2. Data Collection and Graphs (D)
  3. Lesson 7: Organizing Numerical Data Into Intervals

Answers and Solutions


    1. Stem Leaf
      \(0\) \(~5\)
      \(1\) \(~2\) \(~6\) \(~8\)
      \(2\) \(~1\) \(~2\) \(~2\) \(~5\) \(~6\) \(~7\)
      \(3\) \(~0\) \(~0\) \(~6\) \(~8\) \(~9\)
      Key: \(1 | 2 = 12\)
    2. Range Frequency
      Less than \(20\) \(4\)
      \(20\) or greater \(11\)
    3. Range Frequency
      \(0\) – \(8\) \(1\)
      \(8\) – \(16\) \(1\)
      \(16\) – \(24\) \(5\)
      \(24\) – \(32\) \(5\)
      \(32\) – \(40\) \(3\)
    1. Range Frequency
      \(0\) – \(7\) \(3\)
      \(7\) – \(14\) \(1\)
      \(14\) – \(21\) \(0\)
      \(21\) – \(28\) \(5\)
      Answers will vary. One possible data set is \(1,~2,~3,~8,~22,~23,~24,~25,~26\).
    2. Range Frequency
      \(6\) – \(9\) \(0\)
      \(9\) – \(12\) \(1\)
      \(12\) – \(15\) \(2\)
      \(15\) – \(18\) \(3\)
      \(18\) – \(21\) \(4\)
      Answers will vary. One possible data set is \(\dfrac{21}{2},~13,~ 14.5, ~16,~16,~16,~18.9, ~19, ~20.4,~20.8\).
    3. The size of the given interval is \(215 - 213.5 = 1.5\) and so we have the following table:
      Range Frequency
      \(210.5\) – \(212\) \(2\)
      \(212\) – \(213.5\) \(2\)
      \(213.5\) – \(215\) \(2\)
      Answers will vary. One possible data set is \(210.6,~211.6,~212.2,~213.2,~213.9,~214.9\).
  3. Answers will vary. One of the possible answers is given for each part.
    1. The data can be organized into the following ranges: \(0\) – \(5\), \(5\) – \(10\), \(10\) – \(15\), \(15\) – \(20\), \(20\) – \(25\), and \(25\) – \(30\).
    2. The data can be organized into the following ranges: \(10\) – \(30\), \(30\) – \(50\), \(50\) – \(70\), and \(70\) – \(90\).
    3. The data can be organized into the following ranges: \((-10)\) – \(0\), \(0\) – \(10\), \(10\) – \(20\), \(20\) – \(30\), and \(30\) – \(40\).
    1. Answers will vary.
      There are many ways to reclassify the eggs. It likely makes the most sense to name the new classifications "Small," "Medium," and "Large". One possible reclassification is given below:
      Classification Minimum Weight (g)
      Small \(504\) (\(42\) per egg)
      Medium \(672\) (\(56\) per egg)
      Large  \(840\) (\(70\) per egg)
    2. From the original table, there was a \(7\) g difference (per egg) between adjacent classes (for example, medium eggs were \(49\) g each and small eggs were \(42\) g each, \(49-42= 7\)). 
      The reclassification above changes the difference (per egg) between adjacent classes to \(14\). If the price of cartons labelled "Small," "Medium," and "Large" stays the same after the reclassification, then customers will often be getting a larger carton for the same price. However, this larger range of weights could allow the store to maximize profits by making sure that they only stock eggs that are in the lower end of the weight intervals.
    1. Stem Leaf
      \(9\) \(~1.2\) \(~1.9\) \(~2.9\) \(~3.2\) \(~5.0\)
      \(10\) \(~3.5\) \(~3.5\) \(~4.0\) \(~4.1\) \(~5.1\) \(~5.3\) \(~5.4\) \(~6.3\) \(~6.8\) \(~9.0\) \(~9.0\) \(~9.2\)
      \(11\) \(~2.5\) \(~2.7\) \(~3.5\) \(~4.5\) \(~4.6\) \(~6.1\) \(~6.6\)
      \(12\) \(~0.0\)
      Key: \( 9 | 1.2 = 91.2\)
    2. \(120.0-91.2=28.8\)
    3. Answers may vary. One option is given below.
      Range Frequency
      \(91\) – \(97\) \(5\)
      \(97\) – \(103\) \(0\)
      \(103\) – \(109\) \(9\)
      \(109\) – \(115\) \(8\)
      \(115\) – \(121\) \(3\)
  6. Senior Care recently completed a survey on the residents of a residence and recorded the residents' ages in a tablular way.
    \(97\) \(92\) \(78\) \(80\) \(76\) \(94\) \(78\) \(94\) \(80\) \(69\)
    \(91\) \(99\) \(79\) \(66\) \(70\) \(102\) \(96\) \(99\) \(70\) \(67\)
    \(100\) \(96\) \(90\) \(92\) \(100\) \(102\) \(91\) \(66\) \(85\) \(82\)
    \(80\) \(82\) \(93\) \(69\) \(96\) \(102\) \(95\) \(72\) \(83\) \(89\)
    1. Generally people report their ages as whole numbers which would make it discrete data. However, someone's age could be considered continuous data as you could measure age much more accurately in months, or days, or seconds, etc. 
    2. Answers will vary. One option is given below.
      Age Range Frequency
      \(65\) – \(85\) \(18\)
      \(85\) – \(105\) \(22\)
    3. Answers will vary. One option is given below.
      Age Range Frequency
      \(65\) – \(73\) \(8\)
      \(73\) – \(81\) \(7\)
      \(81\) – \(89\) \(4\)
      \(89\) – \(97\) \(13\)
      \(97\) – \(105\) \(8\)
    4. The table from part b) suggests that there is nearly an even split between younger seniors (\(65\) – \(85\)) and older seniors (\(85\) and above) but a few more older seniors. The table from part c) shows that residents with ages near the early-to-mid nineties are the most common, while residents with ages near the early-to-mid eighties are the least common.
    1. Answers will vary. You are collecting continuous data as heights can be any numerical value within a range. If you look up the tallest person and shortest person in history, you can be quite sure your data will fall into this range.
    2. Answers will vary. Since measurements of length can, in theory, be very close to \(0\), it might make sense to start your first interval at \(0\). However, it might make more sense to start your interval closer to the smallest collected piece of data. Your last interval needs to end after your largest collected piece of data. You can choose a convenient place based on the length of interval that you choose to use.
    3. Answers will vary. If you start your first interval at \(0\), instead of closer to the smallest collected piece of data, then likely more than one frequency in your table will be equal to zero. A similar idea applies to the placement of the last interval.
  8. \(51.0\) \(29.0\) \(26.5\) \(37.0\) \(40.0\) \(31.5\) \(38.0\)
    \(22.0\) \(38.5\) \(38.0\) \(18.0\) \(11.0\) \(35.0\) \(33.5\)
    \(45.0\) \(30.5\) \(24.0\) \(26.0\) \(20.0\) \(35.0\) \(20.0\)
    \(5.0\) \(9.0\) \(15.5\) \(23.0\) \(21.0\) \(22.0\) \(30.0\)
    \(35.0\) \(25.0\) \(24.0\) \(30.0\) \(19.5\) \(32.0\) \(14.5\)
    1. Here \(N = 35\). Since \(5^{2} = 25\) and \(6^{2}=36\), the square root of \(N\) is between \(5\) and \(6\) and much closer to \(6\). This means we will be using \(6\) intervals.
    2. The largest value is \(51.0\) and the smallest value is \(5.0\) which makes the range \(51.0 - 5.0 = 46.0\). Since \(46 \div 6 \approx 7.7\), we will round up and use intervals of length \(8\).
    3. Answers may vary. We need to start our first interval somewhere between \(0\) and \(5\). Starting at \(5\) and using \(6\) intervals of length \(8\) will bring us to an upper limit of \(5 + 6\times 8 = 53\) which will cover all of the data.
    4. Answers will vary.
      Range Frequency
      \(5-13\) \(3\)
      \(13-21\) \(6\)
      \(21-29\) \(9\)
      \(29-37\) \(10\)
      \(37-45\) \(5\)
      \(45-53\) \(2\)
      This frequency table shows there is a gradual increase of data pieces as you near the centre of the data. Most of the data falls into the two middle intervals, \(21-29\) and \(29-37\). This may have been missed if fewer or more intervals were used. Experiment with less or more intervals to see if this is the case.
    5. Answers will vary.