Quartiles, box plots and outliers Problem set B One question at a time

Find the mean of 46, 52, 41, 49, 55, 44, 51. Give your answer to 2 decimal places.

• Sum = 338
• Mean = 338 ÷ 7 = 48.29

Find the median of 42, 37, 51, 45, 39, 48, 35, 44, 50.

• In order: 35, 37, 39, 42, 44, 45, 48, 50, 51
• There are 9 values, so the median is the 5th: 44

For the data 34, 41, 29, 52, 38, 45, 31, 47, 36, 50, 43:

1. (a) write down the five-number summary;

   • In order: 29, 31, 34, 36, 38, 41, 43, 45, 47, 50, 52
   • From the calculator: min 29, Q1 34, median 41, Q3 47, max 52
   • n = 11 is odd: the middle value is left out of both halves

2. (b) find the interquartile range.

   • IQR = Q3 − Q1 = 47 − 34 = 13

Two data sets are given. The first is 66, 58, 75, 71, 90, 62, 80, 69, 84, 73. The second has five-number summary 6, 13, 18, 26, 35.

1. (a) Find the five-number summary of the first data set.

   • In order: 58, 62, 66, 69, 71, 73, 75, 80, 84, 90
   • From the calculator: min 58, Q1 66, median 72, Q3 80, max 90
   • n = 10 is even: each half has 5 values

2. (b) State the IQR of the second data set.

   • IQR = Q3 − Q1 = 26 − 13 = 13

Find the population standard deviation of 4, 6, 9, 12, 14, by hand. Give your answer to 3 significant figures.

• Mean = 45 ÷ 5 = 9
• Deviations: −5, −3, 0, 3, 5; squared: 25, 9, 0, 9, 25
• Variance = 68 ÷ 5 = 13.6
• σ = √13.6 = 3.69

Ten days have rainfall in millimetres: 41.3, 38.9, 44.2, 40.6, 43.5, 39.8, 42.7, 45.1, 37.4, 42.1. Use a spreadsheet to find the mean and the population standard deviation, each to 2 decimal places.

• Enter the 10 values in cells A1 to A10
• =AVERAGE(A1:A10) gives 41.56
• =STDEV.P(A1:A10) gives 2.31

A data set has Q1 = 28 and Q3 = 44.

1. (a) Find the lower and upper fences for outliers.

   • IQR = 44 − 28 = 16
   • 1.5 × 16 = 24
   • Lower: 28 − 24 = 4. Upper: 44 + 24 = 68

2. (b) Which of 3, 66 and 70 are outliers?

   • Outliers are below 4 or above 68
   • 3: an outlier
   • 66: not an outlier
   • 70: an outlier

Use the 1.5 × IQR rule to identify any outliers in 57, 49, 18, 56, 54, 95, 53, 47, 52, 50.

• From the calculator: Q1 = 49, Q3 = 56
• IQR = 7; fences 38.5 and 66.5
• Outliers are 18 and 95: each is beyond a fence

The numbers of emails an office receives on seven days are 24, 31, 27, 29, 22, 26, 140.

1. (a) Find the mean, to 2 decimal places, and the median.

   • Mean = 299 ÷ 7 = 42.71
   • In order: 22, 24, 26, 27, 29, 31, 140; median = 27

2. (b) Which is the better measure of the centre? Give a reason.

   • The value 140 is far above the others and pulls the mean above all but one value
   • The median 27 is close to the typical value, so it is the better measure

Box plots show the time, in minutes, that two classes took to finish a puzzle. Class P: 2, 6, 9, 14, 18. Class Q: 4, 7, 9, 11, 15 (five-number summaries).

1. (a) Find the IQR of each data set.

   • Class P: 14 − 6 = 8
   • Class Q: 11 − 7 = 4

2. (b) Compare the two data sets, using the medians and the IQRs.

   • Both medians are 9, so the typical value is the same
   • The IQR of Class P is larger, so its middle half is more spread out
   • Class Q is more consistent

A baker weighs 11 loaves and records their masses in grams: 402, 398, 405, 401, 399, 404, 400, 397, 403, 406, 455.

1. (a) Use the 1.5 × IQR rule to find any outliers.

   • From the calculator: Q1 = 399, Q3 = 405
   • IQR = 6; fences 390 and 414
   • 455 > 414, so it is an outlier

2. (b) Suggest a possible reason for the unusual value.

   • The value 455 g is far from the rest. A possible reason: the loaf may have been made with too much dough, or the mass may have been recorded wrongly

A gym records the number of visitors in each of 12 hours: 64, 112, 58, 70, 66, 60, 68, 55, 52, 63, 57, 61. Show that there is an outlier, then draw a box plot.

• From the calculator: min 52, Q1 57.5, median 62, Q3 67, max 112
• IQR = 9.5; fences 43.25 and 81.25; 112 is above 81.25: an outlier
• Upper whisker ends at 70, the largest value inside the fence
• Draw the box from 57.5 to 67, the median line at 62, whiskers to 52 and 70, a cross at 112

A bakery counts the burnt loaves in each of 32 batches.

1. (a) Find the median.

   • \(n=32\), so the median is between the 16th and 17th values.
   • Running totals: 9, 15, 25, 26, 26, 27, 31, 32
   • Median = 2

2. (b) Find the lower quartile, the upper quartile and the interquartile range.

   • Q1 is between the 8th and 9th values and Q3 between the 24th and 25th values (check with the cumulative frequency).
   • From the calculator: Q1 = 0, Q3 = 2
   • IQR = 2 − 0 = 2

3. (c) Show that a batch with 6 burnt loaves is an outlier.

   • \(Q_3+1.5\times IQR=2+1.5\times 2=5\)
   • \(6>5\), so it is an outlier.

Box plots show the masses, in grams, of two batches of fish. The cross on Batch 1 is at 55.

1. (a) Write down the median and the IQR of Batch 1.

   • Median = 26
   • IQR = 31 − 24 = 7

2. (b) Show that the cross on Batch 1 is an outlier.

   • 1.5 × 7 = 10.5
   • Upper fence = 31 + 10.5 = 41.5
   • 55 > 41.5, so it is an outlier

3. (c) Compare the masses of the two, using the median and the IQR.

   • Medians: Batch 2 28, Batch 1 26
   • IQRs: Batch 2 6, Batch 1 7
   • Batch 2 has the higher typical value and a smaller spread

4. (d) Describe the skew of Batch 1 and say how its mean and median compare.

   • The median is nearer Q1 and the upper whisker and the outlier are far longer than the lower whisker
   • The data are skewed to the right (positive skew)
   • The outlier pulls the mean above the median

The nine values 63, 70, 65, 67, 72, 61, 75, 58, 54 are given. A tenth value x is added, and x is larger than all nine. Find the smallest whole number x for which x is an outlier, by the 1.5 × IQR rule.

• With x added, n = 10 and x is the largest value, so Q1 = 61 and Q3 = 72 do not depend on x
• IQR = 11; upper fence = 72 + 16.5 = 88.5
• x must be greater than 88.5: the smallest whole number is 89