--- title: "Five number summary and box plot of guests" subject: "Probability and Statistics" chapter: 1 key: "semester_5/probability_and_statistics:five-number-summary-guests" source: https://www.bctengineer.com/q/semester_5/probability_and_statistics/five-number-summary-guests site: BCT Engineering --- # Five number summary and box plot of guests Probability and Statistics · Chapter 1: Descriptive Statistics and Basic Probability - Asked in: 2076 Asoj, 6 marks - Video: [The Five Number Summary, Boxplots, and Outliers (1.6)](https://www.youtube.com/watch?v=tpToLyZibKM) — Simple Learning Pro - Answer checked by Claude Sonnet 5.5 on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_5/probability_and_statistics/five-number-summary-guests - Whole chapter: https://www.bctengineer.com/md/semester_5/probability_and_statistics/chapter-1 ## Question List Five Number summary and prepare the box plot for numbers of guest registered each of 60 randomly selected days. 108 94 188 116 165 181 106 133 176 110 169 134 129 109 85 124 119 165 153 135 105 180 105 91 117 148 83 96 101 123 128 143 136 99 169 133 89 90 174 144 151 168 103 116 106 107 179 113 172 120 179 183 99 94 87 120 154 159 103 139 ## Short answer Five-number summary: Min = 83, Q₁ = 105, Median = 123.5, Q₃ = 156.5, Max = 188. IQR = 51.5; no outliers. Boxplot: box from Q₁ to Q₃ with median line inside, whiskers to min and max. Slight right skew. ## Answer #### Five number summary The five numbers are the minimum, first quartile Q₁, median, third quartile Q₃ and maximum. Arrange the 60 values in order: ```text 83 85 87 89 90 91 94 94 96 99 99 101 103 103 105 105 106 106 107 108 109 110 113 116 116 117 119 120 120 123 124 128 129 133 133 134 135 136 139 143 144 148 151 153 154 159 165 165 168 169 169 172 174 176 179 179 180 181 183 188 ``` - Minimum = 83, maximum = 188. - Median: average of the 30th and 31st values = (123 + 124)/2 = **123.5**. - Q₁ is the median of the lower 30 values: average of the 15th and 16th = (105 + 105)/2 = **105**. - Q₃ is the median of the upper 30 values: average of the 45th and 46th = (154 + 159)/2 = **156.5**. | Minimum | Q₁ | Median | Q₃ | Maximum | |---|---|---|---|---| | 83 | 105 | 123.5 | 156.5 | 188 | #### Outlier check IQR = 156.5 − 105 = 51.5, and 1.5 × IQR = 77.25. - Lower fence = 105 − 77.25 = 27.75 (no value below it) - Upper fence = 156.5 + 77.25 = 233.75 (no value above it) So there are no outliers, and the whiskers go to 83 and 188. #### Box plot ```text 80 100 120 140 160 180 200 | | | | | | | 83-----[105====123.5====156.5]-----188 min Q1 median Q3 max ``` The box runs from Q₁ to Q₃ with a line at the median. Whiskers extend to the minimum and maximum. The median is closer to Q₁ than Q₃ and the right whisker is longer, so the distribution of daily guests is slightly right-skewed. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. Which are the five numbers? A. Minimum, Q1, median, Q3, maximum B. Mean, mode, median, range, SD C. Min, mean, median, mode, max D. Q1, Q2, Q3, IQR, range Answer: A. They summarise the ordered data and give the box plot. (Key point: Five numbers) 2. How is the median found? A. The 30th value only B. Average of 30th and 31st values = 123.5 C. The 60/2 + 1 value D. The mean of all Answer: B. With even n, average the two middle values. (Key point: Median of 60) 3. How are outliers checked? A. Beyond min and max B. Beyond mean +/- SD C. Beyond Q1 - 1.5 IQR or Q3 + 1.5 IQR (27.75 to 233.75): none D. Beyond Q1 and Q3 Answer: C. No value lies outside the fences, so whiskers reach 83 and 188. (Key point: Outlier fences) 4. Why is the distribution slightly right-skewed? A. Median is nearer Q3 B. Both whiskers are equal C. There are outliers D. Median is nearer Q1 and the right whisker is longer Answer: D. Longer right side shows right skew. 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