Presentation of Data

Solution

(a) Frequency table

Count each observation once. Include zero as a possible observed value.

Check: \(1+4+6+5+3+1=20\), so all observations are accounted for.

(b) Mode

The mode is the value with the largest frequency. The largest frequency is 6, so the mode is 2 books, not 6 books.

(c) Percentage

“At least three” includes 3, 4 and 5. Their total frequency is \(5+3+1=9\). \[\frac{9}{20}\times100\%=45\%.\]

Solution

(a) Locate the middle, then find the median and quartiles

There are \(n=8\) observations, so \[\text{middle position}=\frac{n+1}{2}=\frac{8+1}{2}=4.5.\] Position \(4.5\) lies between the 4th and 5th observations, whose values are 7 and 8. Average these values, not their position numbers.

Each half has 4 values. Its middle position is \((4+1)/2=2.5\), so average the second and third values of that half to find its median. \[\begin{aligned} \text{Median}&=\frac{7+8}{2}=7.5,\\ Q_1&=\frac{4+5}{2}=4.5,\\ Q_3&=\frac{10+12}{2}=11. \end{aligned}\]

(b) Range and interquartile range

Range \(=16-2=\boxed{14}\); IQR \(=11-4.5=\boxed{6.5}\).

Solution

(a) Locate the middle, then find the median and quartiles

There are \(n=9\) observations, so \[\text{middle position}=\frac{n+1}{2}=\frac{9+1}{2}=5.\] The 5th observation is 8, so the median is 8. The number 5 tells us where to look; it is not the median score.

Using our median-of-halves convention, leave out the overall median, 8, when forming the two halves: \[\text{lower half: }2,4,5,7,\qquad \text{upper half: }10,12,14,16.\] Each half has 4 values. Its middle position is \((4+1)/2=2.5\), so average the second and third values within that half. \[\begin{aligned} \text{Median}&=8,\\ Q_1&=\frac{4+5}{2}=4.5,\\ Q_3&=\frac{12+14}{2}=13. \end{aligned}\]

(b) Range and interquartile range

Range \(=16-2=\boxed{14}\); IQR \(=13-4.5=\boxed{8.5}\).

Solution

(a) Histogram

Each interval has width 10 minutes. Mark \(0,10,\ldots,70\) at the edges of the bars; use the frequencies as heights.

(b) Modal class

The greatest frequency is 16, so the modal class is \(\boxed{30\le t<40}\) minutes.

(c) Number of students

The interval \(20\le t<50\) includes three complete classes. Add their frequencies: \[13+16+13=\boxed{42\text{ students}}.\]

(d) Shape and limitations

The histogram is symmetric about 35 minutes and has one central peak. The grouped table does not give the exact median: we do not know the positions of the observations inside their intervals. A cumulative-frequency graph will let us estimate it.

Solution

(a) Calculate each running total

Work from left to right across the frequency row. Start with zero, then add the frequency for each class to the previous running total: \[\text{Cumulative frequency}=\text{previous cumulative frequency}+\text{frequency}.\]

For example, the total below 40 minutes includes the first four classes: \(2+7+13+16=38\). Adding the previous total, 22, to the next frequency, 16, is a shorter way of doing the same calculation.

Check: the last cumulative frequency is 60, matching the total number of students. The completed row and graph follow on the next steps below.

The completed table for the 60 travel times is repeated here. Each cumulative frequency counts students below that class’s upper boundary.

(b) Turn each table column into a coordinate pair

The upper boundary is the right-hand endpoint of a class interval. For example, \(30\le t<40\) has upper boundary 40. Read down that column: 16 students are in this class, but 38 students are below 40 minutes, counting this class and all earlier classes.

Use (class upper boundary, cumulative frequency): the first column gives \((10,2)\) and the fourth gives \((40,38)\).

Also include \((0,0)\): 0 is the lower boundary of the first class, not an upper boundary. It gives the starting point because none of these travel times is below 0. All the coordinates are \[\begin{aligned} &(0,0),\quad(10,2),\quad(20,9),\quad(30,22),\\ &(40,38),\quad(50,51),\quad(60,58),\quad(70,60). \end{aligned}\] The first point is the extra starting point; the other seven points come from the seven classes.

(c) Draw and interpret the graph

Label the horizontal axis travel time (minutes) and the vertical axis cumulative frequency. Choose scales covering 0–70 minutes and 0–60 students. Plot the points accurately and draw a smooth, non-decreasing curve through them.

The point \((40,38)\) means 38 students travel for less than 40 minutes. Do not confuse this running total with 16, which counts only the students in \(30\le t<40\).

Final checks: the graph starts at \((0,0)\), ends at \((70,60)\) and never decreases. Its final height is \(n=60\).

Solution

Read the grid: each small horizontal division is 1 minute; each small vertical division is 1 student.

(a) Lower quartile: at \(60/4=15\), read \(\boxed{Q_1\approx25\text{ min}}\).

(b) Median: at \(60/2=30\), read \(\boxed{\text{median}\approx35\text{ min}}\).

(c) Upper quartile: at \(3(60)/4=45\), read \(\boxed{Q_3\approx45\text{ min}}\).

(d) IQR: using these readings, \(45-25=\boxed{20\text{ min}}\) approximately.

(e) 80th percentile: at \(0.80(60)=48\), read \(\boxed{P_{80}\approx47\text{ min}}\).

Solution

(a) Find the level on the vertical axis

The lower-quartile level is \(60/4=15\) students. This is a count, not the time we are trying to find.

(b) Read across, then down

Start at 15 on the vertical axis. Draw a horizontal guide to the curve, then a vertical guide down to the time axis.

The vertical guide reaches the axis very close to 25 minutes. To the requested nearest minute, \[\boxed{Q_1\approx25\text{ minutes}}.\] Approximately one quarter of the students have travel times at or below 25 minutes. The curve does not recover the unknown individual times inside each class.

Solution

(a) Read the number of students

Locate 42, two small divisions to the right of 40. Follow the vertical guide up to the curve. Now follow the horizontal guide left: the reading is just above 40, giving \(\boxed{41\text{ students approximately}}\).

(b) Convert the count to a percentage

About 41 out of 60 students meet the condition. Divide by the total, then multiply by 100: \[\text{percentage}\approx\frac{41}{60}\times100\%\approx\boxed{68\%}.\]

(c) Convert the same count to a probability

Each of the 60 students is equally likely to be chosen, so use the same fraction, without multiplying by 100: \[\text{probability}\approx\frac{41}{60}\approx\boxed{0.68}.\] The count is a graph estimate, so the percentage and probability are estimates too. 41 is a count; 68% is a percentage; 0.68 is a probability.

(d) At least 50 minutes: remove those below 50

From 50 on the horizontal axis, go up to the curve and across: 51 students are below 50 minutes. Everyone else travels for at least 50 minutes, including exactly 50: \[\text{number at least 50}=60-51=\boxed{9\text{ students}}.\]

(e) Between 20 and 50 minutes: subtract the earlier running total

Read 51 below 50 minutes and 9 below 20 minutes. The first count includes the second. Remove those below 20, leaving exactly those with \(20\le t<50\): \[\text{number with }20\le t<50=51-9=\boxed{42\text{ students}}.\]

(f) One class: find the increase in cumulative frequency

There are 38 students below 40 minutes and 22 below 30 minutes. The increase is the number in the class \(30\le t<40\): \[\text{class frequency}=38-22=\boxed{16\text{ students}}.\] These boundary counts are given in the table, so (d)–(f) are exact for this dataset. The reading at 42 minutes was between boundaries, so (a)–(c) are estimates.

The curve is steeper where more observations occur in an equal-width interval, and flatter where fewer occur. It need not be a straight line or have one fixed shape.

Solution

(a) Diagram

Choose a uniform scale. Draw the box from 18 to 31, median line at 24, and whiskers ending at 12 and 45.

(b) Range

\(45-12=\boxed{33}\).

(c) Interquartile range

\(31-18=\boxed{13}\).

Solution

(a) Use the range to find the minimum

The maximum is 58 and the full spread is 48, so \[48=58-a\quad\Rightarrow\quad a=58-48=\boxed{10}.\]

(b) Use the IQR to find the upper quartile

The box starts at \(Q_1=20\) and has width 24, so \[24=c-20\quad\Rightarrow\quad c=20+24=\boxed{44}.\]

(c) Use the stated position of the median

We are told that the median is halfway between the quartiles. Therefore \[b=\frac{20+44}{2}=\boxed{32}.\]

(d) Construct the completed plot

The five-number summary is \(10,20,32,44,58\). Draw whiskers ending at 10 and 58, the box from 20 to 44, and the median line at 32.

Check: \(58-10=48\) and \(44-20=24\), as required. The values are in increasing order.

Solution

(a) Read the three quartile levels

The levels are \(15,30,45\). Read \(Q_1\approx25\), median \(\approx35\) and \(Q_3\approx45\) minutes from the curve.

(b) Separate exact from estimated spread

Range \(=67-3=\boxed{64\text{ minutes}}\) exactly; IQR \(\approx45-25=\boxed{20\text{ minutes}}\).

(c) Use one common scale

Solution

(a) IQR and fences

\[\begin{aligned} \operatorname{IQR}&=30.5-16=14.5,\\ \text{Lower fence}&=16-1.5(14.5)=-5.75,\\ \text{Upper fence}&=30.5+1.5(14.5)=52.25. \end{aligned}\]

(b) Outlier

\(55>52.25\), so 55 is the only outlier. No observation is below \(-5.75\).

(c) Modified box plot

The lower whisker ends at 10; the upper whisker ends at 36, the largest non-outlier.

(d) Full range

\(55-10=\boxed{45}\). The outlier remains part of the dataset; the whisker-to-whisker distance is not the full range.

Solution

(a) Typical performance

Class B’s median is higher: \(68>65\). A typical score is slightly higher in Class B.

(b) Middle spread and full spread

\[\begin{aligned} \operatorname{IQR}_A&=74-56=18,&\operatorname{range}_A&=88-42=46,\\ \operatorname{IQR}_B&=72-60=12,&\operatorname{range}_B&=82-18=64. \end{aligned}\] Class B’s middle 50% is less spread out, but its full range is larger because of the low outlier.

(c) Shape, unusual values and conclusion

Class A is approximately symmetric: both halves of its box have width 9 and both whiskers have length 14. Class B is not symmetric and has an unusually low score of 18.

Conclusion: Class B typically performed slightly better and had a more consistent middle half, but its low outlier made its overall spread greater.

Solution

(a) Spread

Group A: IQR \(=66-44=22\); range \(=78-32=46\).

Group B: IQR \(=60-40=20\); range \(=90-20=70\).

(b) Compare in context

The medians are similar, with B slightly higher (\(56>55\)). B’s middle half is a little less spread out, but its overall range is much larger. Do not say B is less variable in every sense.

(c) Symmetry is evidence, not proof

Group A has equal box halves (\(55-44=66-55=11\)) and equal whiskers (\(44-32=78-66=12\)). Its box plot therefore gives stronger evidence of symmetry. The data may be approximately normal, but the box plot alone cannot establish a bell-shaped distribution.