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Measures of Dispersion for Ungrouped Data · Form 4

Measures of Dispersion for Ungrouped Data: Worked Examples (KBAT)

Real situations where the dispersion idea is hidden, choosing the more reliable worker, recovering a missing value, and moderating marks. Helps students decide which measure the question is really asking for.

Written by the spmmath.com.my editorial team.

Worked example 1

A courier company times two riders on the same route (minutes). Ali: 20, 24, 22, 26, 28.

Ben: 18, 30, 24, 20, 28. The manager wants the more reliable rider.

Who should be chosen, and why?

  1. Mean Ali = 120/5 = 24 min; Mean Ben = 120/5 = 24 min, equal, so the mean alone cannot decide; compare the spread.
  2. Ali: Σx² = 400+576+484+676+784 = 2920; variance = 2920/5 − 24² = 584 − 576 = 8; sd = √8 ≈ 2.83 min.
  3. Ben: Σx² = 324+900+576+400+784 = 2984; variance = 2984/5 − 24² = 596.8 − 576 = 20.8; sd = √20.8 ≈ 4.56 min.
  4. Ali's standard deviation is smaller, so his delivery times vary less and are easier to predict.

Worked example 2

Five parcels have a mean mass of 10 kg. Four of them weigh 8, 10, 12 and 16 kg.

Find the mass of the fifth parcel, then the range and the standard deviation of all five masses.

  1. Total mass = mean × N = 10 × 5 = 50 kg.
  2. Fifth parcel = 50 − (8+10+12+16) = 50 − 46 = 4 kg.
  3. Data are 4, 8, 10, 12, 16. Range = 16 − 4 = 12 kg.
  4. Σx² = 16+64+100+144+256 = 580; variance = 580/5 − 10² = 116 − 100 = 16.
  5. Standard deviation = √16 = 4 kg.

Worked example 3

Six students score 8, 10, 11, 13, 16 and 20 marks on a test. During moderation, the teacher adds 4 marks to every student.

Find the new mean and the new standard deviation, and comment on the spread.

  1. Original mean = 78/6 = 13; original Σx² = 64+100+121+169+256+400 = 1110; variance = 1110/6 − 13² = 185 − 169 = 16; sd = √16 = 4.
  2. Adding 4 to every mark raises each value equally, so the new mean = 13 + 4 = 17.
  3. Because every mark shifts by the same 4, the gaps between marks are unchanged, so the variance and standard deviation stay the same.
  4. Check: new data 12, 14, 15, 17, 20, 24; Σx² = 1830; variance = 1830/6 − 17² = 305 − 289 = 16; sd = √16 = 4.

Worked example 4

The test marks of five pupils are 5, 8, 9, 11 and 12. The teacher decides to double every mark as a scaling.

Find (a) the standard deviation of the original marks and (b) the mean and standard deviation of the doubled marks, and state how doubling affects the standard deviation.

  1. Original mean = (5 + 8 + 9 + 11 + 12) ÷ 5 = 45 ÷ 5 = 9.
  2. Squared deviations: 16, 1, 0, 4, 9; sum = 30; variance = 6.
  3. (a) Standard deviation = √6 ≈ 2.45.
  4. (b) Doubling each mark: new mean = 2 × 9 = 18; new SD = 2 × √6 = √24 ≈ 4.90.
  5. Multiplying every value by 2 multiplies both the mean and the standard deviation by 2.

Worked example 5

The masses, in kg, of four kittens are 4, 6, 8 and one unknown mass. The mean mass of the four kittens is 7 kg.

Find (a) the unknown mass and (b) the standard deviation of the four masses, correct to two decimal places.

  1. (a) Sum of all four masses = mean × n = 7 × 4 = 28.
  2. Unknown mass = 28 − (4 + 6 + 8) = 28 − 18 = 10 kg.
  3. (b) Squared deviations from 7: (4−7)² = 9, (6−7)² = 1, (8−7)² = 1, (10−7)² = 9; sum = 20.
  4. Variance = 20 ÷ 4 = 5; standard deviation = √5 ≈ 2.24 kg.

Worked example 6

A factory checks two machines that fill 500 g packets. Five samples (in g) from Machine A are 498, 500, 502, 500, 500, and five from Machine B are 495, 505, 500, 490, 510.

Determine which machine is more reliable by comparing the mean and standard deviation of each.

  1. Machine A mean = 2500 ÷ 5 = 500 g; Machine B mean = 2500 ÷ 5 = 500 g (equal means).
  2. Machine A squared deviations: 4, 0, 4, 0, 0; sum = 8; variance = 1.6; SD = √1.6 ≈ 1.26 g.
  3. Machine B squared deviations: 25, 25, 0, 100, 100; sum = 250; variance = 50; SD = √50 ≈ 7.07 g.
  4. Both hit the target mean of 500 g, but Machine A has the much smaller standard deviation, so Machine A is more reliable.

Worked example 7

The daily number of customers at a cafe over 5 days was recorded with a mean of 20 and a standard deviation of 6. It was later found that one day's count was wrongly recorded as 30 when it should have been 20.

Find the correct mean and the correct standard deviation of the number of customers.

  1. From the wrong data: Σx = mean × n = 20 × 5 = 100
  2. From the wrong data: variance = 6² = 36, so Σx² = (variance + mean²) × n = (36+400) × 5 = 2180
  3. Correct Σx = 100 − 30 + 20 = 90; correct mean = 90/5 = 18
  4. Correct Σx² = 2180 − 30² + 20² = 2180 − 900 + 400 = 1680
  5. Correct variance = 1680/5 − 18² = 336 − 324 = 12; correct standard deviation = √12 = 3.46 (2 d.p.)

Worked example 8

A student's scores (out of 50) in 5 quizzes were 10, 12, 14, 16, 18. In the sixth quiz, the student scored an unusually high 38.

Find the standard deviation (a) for the first 5 quizzes only, and (b) for all 6 quizzes, and comment on the effect of the sixth score on the spread of the data.

  1. (a) Mean of first 5 = (10+12+14+16+18)/5 = 70/5 = 14
  2. Σx² (first 5) = 100+144+196+256+324 = 1020; variance = 1020/5 − 14² = 204 − 196 = 8; standard deviation = √8 = 2.83 (2 d.p.)
  3. (b) Mean of all 6 = (70+38)/6 = 108/6 = 18
  4. Σx² (all 6) = 1020 + 38² = 1020 + 1444 = 2464; variance = 2464/6 − 18² = 410.67 − 324 = 86.67; standard deviation = √86.67 = 9.31 (2 d.p.)
  5. The standard deviation rises sharply from 2.83 to 9.31, showing the unusually high sixth score greatly increases the spread of the data.

Source:DSKP KSSM Mathematics Form 4 and 5 (Versi English)

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Frequently asked questions

What kind of KBAT questions come up for measures of dispersion?

KBAT questions usually give two real data sets, like the scores of two students or the output of two machines, and ask you to decide which is more consistent or better overall. You must calculate the mean and standard deviation for both, then use them together to justify a decision.

What does the examiner actually reward in a dispersion KBAT answer?

Marks go to a written conclusion that links your calculated values to the question, for example, stating which data set has a higher mean and which has a smaller standard deviation, then explaining what that means in context. Correct numbers alone, without an interpretive sentence, will not earn full marks.

What's the most common pitfall in KBAT dispersion questions?

Many students calculate the mean and standard deviation correctly but stop there, without writing a conclusion that answers the actual question asked. Others compare only the mean or only the standard deviation instead of both.

Always finish with a full sentence stating which set is more consistent and why.

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