TopicName Test
WorkSHEET 10.2 Summary statistics Name: ___________________________
| |In order to compare two textbooks, a teacher recommends one book to |Text A [pic]= 58.6 ( = 25.1 |6 |
| |one class and another book to another class. At the end of the year |Text B [pic]= 62.55, ( = 11.8 | |
| |the classes are each tested; the results are detailed below. | | |
| | |The class that used Text B, because of the higher mean. | |
| |Text A | | |
| |44 52 95 76 13 94 83 72 55 81 22 25 64 |The class that used Text B, because of the lower standard deviation. | |
| |72 35 48 56 59 84 98 84 21 35 69 28 | | |
| | | | |
| |Text B | | |
| |65 72 48 63 68 59 68 62 75 79 81 72 64 53 58 59 64 66| | |
| |68 42 37 39 55 58 52 82 79 55 | | |
| | | | |
| |Calculate the mean and standard deviation for each group. | | |
| |Which class performed better? | | |
| |Which class was more consistent? | | |
| | | | |
| |The box-and-whisker plot drawn below shows the marks achieved by | |4 |
| |students in a class, on their end of year exam. | | |
| | |Median = 72 | |
| |[pic] |Lower quartile = 63 | |
| | |Upper quartile = 77 | |
| |State the median. |Interquartile range = 77 – 63 | |
| |Find the interquartile range. |= 14 | |
| |What was the highest mark in the class? |Top mark = 92 | |
| | | | |
| |The figures below show the number of vehicles that pass through a |Lower limit = 82 |4 |
| |particular intersection between 4:00 pm and 5:00 pm over a two-week |Lower quartile = 84 | |
| |period |Median = 89 | |
| | |Upper quartile = 95 | |
| |88 92 114 82 94 83 84 |Upper limit = 114 | |
| |85 85 90 95 82 95 103 | | |
| | |[pic] | |
| | | | |
| |The data below show monthly rainfall in millimetres. | |5 |
| | | | |
| |Jan. 10 | | |
| |Feb. 12 |[pic] | |
| |Mar. 21 | | |
| |Apr. 23 | | |
| |May 39 | | |
| |June 22 | | |
| |July 15 | | |
| |Aug. 11 | | |
| |Sept. 22 | | |
| |Oct. 37 | | |
| |Nov. 45 | | |
| |Dec. 30 | | |
| | | | |
| |Draw a box-and-whisker plot of the data. | | |
| | | | |
| |The number of points scored in each match by two Rugby Union teams are|Key 3 | 4 = 34 |3 |
| |shown below. | | |
| | |Team 1 Team 2 | |
| |Team 1 : 34 32 24 25 8 18 17 23 29 40 19 42 |8 | 0 | 7 9 | |
| | |9 8 7 | 1 | | |
| |Team 2 : 23 20 35 21 46 7 9 24 27 38 41 30 |9 5 4 3 | 2 | 0 1 3 4 7 | |
| | |4 2 | 3 | 0 5 8 | |
| |Display these sets of data in a back-to-back stem-and-leaf plot. |2 0 | 4 | 1 6 | |
| | | | |
| |In a class of 30 students there are 15 boys and 15 girls. Their |Key 1.6 | 5 = 1.65 |3 |
| |heights are measured and are listed below. | | |
| | |Boys Girls | |
| |Boys: 1.65, 1.71, 1.59, 1.74, 1.66, 1.69, 1.72 1.66, 1.65, 1.64, 1.68,|9 9 7 | 1.5 | 1 2 5 6 7 8 8 | |
| |1.74, 1.57, 1.59 1.60 |9 8 6 6 5 5 4 0 | 1.6 | 4 4 6 7 8 9 9 | |
| | |4 4 2 1 | 1.7 | 0 | |
| |Girls: 1.66, 1.69, 1.58, 1.55, 1.51, 1.56, 1.64, 1.69, 1.70, 1.57, | | |
| |1.52, 1.58, 1.64, 1.68 1.67 | | |
| | | | |
| |Display this information in a back-to-back stem-and-leaf plot. | | |
| | | | |
| |The boxplot below shows Emma’s performance in her physics and |Physics [pic]= 71 |8 |
| |chemistry exams. |Chemistry [pic] = 72 | |
| | |Physics range = 90 – 50 | |
| |[pic] |= 40 | |
| | |Chemistry range = 83 – 30 | |
| |State the median mark for each subject. |= 53 | |
| |Find the range of marks in each subject. |Physics interquartile range = 76 – 66 | |
| |Find the interquartile range for each subject. |= 10 | |
| |In which subject did Emma perform better? Explain your answer. |Chemistry interquartile range = 74 – 71 | |
| | |= 3 | |
| | |Emma performed slightly better in Chemistry, as indicated by the | |
| | |slightly higher median and greater consistency as shown by the low | |
| | |interquartile range | |
| | | | |
| |The stem-and-leaf plot below is used to display the number of vehicles|Ford median = 15 |6 |
| |sold by the Ford and Holden dealerships in a Sydney suburb each week |Holden median = 16 | |
| |for a three month period. | | |
| | |Ford range = 30 – 4 | |
| |Key: 1 | 5 = 15 |= 26 | |
| |Ford Holden |Holden range = 35 – 3 | |
| |7 4 | 0 | 3 9 |= 33 | |
| |9 5 2 2 1 0 | 1 | 1 1 1 6 6 8 | | |
| |8 5 4 4 | 2 | 2 2 7 9 |Ford : Lower quartile = 1.05 | |
| |0 | 3 | 5 |Upper quartile = 2.45 | |
| | |Interquartile range = 2.45 – 1.05 | |
| |State the median of both distributions. |= 1.4 | |
| |Calculate the range of both distributions. | | |
| |Calculate the interquartile range of both distributions. |Holden : Lower quartile = 1.1 | |
| |Show both distributions on a box-and-whisker plot. |Upper quartile = 2.45 | |
| | |Interquartile range = 2.45 – 1.1 | |
| | |= 1.35 | |
| | | | |
| | | | |
| | |[pic] | |
| | | | |
| |The data below show the weekly incomes for a sample of Year 11 boys | |8 |
| |and girls. |Boys [pic]= $106.36 | |
| | |Girls [pic] = $97.78 | |
| |Boys : $80 $110 $75 $130 $90 $125 | | |
| |$100 $100 $95 $115 $150 |Boys s = 22.48 | |
| | |Girls s = 59.27 | |
| |Girls : $50 $80 $75 $90 $60 $250 | | |
| |$80 $100 $95 |Boys have a higher average weekly income as indicated by the higher | |
| | |mean. | |
| |Find the mean of each data set. | | |
| |Find the standard deviation of each data set. |Boys have a more consistent weekly income as indicated by the lower | |
| |Discuss whether boys or girls have a higher average weekly income. |standard deviation. | |
| |Discuss whether boys or girls have a more consistent weekly income. | | |
| | | | |
| |Use the data in question 9 to draw a parallel boxplot of the data | |4 |
| |clearly showing any outliers. |[pic] | |
| | | | |
| | |Boys: Lowest score = 75 | |
| | |Lower quartile = 90 | |
| | |Median = 100 | |
| | |Upper quartile = 115 | |
| | |Highest score = 150 | |
| | | | |
| | |Girls: Lowest score = 50 | |
| | |Lower quartile = 67.5 | |
| | |Median = 80 | |
| | |Upper quartile = 97.5 | |
| | |Highest value = 100 (ignoring outlier) | |
| | |Outlier = 250 | |
| | | | |
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