Questions on measures of location and spread

Question 1

For the data 4, 7, 7, 9, 13, write down the mean, median and mode.

Accepted Answer

Mean 40/5 = 8. Median 7, the middle of the ordered list. Mode 7, the most frequent value. B1 B1 B1 for the three measures. Two different answers from three measures on one small set, because the three measures genuinely measure different things. The 13 drags the mean above the median.

Question 2

For 25 values, Σx = 450 and Σx² = 8500. Find the mean and the standard deviation.

Accepted Answer

The mean is 450/25 = 18. The variance is 8500/25 − 18² = 340 − 324 = 16, so σ = √16 = 4. B1 for the mean, M1 for the variance formula, A1 for σ = 4. Take the mean of the squares and subtract the square of the mean, in that order. Reversing the two gives a negative variance, which is the instant sign that the formula has been written the wrong way round.

Question 3

The eight values 5, 8, 8, 9, 11, 12, 12, 15 have Σx = 80. Write down the mean and any modes.

Accepted Answer

The mean is 80/8 = 10. B1. Two values occur twice and none occurs more often than that, so the set has two modes, 8 and 12, and both must be given. B1 for the pair. A set may have two modes, or none at all when every value appears once, which is why the mode is the weakest of the three averages for measured data. Neither mode here is the mean.

Question 4

For 12 values, Σx = 96 and Σx² = 848. Find S_xx and the standard deviation.

Accepted Answer

S_xx = Σx² − (Σx)²/n = 848 − 96²/12 = 848 − 768 = 80. M1 for the formula with both sums placed correctly, A1 for 80. The standard deviation is √(S_xx/n) = √(80/12) = √6.667 = 2.58 to 3 significant figures. A1 for 2.58. S_xx is the total squared deviation from the mean, so it can never come out negative; a negative value means (Σx)² and Σx² have changed places in the formula.

Question 5

Forty journey times are grouped: 8 in 0 ≤ t < 10, 14 in 10 ≤ t < 20, 12 in 20 ≤ t < 30, 6 in 30 ≤ t < 40. Use interpolation to estimate the median.

Accepted Answer

The median position is n/2 = 20. The first class holds 8, so the 20th value sits 12 into the second class, which holds 14 spread across a width of 10. Interpolating, the median is 10 + (12/14) × 10 = 18.6 minutes. M1 for the position n/2 = 20, M1 for a correct interpolation fraction, A1 for 18.6. The answer has to land inside 10 to 20, and if it does not, the running total has gone astray. For grouped continuous data Edexcel uses n/2 rather than (n + 1)/2, so use 20 here, not 20.5.

Question 6

Data are coded by y = (x − 100)/5, and the coded values have mean 4 and standard deviation 2. Find the mean and standard deviation of the original data.

Accepted Answer

Uncode in reverse, so x = 100 + 5y. The mean becomes 100 + 5 × 4 = 120 and the standard deviation becomes 5 × 2 = 10. B1 for the mean 120, M1 for scaling the standard deviation, A1 for 10. Subtracting 100 slid the whole data set along without stretching it, so only the division by 5 touched the spread.

Question 7

Two machines fill bags with mean mass 500 g each; machine A's masses have σ = 2 g and machine B's σ = 9 g. Compare the machines in context.

Accepted Answer

The centres agree, since both average 500 g. The spreads do not. Machine A's bags stay within a few grams of the target while machine B's stray four to five times as far, so machine A is much the more consistent filler. B1 for comparing the means, B1 for comparing the spreads, B1 for the context and units.

Question 8

The ordered data set 3, 5, 6, 8, 9, 11, 12, 14, 15, 18, 21, 24 holds twelve values. Find the median and the interquartile range, using the position rules for listed data.

Accepted Answer

Positions first, values second. With n = 12, the median position n/2 = 6 is a whole number, so the median is the midpoint of the 6th and 7th values: (11 + 12)/2 = 11.5. Q₁ sits at position n/4 = 3, also whole, so Q₁ = (6 + 8)/2 = 7; Q₃ sits at 3n/4 = 9, so Q₃ = (15 + 18)/2 = 16.5. The interquartile range is 16.5 − 7 = 9.5. M1 for the position rules applied with whole positions taken as midpoints, A1 for the median, A1 for both quartiles, A1 for 9.5. These are Edexcel's rules for listed data: a position that is not whole rounds up instead, so state which convention is in use before reading the value off.

Question 9

The seven values 4, 6, 6, 9, 10, 12, 15 have Σx = 62 and Σx² = 638. Find the mean and the standard deviation, each to 3 significant figures.

Accepted Answer

The mean is 62/7 = 8.86 to 3 significant figures. M1 for the mean. The variance is Σx²/n − x̄² = 638/7 − (62/7)² = 91.14 − 78.45 = 12.69, so the standard deviation is √12.69 = 3.56. M1 for the variance formula with both terms, A1 for 3.56. Keep the mean unrounded inside the variance. Using 8.86 in place of 62/7 gives a variance of 12.64 rather than 12.69; here both still round to 3.56, so nothing is visible, and on data with a longer mean the third figure moves.

Question 10

Sixty masses, in grams, are grouped: 7 in 0 ≤ m < 20, 18 in 20 ≤ m < 40, 21 in 40 ≤ m < 60 and 14 in 60 ≤ m < 80. Use linear interpolation to estimate the 90th percentile and the interquartile range.

Accepted Answer

Write the running totals down first: 7, 25, 46, 60. The 90th percentile sits at position 0.9 × 60 = 54, which falls in the fourth class, 54 − 46 = 8 values into a class of 14 spread over a width of 20, so it is 60 + (8/14) × 20 = 71.4 g. M1 for a position with its interpolation, A1 for 71.4. For the quartiles, Q₁ is at position 15, which is 8 into the second class of 18, giving 20 + (8/18) × 20 = 28.89; Q₃ is at position 45, which is 20 into the third class of 21, giving 40 + (20/21) × 20 = 59.05. M1 for both quartiles. The interquartile range is 59.05 − 28.89 = 30.2 g. A1 for 30.2.

Question 11

Data x are coded by y = (x − 250)/8. The original data have mean 306 and standard deviation 24. Find the mean and the standard deviation of the coded data.

Accepted Answer

Code in the order the formula gives. A mean codes exactly as a value does: ȳ = (306 − 250)/8 = 56/8 = 7. B1 for 7. The standard deviation ignores the subtraction, because sliding every value along by 250 moves them together and alters no distance between them, so only the division by 8 touches the spread: 24/8 = 3. M1 for dividing the standard deviation by 8 alone, A1 for 3. Subtracting 250 from the standard deviation as well would give −226, and a standard deviation cannot be negative.

Question 12

The cumulative frequency diagram shows the times that 160 customers waited to be served at a post office. Use the diagram to estimate the median waiting time, the interquartile range of the waiting times, and the 90th percentile.

Accepted Answer

A cumulative frequency curve is entered on the vertical axis. The median is read at n/2, the lower quartile at n/4 and the upper quartile at 3n/4, so with n = 160 the three heights are 80, 40 and 120. M1 for all three positions. Going across from 80 to the curve and down to the time axis gives a median of 15 minutes. A1 for 15. The same reading taken at 40 gives Q₁ = 10 minutes and at 120 gives Q₃ = 20 minutes. M1 for both quartiles read from the curve. The interquartile range is 20 − 10 = 10 minutes. A1 for 10. The 90th percentile is entered at 0.90 × 160 = 144, and the curve stands at 144 above t = 25, so the 90th percentile is 25 minutes. B1 for 25 minutes. The curve is read at n/2 rather than at (n + 1)/2 because it plots a running total against the upper end of a class and not a list of ranked values, so there is no ranked position to correct for.

Question 13

A set of 20 values has mean 15 and standard deviation 3. A further value of 36 is added to the set. Find the new mean and the new standard deviation.

Accepted Answer

Work back to the sums, because those are what combine. Σx = 20 × 15 = 300. For the squares, σ² = Σx²/n − x̄² rearranges to Σx² = n(σ² + x̄²) = 20(9 + 225) = 4680. Adding the new value: Σx becomes 300 + 36 = 336 and Σx² becomes 4680 + 1296 = 5976, with n now 21. The new mean is 336/21 = 16. The new variance is 5976/21 − 16² = 284.57 − 256 = 28.57, so the new standard deviation is 5.35. M1 for recovering Σx, M1 for recovering Σx², M1 for updating both sums, A1 for the new mean 16, M1 for the new variance, A1 for 5.35. A single value 7 standard deviations out has lifted the mean by 1 and nearly doubled the spread.

Question 14

Explain which of the mean and the median better represents a typical member of the data set 2, 3, 3, 4, 5, 5, 6, 7, 41, supporting your answer by finding both.

Accepted Answer

The median: the middle of the nine ordered values is the 5th, which is 5. B1 for the median. The mean is 76/9 = 8.44 to three significant figures. B1 for the mean. The single wild reading of 41 drags the mean above every other value in the set, so the mean describes nobody; the median never looks at the outlier's size and stays among the values where the data actually sit, so it is the better representative here. B1 for the comparison made through the outlier.

Question 15

A group of 30 men has mean height 178 cm and standard deviation 7 cm. A group of 20 women has mean height 165 cm and standard deviation 6 cm. Find the mean and the standard deviation of the heights of all 50 people.

Accepted Answer

Work back to the sums, because sums are what combine. For the men Σx = 30 × 178 = 5340 and for the women Σx = 20 × 165 = 3300, so the combined Σx = 8640 and the mean is 8640/50 = 172.8 cm. M1 for a group total, A1 for 172.8. For the squares, σ² = Σx²/n − x̄² rearranges to Σx² = n(σ² + x̄²), giving 30(49 + 31684) = 951990 and 20(36 + 27225) = 545220, so the combined Σx² = 1497210. M1 for the rearrangement, A1 for 1497210. The combined variance is 1497210/50 − 172.8² = 29944.2 − 29859.84 = 84.36, so the standard deviation is √84.36 = 9.18 cm. M1 for the variance, A1 for 9.18. The combined spread of 9.18 cm is larger than either group's, and it has to be: the 13 cm gap between the two group means is itself a source of variation that neither group sees inside itself.

Question 16

A set of 10 values has mean 14 and standard deviation 5. One of the values, 26, was recorded in error and is removed from the set. Find the mean and the standard deviation of the remaining 9 values.

Accepted Answer

Recover both sums before anything is removed. Σx = 10 × 14 = 140, and Σx² = n(σ² + x̄²) = 10(25 + 196) = 2210. M1 for Σx, M1 for Σx². Removing the value takes 26 off the first and 26² = 676 off the second, so Σx becomes 114 and Σx² becomes 1534, with n now 9. M1 for updating both sums. The new mean is 114/9 = 12.7 to 3 significant figures. A1. The new variance is 1534/9 − (114/9)² = 170.44 − 160.44 = 10, so the new standard deviation is √10 = 3.16. A1 for 3.16. The value removed lay (26 − 14)/5 = 2.4 standard deviations above the old mean, and taking it out has cut the standard deviation from 5 to 3.16 while shifting the mean by only 1.3.