}{60} \approx 55.67$. $\sigma = \sqrt{\frac{200300}{60} - \left(\frac{3340}{60}\right)^2} = \sqrt{3338.33 - 3098.78} = \sqrt{239.56} \approx$ 15.48 marks.
• c.C.V. $= \frac{15.48}{55.67} \times 100\% \approx 27.8\%$.

• a.MD $= \dfrac{\Sigma f|m - \overline{X}|}{N}$.
• b.$\Sigma fm = 1850$, $\Sigma fm^2 = 93400$, X = 46.25. $\sigma = \sqrt{\frac{93400}{40} - 46.25^2} = \sqrt{2335 - 2139.06} = \sqrt{195.94} \approx$ 14.00 marks.
• c.The first term’s $\sigma$, 8.75, is much smaller than the second term’s, 14.00, so the first-term marks were more consistent: the class was closer together then, and spread out more in the second term. (Strictly, the C.V. is the fair comparison; with the first term’s mean not given, the standard deviations are compared, which is fair if the two means are close.)

• a.Half the interquartile range, $\frac{Q_3 - Q_1}{2}$: half the width of the middle 50% of the data.
• b.The quartile deviation. The last class, “more than 70,000”, is open-ended: there is no upper limit and so no mid-value, and the mean, the mean deviation and the standard deviation can’t be found. The quartiles don’t need it.
• c.In thousands of rupees: 20–30 (5), 30–40 (20), 40–50 (16), 50–60 (5), 60–70 (2), 70 and over (2). cf: 5, 25, 41, 46, 48, 50.

Q₁: the 12.5th item, in 30–40: $30 + \frac{12.5 - 5}{20} \times 10 = 33.75$.
Q₃: the 37.5th item, in 40–50: $40 + \frac{37.5 - 25}{16} \times 10 = 47.81$.
Q.D. $= \frac{47.81 - 33.75}{2} \approx 7.03$ thousand (about Rs 7,031); coefficient $= \frac{14.06}{81.56} \approx 0.172$.

• a.C.V. $= \dfrac{\sigma}{\overline{X}} \times 100\%$.
• b.Starting at 35:

• c.N = 40, $\Sigma fm = 2130$, $\Sigma fm^2 = 116650$. X = 53.25. $\sigma = \sqrt{\frac{116650}{40} - 53.25^2} = \sqrt{2916.25 - 2835.56} = \sqrt{80.69} \approx 8.983$. C.V. $= \frac{8.983}{53.25} \times 100\% \approx 16.87\%$. (With $\sigma$ rounded to 8.98 first, 16.86%.)
• d.B’s, because B’s C.V., 12%, is smaller than A’s, 21%: B’s marks vary less compared with their mean.

a. With 11 values, Q₁ is the 3rd, the median the 6th and Q₃ the 9th.
Mathematics, in order 20, 23, 44, 58, 64, 68, 75, 82, 87, 93, 98: 20, 44, 68, 87, 98.
Science, in order 15, 19, 25, 31, 35, 38, 45, 57, 65, 71, 80: 15, 25, 38, 65, 80.
b.

c. Interquartile range: mathematics 87 − 44 = 43, science 65 − 25 = 40; range: mathematics 78, science 65. Science’s marks are slightly more consistent: its box and its whole spread are a little narrower. (Mathematics has the higher median, 68 against 38: the class did better in it, but more unevenly.)