Answers and Solutions

1. Typically, the following can be read from a histogram: the data type, the frequency of each interval, the most frequent interval, the approximate minimum and maximum data values, and the least frequent interval. 
   The raw data, the range of the data, and the mean of the data cannot be read from a histogram. 
2. 1. If people only report books that are completely finished, then this would be discrete numerical data that has been collected. It is possible that people reported that they are only part way through a book (for example, 6.5 books in a month) making this continuous data.
   2. The most frequent interval is \(4\) – \(6\) books.
   3. There were \(40\) people were surveyed in total.
   4. \(3\) people read at least 6 but less than 8 books.
   5. \(\dfrac{2}{40}= \dfrac{1}{20} =5\% \)
3. Answers may vary.
   Histograms allow you to see the distribution of the data that might be hard to see if you only had a list of the raw data. This is particularly true of large data sets.
   Histograms do not allow you to see the raw data and so you cannot calculate the range of the data, or the mean of the data, among other things.
4. The histogram in part c) seems to have four bars that are all very close in size, so it looks a lot like a tie between you and your friend. It is also hard to count the exact frequencies in this graph.
   The histograms in parts a) and b) show more clearly that you had more votes than your friend. The histogram in part a) most easily allows you to calculate the totals (over all ages) to be \(62\) (for you) and \(55\) (for your friend). This would make you the clear winner by \(7\) votes.
5. 1. A histogram is likely the best way to show that a large percentage of days are hot enough for an air conditioner. A line graph is more useful to show the trends in temperature over the year. Since there are \(60\) data points, making an accurate line graph might be time consuming. A small portion of the graph is shown.
      
   2. 1. 
      2. 

         Expand/Collapse Alternative Format for Question 5b) Part ii

         Daily High Temperatures

         Temperature (\(^\circ \)C)	Number of Days
         \((-12)\) – \((-7)\)	\(4\)
         \((-7)\) – \((-2)\)	\(5\)
         \((-2)\) – \(3\)	\(8\)
         \(3\) – \(8\)	\(5\)
         \(8\) – \(13\)	\(10\)
         \(13\) – \(18\)	\(2\)
         \(18\) – \(23\)	\(6\)
         \(23\) – \(28\)	\(17\)
         \(28\) – \(33\)	\(3\)

      3. 

         Expand/Collapse Alternative Format for Question 5b) Part iii

         Daily High Temperatures

         Temperature (\(^\circ\)C)	Number of Days
         \((-12)\) – \((-1)\)	\(11\)
         \((-1)\) – \(10\)	\(15\)
         \(10\) – \(21\)	\(11\)
         \(21\) – \(32\)	\(23\)

      Answers will vary. The histogram in part iii) shows the most days in the "hottest" interval, so might be most convincing.
6. Assuming that all of the data was plotted in the histogram, we can be sure that no student was shorter than \(135\) cm and that all students were shorter than \(165\) cm. We also know that one student was shorter than \(140\) cm and one student was \(160\) cm in height or taller.
   1. If the shortest student was just under \(140\) cm tall and the tallest student was \(160\) cm tall, then the range would be just over \(20\) cm. This would be the smallest possible height range.
   2. If the shortest student was \(135\) cm tall and the tallest student was \(165\) cm tall, then the range would be \(30\) cm. According to the histogram, no student reached \(165\) cm, and so the largest possible height range is just under \(30\) cm.
   3. Adding the frequencies in the graph, we see that there are \(80\) students in total. This means that the median height will be the mean of the \(39^{th}\) and \(40^{th}\) heights in the list, when ordered. This height will fall somewhere in the interval \(145\) – \(150\). 
      We cannot determine the median height exactly, but we can be sure that it is larger than \(145\) cm and at most \(150\) cm.
   4. This cannot be determined from the histogram. We can see from the histogram that there are \(26\) students that are at least \(150\) cm tall, which represents \(\frac{26}{80} = 32.5\%\), but we cannot be sure how many of them are more than \(150\) cm tall.
   5. This cannot be determined from the histogram.
7. 1. Answers will vary.
      All measured days in St. John's, NL had wind speeds of at least \(25\) km/h while all measured days in Victoria, BC had wind speeds less than \(55\) km/h. 
      St. John's had the highest measured wind speed. From the histograms, St. John's must have had many measurements at \(55\) km/h or above, but all of Victoria's measurements are smaller than \(55\) km/h.
      Victoria had the lowest measured wind speed.
   2. Answers will vary.
      Since you have the raw data, you can now determine the mean, median, and mode of each data set, along with the minimum and maximum wind speeds and the ranges. 
8. Since we do not have the raw data that makes up the histogram in Question 6, we cannot calculate the mean of the data exactly. However, we can determine a range in which the mean must lie.
   First, let's assume that all of the measured heights are as small as possible. This means that we assume that the \(1\) height in the interval \(135\) – \(140\) is actually \(135\) cm, the \(9\) heights in the interval \(140\) – \(145\) are all \(140\) cm, and so on. If we calculate the mean of all of these \(80\) numbers, then we get\[ \frac{1 \times 135 + 9 \times 140 + 44 \times 145 + 20 \times 150 + 5 \times 155 + 1 \times 160}{80} = \frac{11~710}{80} = 146.375\]From this calculation, we conclude that the mean of the \(80\) actual heights must be \(146.375\) cm or larger.
   
   Second, let's assume that all of the measured heights are as large as possible. (This is a bit trickier as there is actually no "largest possible height" in each interval! What is the largest possible value of the \(1\) height in the interval \(135\) – \(140\)? We actually can't answer this question.) Since we know that the \(1\) height in the interval \(135\) – \(140\) must be smaller than \(140\), let's just take it to be \(140\) cm. Similarly, we take the \(9\) heights in the interval \(140\) – \(145\) to all be \(145\), and so on. If we calculate the mean of these \(80\) numbers, then we get \[ \frac{1 \times 140 + 9 \times 145 + 44 \times 150 + 20 \times 155 + 5 \times 160 + 1 \times 165}{80} = \frac{12~110}{80} = 151.375\]

   From this calculation, we conclude that the mean of the \(80\) actual heights must be smaller than \(151.375\) cm.