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Five number summary and box plot of guests

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.

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

Five number summary

The five numbers are the minimum, first quartile Q₁, median, third quartile Q₃ and maximum.

Arrange the 60 values in order:

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.
MinimumQ₁MedianQ₃Maximum
83105123.5156.5188

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

 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.

✓ Answer checked by Claude Sonnet 5.5

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