Exam practice

An individual series lists each observation separately, one value per item: the marks of 8 students, 25, 48, 32, 52, 21, 64, 29, 57.
A discrete series gives each distinct value with its frequency, the number of times it occurs, usually in a table: 3 households sold 5 pathi of rice, 7 sold 15 pathi, 15 sold 25 pathi, and so on.

Quartiles are the three values that divide data, arranged in ascending order, into four equal parts. Q₁ (lower quartile) has about 25% of the data below it, Q₂ (the median) 50%, and Q₃ (upper quartile) 75%. Example: for 3, 5, 7, 8, 9, 10, 12, 13, 15, 17, 19, Q₁ = 7, Q₂ = 10, Q₃ = 15.

Percentiles are the 99 values that divide data, arranged in ascending order, into 100 equal parts. They are written $P_1, P_2, \ldots, P_{99}$; about n% of the data lies below $P_n$. For example, $P_{60}$ of the 19 heights in 5.1.4 is 141 cm.

With N items in ascending order: $Q_1 = \left(\frac{N + 1}{4}\right)^{\text{th}}$ item, $Q_2 = 2\left(\frac{N + 1}{4}\right)^{\text{th}} = \left(\frac{N + 1}{2}\right)^{\text{th}}$ item, $Q_3 = 3\left(\frac{N + 1}{4}\right)^{\text{th}}$ item.

When the data are arranged in ascending order, three values divide them into four equal parts: the first quartile, the second quartile (the median) and the third quartile. For 11 values they are the 3rd, 6th and 9th; in general the $\frac{N + 1}{4}$th, $\frac{2(N + 1)}{4}$th and $\frac{3(N + 1)}{4}$th.

Cutting a rope at 99 equally spaced points makes 100 equal pieces. In the same way 99 values, the percentiles P₁ to $P_{99}$, divide ordered data into 100 equal parts, and $P_n$ is the $\frac{n(N + 1)}{100}$th item.

With $N = \Sigma f$: $Q_1 = \left(\frac{N + 1}{4}\right)^{\text{th}}$ item, $Q_2 = 2\left(\frac{N + 1}{4}\right)^{\text{th}}$ item, $Q_3 = 3\left(\frac{N + 1}{4}\right)^{\text{th}}$ item; each is the value whose cumulative frequency is equal to or just greater than the place.

Any real data will do. For example:

Individual: the minutes your school bus was late on 9 days: 2, 0, 5, 3, 12, 1, 4, 0, 6.

Discrete: the number of people in the households of your lane: 2 people in 3 houses, 3 in 5, 4 in 9, 5 in 6, 6 in 2.

It is a cumulative frequency curve (an ogive) of the marks of 40 students: above each mark it shows how many students scored less than that mark. The curve rises from 0 at 10 marks to all 40 at 80 marks, steepest between 30 and 50, where most of the marks are.
Reading across from a quarter, a half and three quarters of 40 (10, 20 and 30 students) to the curve and down to the marks:

• lower quartile Q₁ = 36: a quarter of the students scored under 36;
• median = 44: half scored under 44;
• upper quartile Q₃ = 52: three quarters scored under 52.

So the middle half of the class scored between 36 and 52. (Drawing such curves for grouped data comes in Grade 10.)

The 20 students are arranged by their marks, lowest first. You are 17th, and the 16 students before you are $\frac{16}{20} = 80\%$ of the group: 80% of them scored less than you. So your mark is at about the 80th percentile, $P_{80}$; only 3 students, 15%, scored more. It says nothing about your mark itself, only your place in the group.

In order: 21, 25, 29, 32, 48, 52, 57, 64; N = 8, $\frac{N + 1}{4} = 2.25$.
$Q_1 = 2.25^{\text{th}}$ item = 25 + 0.25(29 − 25) = 26
$Q_2 = 4.5^{\text{th}}$ item = 32 + 0.5(48 − 32) = 40
$Q_3 = 6.75^{\text{th}}$ item = 52 + 0.75(57 − 52) = 55.75
The median is the $\frac{8 + 1}{2} = 4.5^{\text{th}}$ item = 40. Q₁ = 26, Q₂ = 40, Q₃ = 55.75. Yes, she is right: $Q_2 = M_d = 40$, because $2 \times \frac{N + 1}{4} = \frac{N + 1}{2}$, the same place.

Already in order; N = 9, $\frac{N + 1}{4} = 2.5$.
$Q_1 = 2.5^{\text{th}}$ item = 20 + 0.5(21 − 20) = 20.5
$Q_2 = 5^{\text{th}}$ item = 23
$Q_3 = 7.5^{\text{th}}$ item = 25 + 0.5(27 − 25) = 26
The median is the $\frac{9 + 1}{2} = 5^{\text{th}}$ item = 23. Q₁ = 20.5, Q₂ = 23, Q₃ = 26. She is right again: the second quartile and the median are at the same place.

The cfs are 3, 64, 196, 349, 489, 540, 543; N = 543.
$Q_1 = \left(\frac{543 + 1}{4}\right)^{\text{th}} = 136^{\text{th}}$ item; the cf just greater than 136 is 196, so Q₁ = 40 years.
$Q_3 = 3 \times 136 = 408^{\text{th}}$ item; the cf just greater than 408 is 489, so Q₃ = 60 years.

The cfs are 3, 10, 25, 30, 38, 40; N = 40, $\frac{N + 1}{4} = 10.25$.

• a.Q₁ is the 10.25th item; the cf just greater than 10.25 is 25: Q₁ = 25 pathi.
• b.Q₃ is the $3 \times 10.25 = 30.75^{\text{th}}$ item; the cf just greater is 38: Q₃ = 45 pathi.
• c.Q₂ is the $2 \times 10.25 = 20.5^{\text{th}}$ item, and the median the $\frac{40 + 1}{2} = 20.5^{\text{th}}$ item: the same place. The cf just greater than 20.5 is 25, so $Q_2 = M_d = 25$ pathi.

In order: 20, 21, 24, 25, 28, 30, 30, 33, 35, 37, 37, 42, 45, 45, 47, 47, 50, 52, 54; N + 1 = 20.

• a.$P_{40} = \left(\frac{40 \times 20}{100}\right)^{\text{th}} = 8^{\text{th}}$ item: $P_{40} = 33$.
• b.$P_{80} = \left(\frac{80 \times 20}{100}\right)^{\text{th}} = 16^{\text{th}}$ item: $P_{80} = 47$.
• c.$P_{50} = \left(\frac{50 \times 20}{100}\right)^{\text{th}} = 10^{\text{th}}$ item = 37; $Q_2 = 2\left(\frac{20}{4}\right)^{\text{th}} = 10^{\text{th}}$ item = 37. $P_{50} = Q_2 = 37$.

The cfs are 3, 10, 25, 30, 38, 40; N = 40, N + 1 = 41.

• a.$P_{30} = \left(\frac{30 \times 41}{100}\right)^{\text{th}} = 12.3^{\text{th}}$ item; the cf just greater is 25: $P_{30} = 30$.
• b.$P_{85} = \left(\frac{85 \times 41}{100}\right)^{\text{th}} = 34.85^{\text{th}}$ item; the cf just greater is 38: $P_{85} = 50$.
• c.$P_{50} = \left(\frac{50 \times 41}{100}\right)^{\text{th}} = 20.5^{\text{th}}$ item and $Q_2 = 2\left(\frac{41}{4}\right)^{\text{th}} = 20.5^{\text{th}}$ item: the same place, with cf 25 just greater. $P_{50} = Q_2 = 30$.

One possible answer. The number of eggs laid in a week by the 25 hens of a farm in Chitwan:

The cfs are 2, 7, 15, 21, 25; N = 25, $\frac{N + 1}{4} = 6.5$.

• a.Q₁ is the 6.5th item; cf just greater 7: Q₁ = 4 eggs.
• b.Q₃ is the 19.5th item; cf just greater 21: Q₃ = 6 eggs.
• c.Q₂ is the 13th item and the median the $\frac{25 + 1}{2} = 13^{\text{th}}$: cf just greater 15, so $Q_2 = M_d = 5$ eggs.

One possible answer. The minutes 9 friends spent on their phones last evening: 45, 20, 90, 30, 60, 15, 75, 40, 50.
In order: 15, 20, 30, 40, 45, 50, 60, 75, 90; N + 1 = 10.

• a.$P_{20} = \left(\frac{20 \times 10}{100}\right)^{\text{th}} = 2^{\text{nd}}$ item: 20 minutes.
• b.$P_{60} = \left(\frac{60 \times 10}{100}\right)^{\text{th}} = 6^{\text{th}}$ item: 50 minutes.
• c.$P_{50} = 5^{\text{th}}$ item = 45, and $Q_2 = 2\left(\frac{10}{4}\right)^{\text{th}} = 5^{\text{th}}$ item = 45: $P_{50} = Q_2 = 45$ minutes.

Dispersion is the degree to which the values of a set of data are scattered or spread out from their middle (the mean or median). For example, the monthly fine dust in Lalitpur’s air in 2023 went from 13.7 to 80.1 µg/m³, a large dispersion, while the heights of the players in a football team are close together, a small dispersion. It is measured by the range, quartile deviation, mean deviation and standard deviation.

The quartile deviation (semi-interquartile range) is half the difference between the third and first quartiles, $\text{Q.D.} = \frac{Q_3 - Q_1}{2}$. It uses only the first and third quartiles: between them lies the middle half of the data, and the quartile deviation measures the width of that half. The median, Q₂, lies inside it and doesn’t change its width, so it isn’t needed.

The interquartile range is the difference between the third (upper) and first (lower) quartiles, Q₃ − Q₁: the width of the middle 50% of the data. It is twice the quartile deviation.

$\text{Q.D.} = \frac{Q_3 - Q_1}{2} = \frac{38 - 24}{2} = 7$.
$\text{Coefficient} = \frac{38 - 24}{38 + 24} = \frac{14}{62} \approx 0.226$.
Q.D. = 7, coefficient ≈ 0.226. (The shaded pieces each hold a quarter of the data; the middle two together, from 24 to 38, are the middle half.)

• a.$\text{Q.D.} = \frac{40 - 30}{2} = 5$.
• b.$\text{Coefficient} = \frac{40 - 30}{40 + 30} = \frac{10}{70} \approx 0.143$.

• a.$\frac{Q_3 - 35}{2} = 20$, so Q₃ − 35 = 40 and Q₃ = 75.
• b.$\text{Coefficient} = \frac{75 - 35}{75 + 35} = \frac{40}{110} \approx 0.364$.

• a.Q₃ − Q₁ = 2 × 14 = 28. Then $\frac{28}{Q_3 + Q_1} = \frac{7}{20}$, so Q₃ + Q₁ = 80. Subtracting, 2Q₁ = 80 − 28 = 52: Q₁ = 26 (and Q₃ = 54).
• b.$\frac{15 - Q_1}{15 + Q_1} = \frac{1}{4}$, so 60 − 4Q₁ = 15 + Q₁, 5Q₁ = 45: Q₁ = 9, and $\text{Q.D.} = \frac{15 - 9}{2} = 3$.
• c.Q₃ − 12 = 2 × 2 = 4: Q₃ = 16; coefficient $= \frac{4}{28} \approx 0.143$.

In order: 150, 160, 200, 210, 250, 400, 420; N = 7, $\frac{N + 1}{4} = 2$.
$Q_1 = 2^{\text{nd}}$ item = 160; $Q_3 = 6^{\text{th}}$ item = 400.
Q.D. $= \frac{400 - 160}{2} =$ Rs 120; coefficient $= \frac{240}{560} \approx$ 0.429.

In order: 10, 12, 13, 13, 18, 21, 22, 25, 27, 30, 40; N = 11, $\frac{N + 1}{4} = 3$.
$Q_1 = 3^{\text{rd}}$ item = 13; $Q_3 = 9^{\text{th}}$ item = 27.
Q.D. $= \frac{27 - 13}{2} =$ 7 ∘°C; coefficient $= \frac{14}{40} =$ 0.35.

In order: 6, 9, 12, 18, 20, 21, 25, 28, 35, 42; N = 10, $\frac{N + 1}{4} = 2.75$.
Q₁ = 9 + 0.75(12 − 9) = 11.25; Q₃ at the 8.25th place = 28 + 0.25(35 − 28) = 29.75.
Q.D. $= \frac{29.75 - 11.25}{2} =$ 9.25 kg; coefficient $= \frac{18.5}{41} \approx$ 0.451.

In order: 50, 70, 75, 80, 85, 90, 100, 105, 110, 120, 130; N = 11, $\frac{N + 1}{4} = 3$.
$Q_1 = 3^{\text{rd}}$ item = 75; $Q_3 = 9^{\text{th}}$ item = 110.
Q.D. $= \frac{110 - 75}{2} =$ Rs 17.50; coefficient $= \frac{35}{185} \approx$ 0.189.

The cfs are 4, 12, 17, 20, 23, 28; N = 28, $\frac{N + 1}{4} = 7.25$.
Q₁: the 7.25th item; cf just greater 12, so Q₁ = 47 kg.
Q₃: the 21.75th item; cf just greater 23, so Q₃ = 53 kg.
Q.D. $= \frac{53 - 47}{2} =$ 3 kg; coefficient $= \frac{6}{100} =$ 0.06.

The cfs are 3, 10, 20, 28, 30; N = 30, $\frac{N + 1}{4} = 7.75$.
Q₁: the 7.75th item; cf just greater 10: Q₁ = 15 ∘°C.
Q₃: the 23.25th item; cf just greater 28: Q₃ = 25 ∘°C.
Q.D. $= \frac{25 - 15}{2} =$ 5 ∘°C; coefficient $= \frac{10}{40} =$ 0.25.

The cfs are 5, 12, 20, 26, 30, 36, 39; N = 39, $\frac{N + 1}{4} = 10$.
Q₁: the 10th item; cf just greater 12: Q₁ = 45.
Q₃: the 30th item; the cf of 60 is exactly 30: Q₃ = 60.
Q.D. $= \frac{60 - 45}{2} =$ 7.5; coefficient $= \frac{15}{105} \approx$ 0.143.

The cfs are 8, 10, 14, 20, 23, 27, 34, 35, 39; N = 39, $\frac{N + 1}{4} = 10$.
Q₁: the 10th item; the cf of 155 is exactly 10: Q₁ = 155 cm.
Q₃: the 30th item; cf just greater 34: Q₃ = 165 cm.
Q.D. $= \frac{165 - 155}{2} =$ 5 cm; coefficient $= \frac{10}{320} \approx$ 0.031.

The cfs are 3, 64, 196, 349, 489, 540, 543; N = 543, $\frac{N + 1}{4} = 136$.
Q₁: the 136th item; cf just greater 196: Q₁ = 40 years. Q₃: the 408th item; cf just greater 489: Q₃ = 60 years.
Interquartile range = 60 − 40 = 20 years; coefficient $= \frac{60 - 40}{60 + 40} = \frac{20}{100} =$ 0.2. (The quartile deviation is half the interquartile range, 10 years.)
Some answer keys give 10 years here: that is the quartile deviation, not the interquartile range.

One possible answer. The minutes it took 11 students to walk to school: 5, 12, 25, 8, 18, 30, 15, 10, 40, 20, 22.
In order: 5, 8, 10, 12, 15, 18, 20, 22, 25, 30, 40; $\frac{N + 1}{4} = 3$.

• a.$Q_1 = 3^{\text{rd}}$ item = 10 minutes. b. $Q_3 = 9^{\text{th}}$ item = 25 minutes.
• c.Q.D. $= \frac{25 - 10}{2} =$ 7.5 minutes; coefficient $= \frac{15}{35} \approx$ 0.429.

One possible answer. The shoe sizes of 31 students of a class:

The cfs are 3, 11, 21, 28, 31; $\frac{N + 1}{4} = 8$.

• a.The 8th item; cf just greater 11: Q₁ = 5. b. The 24th item; cf just greater 28: Q₃ = 7.
• c.Q.D. $= \frac{7 - 5}{2} =$ 1 size; coefficient $= \frac{2}{12} \approx$ 0.167.
• d.The interquartile range, Q₃ − Q₁, is the width of the middle half of the data, and only its two ends are needed for that; the median lies inside it and doesn’t change its width.

The mean deviation is the average of the absolute deviations of the items from a measure of central tendency (the mean or the median): $\text{M.D.} = \frac{\Sigma |X - \overline{X}|}{N}$ or $\frac{\Sigma |X - M_d|}{N}$. It says how far, on average, the values are from their middle.

Find the mean (or median); find each value’s distance from it, ignoring the sign; add the distances (each times its frequency in a discrete series); divide by the number of items N.

Wherever the average size of a difference matters, for example: how far, on average, a factory’s packets are from the weight printed on them; how far a weather forecast’s temperatures were from the real ones; how much a shop’s daily sales swing around their usual level; how steady a player’s or student’s scores are (Pemba and Tashi).

The quartile deviation uses only two values, Q₁ and Q₃, and ignores the quarter of the data at each end. The mean deviation is worked out from every value’s distance from the middle, so a change in any value shows in it, and it describes the whole set.

The mean deviation is an absolute measure, in the data’s own units and size: 12 words and 9 marks can’t be compared, and a deviation of Rs 50 means a lot for oil but little for a sack of rice. The coefficient, $\frac{\text{M.D.}}{\overline{X}}$ or $\frac{\text{M.D.}}{M_d}$, has no units and compares the spread with the size of the values, so sets in different units or of different sizes can be compared.

Find the mean $\overline{X} = \frac{\Sigma X}{N}$ (or $\frac{\Sigma fX}{N}$); write D = |X − X| for each value; add them ($\Sigma D$, or $\Sigma fD$ with frequencies); divide by N. For 3, 6, 6, 7, 8, 11, 16, 15: X = 9, $\Sigma D = 30$, M.D. = 3.75.

Arrange the data in ascending order and find the median, the $\frac{N + 1}{2}$th item; write $D = |X - M_d|$ for each value; add them ($\Sigma fD$ with frequencies); divide by N. For 3, 6, 6, 7, 8, 11, 15, 16: $M_d = 7.5$, $\Sigma D = 28$, M.D. = 3.5.

$\overline{X} = \frac{1792}{7} = 256$. D: 106, 56, 6, 144, 96, 46, 166; $\Sigma D = 620$.
$\text{M.D.} = \frac{620}{7} \approx$ Rs 88.57; coefficient $= \frac{88.57}{256} \approx$ 0.346.

$\overline{X} = \frac{220}{10} = 22$. D: 2, 4, 3, 10, 13, 16, 0, 23, 13, 6; $\Sigma D = 90$.
$\text{M.D.} = \frac{90}{10} =$ 9 ∘°C; coefficient $= \frac{9}{22} \approx$ 0.409.

$\overline{X} = \frac{231}{11} = 21$. D: 8, 19, 6, 9, 4, 1, 0, 3, 9, 8, 11; $\Sigma D = 78$.
$\text{M.D.} = \frac{78}{11} \approx$ 7.09 kg; coefficient $= \frac{7.09}{21} \approx$ 0.338.

$\overline{X} = \frac{1012}{11} = 92$. D: 42, 12, 7, 17, 22, 2, 8, 13, 28, 18, 35; $\Sigma D = 204$.
$\text{M.D.} = \frac{204}{11} \approx$ Rs 18.55; coefficient $= \frac{18.55}{92} \approx$ 0.202.

$\overline{X} = \frac{160}{9} \approx 17.78$. D: 3.78, 7.78, 9.78, 5.78, 4.22, 10.22, 1.78, 6.22, 8.22; $\Sigma D \approx 57.78$.
$\text{M.D.} = \frac{57.78}{9} \approx$ 6.42 mm; coefficient $= \frac{6.42}{17.78} \approx$ 0.361.

$\overline{X} = \frac{120}{10} = 12$. D: 5, 2, 3, 5, 1, 3, 6, 6, 1, 2; $\Sigma D = 34$.
$\text{M.D.} = \frac{34}{10} =$ 3.4; coefficient $= \frac{3.4}{12} \approx$ 0.283.

In order: 150, 160, 200, 210, 250, 400, 422; $M_d = 4^{\text{th}}$ item = 210.
D: 60, 50, 10, 0, 40, 190, 212; $\Sigma D = 562$. $\text{M.D.} = \frac{562}{7} \approx$ Rs 80.29; coefficient $= \frac{80.29}{210} \approx$ 0.382.

In order: 6, 9, 12, 18, 20, 22, 25, 28, 35, 45; $M_d = 5.5^{\text{th}}$ item $= \frac{20 + 22}{2} = 21$.
D: 15, 12, 9, 3, 1, 1, 4, 7, 14, 24; $\Sigma D = 90$. M.D. = 9 ∘°C; coefficient $= \frac{9}{21} \approx$ 0.429.

In order: 10, 12, 13, 13, 18, 21, 22, 25, 27, 30, 40; $M_d = 6^{\text{th}}$ item = 21, the same as the mean.
So $\Sigma D = 78$, $\text{M.D.} \approx$ 7.09 kg, coefficient $\approx$ 0.338, as in 8c.

In order: 50, 70, 75, 80, 85, 90, 100, 105, 110, 120, 127; $M_d = 6^{\text{th}}$ item = 90.
D: 40, 20, 15, 10, 5, 0, 10, 15, 20, 30, 37; $\Sigma D = 202$. $\text{M.D.} = \frac{202}{11} \approx$ Rs 18.36; coefficient $= \frac{18.36}{90} \approx$ 0.204.

In order: 8, 10, 12, 14, 16, 22, 24, 26, 28; $M_d = 5^{\text{th}}$ item = 16.
D: 8, 6, 4, 2, 0, 6, 8, 10, 12; $\Sigma D = 56$. $\text{M.D.} = \frac{56}{9} \approx$ 6.22 mm; coefficient $= \frac{6.22}{16} \approx$ 0.389.

In order: 6, 7, 9, 10, 11, 13, 14, 15, 17, 18; $M_d = 5.5^{\text{th}}$ item $= \frac{11 + 13}{2} = 12$, the same as the mean.
So $\Sigma D = 34$, M.D. = 3.4, coefficient $\approx$ 0.283, as in 8f.

• a.$\overline{X} = \frac{55}{10} = 5.5$. D: 4.5, 3.5, 2.5, 1.5, 0.5, 0.5, 1.5, 2.5, 3.5, 4.5; $\Sigma D = 25$. M.D. = 2.5; coefficient $= \frac{2.5}{5.5} \approx$ 0.455.
• b.$M_d = 5.5^{\text{th}}$ item $= \frac{5 + 6}{2} = 5.5$. The distances are the same: M.D. = 2.5, coefficient ≈ 0.455.
• c.No difference, because the mean and the median are equal (5.5): the numbers are spread evenly on both sides of the middle, so the distances are measured from the same point.

• a.A discrete series. b. $N = \Sigma f =$ 39 students.
• c.$\Sigma fX = 2073$, $\overline{X} = \frac{2073}{39} \approx 53.15$.

fD: 5 × 13.15, 7 × 8.15, 8 × 3.15, 6 × 1.85, 4 × 6.85, 6 × 10.85, 3 × 14.85 $\approx$ 65.77, 57.08, 25.23, 11.08, 27.38, 65.08, 44.54; $\Sigma fD \approx 296.15$.
$\text{M.D.} = \frac{296.15}{39} \approx$ 7.59; coefficient $= \frac{7.59}{53.15} \approx$ 0.143.
d. The cfs are 5, 12, 20, 26, 30, 36, 39; the median is the 20th item, and the cf of 50 is exactly 20, so $M_d = 50$.
fD: 50, 35, 0, 30, 40, 84, 54; $\Sigma fD = 293$. $\text{M.D.} = \frac{293}{39} \approx$ 7.51; coefficient $= \frac{7.51}{50} \approx$ 0.150.

• a.A discrete series. b. $\Sigma f =$ 100.
• c.$\Sigma fX = 2878$, X = 28.78. fD: 52.68, 75.6, 18.72, 6.16, 63.3, 36.88, 26.44, 14.22; $\Sigma fD = 294$.

$\text{M.D.} = \frac{294}{100} =$ 2.94 ∘°C; coefficient $= \frac{2.94}{28.78} \approx$ 0.102.
d. The cfs are 6, 26, 50, 78, 93, 97, 99, 100; the median is the 50.5th item, cf just greater 78, so $M_d = 29$.
fD: 54, 80, 24, 0, 60, 36, 26, 14; $\Sigma fD = 294$. M.D. = 2.94 ∘°C; coefficient $= \frac{2.94}{29} \approx$ 0.101.

• a.$M_d =$ the $\left(\frac{N + 1}{2}\right)^{\text{th}}$ item, the value whose cf is equal to or just greater than $\frac{N + 1}{2}$, with $N = \Sigma f$.
• b.N = 40, $\Sigma fX = 1210$, $\overline{X} = \frac{1210}{40} =$ 30.25 mm.
• c.The cfs are 5, 13, 25, 35, 40; the 20.5th item has cf just greater 25: $M_d = 30$ mm.
• d.fD: 51.25, 42, 3, 47.5, 48.75; $\Sigma fD = 192.5$. $\text{M.D.} = \frac{192.5}{40} \approx$ 4.81 mm; coefficient $= \frac{4.81}{30.25} \approx$ 0.159.
• e.fD: 50, 40, 0, 50, 50; $\Sigma fD = 190$. $\text{M.D.} = \frac{190}{40} =$ 4.75 mm; coefficient $= \frac{4.75}{30} \approx$ 0.158.

The standard deviation is the square root of the mean of the squares of the deviations of the items from their mean, $\sigma = \sqrt{\frac{\Sigma (X - \overline{X})^2}{N}}$. It measures, in the data’s own units, how spread out the values are about their mean.

The actual mean method, the direct method and the assumed mean method.

Individual data: actual mean $\sigma = \sqrt{\frac{\Sigma d^2}{N}}$, d = X − X; direct $\sigma = \sqrt{\frac{\Sigma X^2}{N} - \left(\frac{\Sigma X}{N}\right)^2}$; assumed mean $\sigma = \sqrt{\frac{\Sigma d^2}{N} - \left(\frac{\Sigma d}{N}\right)^2}$, d = X − A.
Discrete data: actual mean $\sigma = \sqrt{\frac{\Sigma fd^2}{N}}$, d = X − X; direct $\sigma = \sqrt{\frac{\Sigma fX^2}{N} - \left(\frac{\Sigma fX}{N}\right)^2}$; assumed mean $\sigma = \sqrt{\frac{\Sigma fd^2}{N} - \left(\frac{\Sigma fd}{N}\right)^2}$, d = X − A; with $N = \Sigma f$.

By the actual mean method: find the mean; subtract it from each value to get the deviations d; square each deviation (times its frequency, if any); add the squares; divide by N to get the variance; take the square root. For 6, 9, 10, 12, 13: mean 10, $d^2$ = 16, 1, 0, 4, 9, sum 30, variance 6, $\sigma = \sqrt{6} \approx 2.45$.

For example: factories checking that their products (packets, bolts, medicines) are all close to the right size; weather services saying how unusual a hot or wet month is; doctors’ growth charts and the normal ranges of blood tests; surveys and polls, whose “margin of error” is worked out from it; comparing how consistent two players, students or machines are; investors measuring how risky a share price is by how much it swings.

It is the relative measure of dispersion based on the standard deviation, $\frac{\sigma}{\overline{X}}$: the standard deviation as a fraction of the mean. It has no units, so it compares the uniformity of sets in different units or of different sizes; the smaller it is, the more uniform the data.

The coefficient of variation is the coefficient of standard deviation written as a percentage: $\text{C.V.} = \frac{\sigma}{\overline{X}} \times 100\%$. Find the mean and the standard deviation, divide $\sigma$ by the mean and multiply by 100.

A lower C.V. means the values are close to their mean (uniform, consistent, stable); a higher C.V. means they are scattered. So when comparing two sets, such as two students’ marks, two machines’ outputs or two shops’ daily sales, the one with the lower C.V. is the more consistent, even when their means or units differ.

Direct method: N = 7, $\Sigma X = 1790$, $\Sigma X^2 = 531100$, $\overline{X} \approx 255.71$.
$\sigma = \sqrt{\frac{531100}{7} - 255.71^2} = \sqrt{10481.6} \approx$ Rs 102.38; coefficient $\approx$ 0.400; C.V. $\approx$ 40.04%.

N = 10, $\Sigma X = 216$, $\Sigma X^2 = 5824$, X = 21.6.
$\sigma = \sqrt{582.4 - 466.56} = \sqrt{115.84} \approx$ 10.76 ∘°C; coefficient $\approx$ 0.498; C.V. $\approx$ 49.83%.

N = 11, $\overline{X} = \frac{231}{11} = 21$: the actual mean method is easy. d: −8, 19, 6, 9, 4, 1, 0, −3, −9, −8, −11; $\Sigma d^2 = 834$.
$\sigma = \sqrt{\frac{834}{11}} = \sqrt{75.82} \approx$ 8.71 kg; coefficient $\approx$ 0.415; C.V. $\approx$ 41.46%.

N = 11, $\Sigma X = 1015$, $\Sigma X^2 = 99175$, $\overline{X} \approx 92.27$.
$\sigma = \sqrt{\frac{99175}{11} - 92.27^2} = \sqrt{501.65} \approx$ Rs 22.40; coefficient $\approx$ 0.243; C.V. $\approx$ 24.27%.

N = 9, $\Sigma X = 160$, $\Sigma X^2 = 3280$, $\overline{X} \approx 17.78$.
$\sigma = \sqrt{\frac{3280}{9} - 17.78^2} = \sqrt{48.40} \approx$ 6.96 mm; coefficient $\approx$ 0.391; C.V. $\approx$ 39.13%.

N = 10, X = 12. d: 5, −2, 3, −5, 1, −3, −6, 6, −1, 2; $\Sigma d^2 = 150$.
$\sigma = \sqrt{15} \approx$ 3.87; coefficient $\approx$ 0.323; C.V. $\approx$ 32.27%.

Take A = 1400; N = 50. d: −200, −100, 0, 100, 200; fd: −1600, −1200, 0, 900, 1200, so $\Sigma fd = -700$; $fd^2$: 320000, 120000, 0, 90000, 240000, so $\Sigma fd^2 = 770000$.

• a.$\overline{X} = 1400 + \frac{-700}{50} =$ Rs 1,386.
• b.$\sigma = \sqrt{\frac{770000}{50} - (-14)^2} = \sqrt{15400 - 196} = \sqrt{15204} \approx$ Rs 123.30.
• c.C.V. $= \frac{123.30}{1386} \times 100\% \approx$ 8.90%.

a. A = 13, N = 55. fd: −24, −12, 0, 9, 12, 15, $\Sigma fd = 0$ (so the mean is exactly 13); $fd^2$: 72, 12, 0, 9, 24, 45, $\Sigma fd^2 = 162$.
$\sigma = \sqrt{\frac{162}{55}} = \sqrt{2.945} \approx$ 1.72 years; coefficient $= \frac{1.72}{13} \approx$ 0.132; C.V. $\approx$ 13.20%.
b. A = 55, N = 41. fd: −160, −40, −25, 0, 30, 60, 90, 20, $\Sigma fd = -25$; $fd^2$: 3200, 400, 125, 0, 150, 600, 1350, 400, $\Sigma fd^2 = 6225$.
Mean $= 55 - \frac{25}{41} \approx 54.39$; $\sigma = \sqrt{\frac{6225}{41} - \left(\frac{25}{41}\right)^2} = \sqrt{151.83 - 0.37} \approx$ 12.31 marks; coefficient $\approx$ 0.226; C.V. $\approx$ 22.63%.
c. A = 1100, N = 41. fd: −1400, −450, 0, 400, 500, $\Sigma fd = -950$; $fd^2$: 140000, 22500, 0, 20000, 50000, $\Sigma fd^2 = 232500$.
Mean $= 1100 - \frac{950}{41} \approx 1076.83$; $\sigma = \sqrt{\frac{232500}{41} - \left(\frac{950}{41}\right)^2} = \sqrt{5670.73 - 536.88} \approx$ Rs 71.65; coefficient $\approx$ 0.0665; C.V. $\approx$ 6.65%.

The five-number summary is the smallest value, lower quartile, median, upper quartile and largest value. Here: smallest 20, Q₁ = 40, median 50, Q₃ = 70, largest 90.

Smallest 60 in, Q₁ = 65 in, median 70 in, Q₃ = 75 in, largest 80 in. (60 to 80 inches is about 152 to 203 cm.)

102 kg, the end of the right whisker.

38 kg, the end of the left whisker.

88 kg, the line inside the box.

72 kg is Q₁, with about 25% of the data below it, so about 75% weigh more.

From the median to Q₃: about 25%.

Lower. The left whisker is very long (38 to 72) and the left half of the box is longer too (72 to 88, against 88 to 96), so the lower values are spread far out, while the upper ones are bunched. The few very light weights pull the mean down below the median, which doesn’t feel how far they are.

In order: 175, 176, 180, 187, 187, 187, 188, 190, 191, 193, 196, 199, 200, 205; N = 14, $\frac{N + 1}{4} = 3.75$.
Q₁: 3.75th item = 180 + 0.75(187 − 180) = 185.25. Median: 7.5th item $= \frac{188 + 190}{2} = 189$. Q₃: 11.25th item = 196 + 0.25(199 − 196) = 196.75.
Five-number summary: 175, 185.25, 189, 196.75, 205 pounds.

(205 pounds is about 93 kg.)

In order: 54, 55, 56, 58, 60, 61, 64, 65, 66, 69, 70, 71, 73, 73, 74; $\frac{N + 1}{4} = 4$.
$Q_1 = 4^{\text{th}}$ item = 58; median $= 8^{\text{th}}$ item = 65; $Q_3 = 12^{\text{th}}$ item = 71. Five-number summary: 54, 58, 65, 71, 74.

Store 1 in order: 20, 50, 120, 160, 200, 210, 250, 290, 380, 460, 510, 580. $\frac{N + 1}{4} = 3.25$: Q₁ = 120 + 0.25(160 − 120) = 130; median $= \frac{210 + 250}{2} = 230$; Q₃ (9.75th) = 380 + 0.75(460 − 380) = 440.
Store 2 in order: 70, 80, 140, 180, 210, 260, 380, 420, 440, 500, 520, 630. Q₁ = 140 + 0.25(180 − 140) = 150; median $= \frac{260 + 380}{2} = 320$; Q₃ = 440 + 0.75(500 − 440) = 485.

Store 2 sells more: its median month, 320 computers, is 90 above Store 1’s, and each of its five numbers is higher. Both are very uneven: their boxes are about 310 and 335 computers wide. Store 1’s median sits low in its box, so its middle months lean towards the lower sales.

29 g is the lower quartile, so about 75% of the dates weigh more than 29 g. Yes, it can be read from the box plot: everything to the right of the box’s left edge, the box and the right whisker, is three of the four quarters.