Free math worksheets

- The left edge of the box marks \(Q_1\). - The line drawn inside the box marks the median.

1. The left edge of the box is \(Q_1\), which is at \(7\) minutes. 2. The line inside the box is the median, which is at \(9\) minutes.

\(Q_1=7\,\text{minutes}\) Median \(=9\,\text{minutes}\)

- Identify the five values represented by the whiskers, box edges, and median line. - The third quartile is the value at or below which at least \(75\%\) of the data lie. - The IQR is the width of the box.

1. The left whisker endpoint shows a minimum of \(15\) minutes. 2. The line inside the box shows a median of \(30\) minutes. 3. The right edge of the box is \(Q_3 = 40\) minutes, so at least \(75\%\) of the delivery times are no more than \(40\) minutes. 4. The IQR is \(Q_3 - Q_1 = 40 - 20 = 20\) minutes.

a) \(15\) minutes b) \(30\) minutes c) \(40\) minutes d) \(20\) minutes

- The two edges of the box are the first and third quartiles. - Subtract \(Q_1\) from \(Q_3\) to find the IQR. - Consider what percent of the data lies at or above the first quartile.

1. The box extends from \(Q_1 = 6\) hours to \(Q_3 = 11\) hours. Therefore, the middle \(50\%\) lasted from \(6\) to \(11\) hours. 2. The IQR is \(11 - 6 = 5\) hours. 3. Since \(6\) hours is the first quartile, at least \(75\%\) of the batteries lasted \(6\) hours or more.

a) From \(6\) hours to \(11\) hours b) \(5\) hours c) At least \(75\%\)

- Read the left and right edges of each box as \(Q_1\) and \(Q_3\). - Find each IQR using \(Q_3-Q_1\). - Compare the numerical IQRs with the visible box lengths.

1. Group 1 has \(Q_1=60\) minutes and \(Q_3=180\) minutes, so its IQR is \(180-60=120\) minutes. 2. Group 2 has \(Q_1=100\) minutes and \(Q_3=160\) minutes, so its IQR is \(160-100=60\) minutes. 3. Group 2 has less variability in the middle \(50\%\) because its box is shorter and its IQR is smaller.

a) Group 1: \(120\) minutes; Group 2: \(60\) minutes. b) Group 2 has less variability in the middle \(50\%\) because its box is shorter and its IQR is smaller.

- Which line inside each box represents the median? - Read both values carefully from the scale.

1. Read the median from the line inside each box: City X has a median of \(45\,\text{mm}\), and City Y has a median of \(55\,\text{mm}\). 2. City Y has the higher median. 3. The difference is \(55 - 45 = 10\,\text{mm}\).

City Y has the higher median daily rainfall. The medians differ by \(10\,\text{mm}\).

- The range is the maximum minus the minimum. - Read the minimum and maximum from the ends of each pair of whiskers.

1. Restaurant A has a minimum of \(15\) minutes and a maximum of \(60\) minutes, so its range is \(60 - 15 = 45\) minutes. 2. Restaurant B has a minimum of \(20\) minutes and a maximum of \(45\) minutes, so its range is \(45 - 20 = 25\) minutes. 3. Since \(25 < 45\), Restaurant B delivers within the smaller overall time interval.

Restaurant A has a range of \(45\) minutes. Restaurant B has a range of \(25\) minutes. Restaurant B delivers within the smaller overall time interval.

- Distinguish the line inside the box from the \(\times\). - Use the labeled scale to read both centers. - Compare the two values.

1. The median line is at \(30\) points. 2. The \(\times\) marking the mean is at \(25\) points. 3. Since \(25<30\), the mean is below the median.

a) \(30\) points b) The mean is below the median.

- Match each symbol in the box plot with its meaning. - Read values only from labeled tick marks. - Distinguish the whisker endpoints from the open circles.

1. The whisker endpoints are \(15\) seconds and \(55\) seconds. 2. The quartiles and median are \(Q_1 = 25\), median \(35\), and \(Q_3 = 45\) seconds. 3. The outliers are \(5\) seconds and \(100\) seconds. 4. The mean is \(40\) seconds.

a) \(15, 25, 35, 45, 55\) seconds b) Outliers: \(5\) seconds and \(100\) seconds; mean: \(40\) seconds

- Read the values at the two edges of the box. - The whiskers are not part of the box length. - Recall what percent of the data lies between the quartiles.

1. The left edge of the box is \(Q_1 = 15\) points. 2. The right edge of the box is \(Q_3 = 40\) points. 3. The length of the box is \(40 - 15 = 25\) points. 4. The interval from \(Q_1\) to \(Q_3\) contains the middle \(50\%\) of the results.

The box is \(25\) points long and contains the middle \(50\%\) of the results.

- A box plot divides ordered data into four approximately equal groups. - Find one-fourth of the total number of measurements. - The condition about repeated values avoids ambiguity at the boundaries.

1. The quartiles divide the ordered data into four groups, each containing about one-fourth of the measurements. 2. One-fourth of \(40\) is \(40 \div 4 = 10\). 3. Therefore, each of the four described regions contains about \(10\) measurements.

a) About \(10\) b) About \(10\) c) About \(10\) d) About \(10\)

- Open circles and whisker endpoints use different symbols. - Read each position from the scale. - The \(\times\) is not one of the requested values.

1. The open circles outside the whiskers are at \(10\,\text{dB}\) and \(55\,\text{dB}\). 2. The left whisker endpoint is \(20\,\text{dB}\). 3. The right whisker endpoint is \(50\,\text{dB}\).

The values outside the whiskers are \(10\,\text{dB}\) and \(55\,\text{dB}\). The whisker endpoints are \(20\,\text{dB}\) and \(50\,\text{dB}\).

- Read the box plot from bottom to top. - The edges of the box are the quartiles. - Use the whisker endpoints for the range.

1. The five-number summary is minimum \(4\), \(Q_1 = 10\), median \(14\), \(Q_3 = 20\), and maximum \(26\) centimeters. 2. The range is \(26 - 4 = 22\) centimeters.

Minimum \(4\), \(Q_1 = 10\), median \(14\), \(Q_3 = 20\), maximum \(26\) centimeters; range \(22\) centimeters.

- Read the median, mean, and outliers separately. - Compare the measures with the interval containing the box and whiskers. - Focus on what “typical” means in this context.

1. The median is \(8\) minutes, and the mean is \(10\) minutes. 2. Most displayed values lie between \(4\) and \(12\) minutes. 3. The high outliers at \(30\) and \(36\) minutes pull the mean upward. 4. Therefore, the median better describes a typical group.

The median of \(8\) minutes is more representative because the high outliers pull the mean up to \(10\) minutes.

- Compare the left and right sides of the display around the median. - Locate the mean marker relative to the median line. - Consider whether the display shows a feature that would strongly pull the mean to one side.

1. The box and whiskers are symmetric around \(30\) points. 2. The mean marker and median line are both at \(30\) points. 3. No separated outliers are displayed. 4. Therefore, the mean is an appropriate summary of the center.

Yes. The distribution is symmetric around \(30\) points, the mean and median agree, and no separated outliers are shown.

- Recall the required left-to-right order of the five marked statistics. - Check the ordering before calculating any spread. - Use the box endpoints for one spread and the whisker endpoints for the other.

1. A five-number summary must be ordered as minimum, \(Q_1\), median, \(Q_3\), maximum. 2. Summary A is invalid because \(Q_1=9\) is greater than the median \(8\). 3. Summary B is in nondecreasing order, so it can produce a valid box plot. 4. For Summary B, the interquartile range is \(12-8=4\), and the range is \(20-4=16\).

a) Summary B b) Interquartile range \(4\); range \(16\)

- Identify which two quartiles determine the length of the box. - Work backward from the stated box length. - Use the two outer endpoints for the full spread.

1. The interquartile range satisfies \(Q_3-Q_1=9\). 2. Therefore, \(17-Q_1=9\), so \(Q_1=8\). 3. The complete five-number summary is \(3, 8, 11, 17, 25\). 4. The range is \(25-3=22\).

\(Q_1=8\) Five-number summary: \(3, 8, 11, 17, 25\) Range: \(22\)

- Convert one quarter of the sample size into a count. - Separate horizontal interval length from the number of ordered positions. - Recall what each section between consecutive quartile landmarks represents.

1. Each quarter represents about \(25\%\) of the ordered data. 2. For \(36\) observations, \(25\%\times36=9\) observations correspond to each quarter. 3. Interval length shows how spread out that quarter’s values are, not how many observations it contains. 4. The upper quarter can contain about the same number of observations over a wider numerical interval.

About \(9\) observations belong to each quarter. The longer whisker shows greater numerical spread, not greater frequency.

- Compare the middle landmark first. - Use the same outer endpoints for the full spread. - Use the box endpoints to compare the middle-half spread.

1. Both medians are \(5\). 2. Both ranges are \(10-0=10\). 3. Plot A has IQR \(8-2=6\). 4. Plot B has IQR \(6-4=2\). 5. Plot B has the more concentrated middle half because its IQR is smaller.

Both medians are \(5\), and both ranges are \(10\). Plot A: IQR \(6\). Plot B: IQR \(2\). Plot B has the more concentrated middle half.

- With eight values, split the ordered list into two halves of four values each. - Find the median of the whole list and the medians of the two halves. - Match minimum, \(Q_1\), median, \(Q_3\), and maximum to the five box-plot landmarks.

1. The minimum is \(3\) and the maximum is \(12\). 2. The median is \((7+8)\div2=7.5\). 3. The lower half is \(3,5,6,7\), so \(Q_1=(5+6)\div2=5.5\). 4. The upper half is \(8,9,11,12\), so \(Q_3=(9+11)\div2=10\). 5. Therefore the five-number summary, and the box-plot landmark positions, are \(3,5.5,7.5,10,12\).

Five-number summary: \(3,5.5,7.5,10,12\) Left whisker: \(3\); left box edge: \(5.5\); median line: \(7.5\); right box edge: \(10\); right whisker: \(12\).

- With an odd number of values, use the stated convention and leave the overall median out of both halves. - Find the median of the four lower values and the four upper values. - Match the five-number summary in order to the five ordinary box-plot landmarks.

1. In order, the data are \(4,6,6,8,10,13,13,15,18\). The minimum is \(4\), the median is \(10\), and the maximum is \(18\). 2. The lower half is \(4,6,6,8\), so \(Q_1=(6+6)\div2=6\). 3. The upper half is \(13,13,15,18\), so \(Q_3=(13+15)\div2=14\). 4. The five-number summary is \(4,6,10,14,18\). 5. Those values are the positions of the left whisker, left box edge, median line, right box edge, and right whisker, respectively.

Five-number summary: \(4,6,10,14,18\) Left whisker: \(4\); left box edge: \(6\); median line: \(10\); right box edge: \(14\); right whisker: \(18\).

- Several choices can give the same median, so check more than the center line. - After inserting a candidate, split the eight ordered values into two halves. - Compare both quartiles and the maximum with the box plot.

1. The box plot shows minimum \(1\), \(Q_1=2.5\), median \(4.5\), \(Q_3=6\), and maximum \(7\). 2. All four candidates are inserted after \(5\) and do not change the minimum or maximum, so the median and endpoints do not determine the answer. 3. Adding \(6\) gives \(1,2,3,4,5,6,6,7\). 4. The lower half \(1,2,3,4\) has \(Q_1=(2+3)\div2=2.5\). 5. The upper half \(5,6,6,7\) has \(Q_3=(6+6)\div2=6\). 6. The other candidates give different third quartiles, so \(6\) is the unique choice matching the full box plot.

b) \(6\)

- Read each median from the line inside the box. - Find each range using the two whisker endpoints. - Compare the minimum tested life with the required \(30\) hours.

1. Brand A has a median of \(40\) hours and a range of \(60 - 20 = 40\) hours. 2. Brand B has a median of \(39\) hours and a range of \(45 - 32 = 13\) hours. 3. Brand A has a tested minimum of \(20\) hours, so some tested batteries lasted less than \(30\) hours. 4. Brand B has a tested minimum of \(32\) hours, so every tested battery lasted at least \(30\) hours. Based on this sample, Brand B better meets the requirement.

a) Brand A: median \(40\) hours, range \(40\) hours Brand B: median \(39\) hours, range \(13\) hours b) Brand B, because every tested Brand B battery lasted at least \(32\) hours.

- Read \(Q_3\) from the right edge of each box. - Read the maximum from the right whisker endpoint. - Interpret the median as the middle recorded delay. - Compare the largest observed delays for the connection decision.

1. Route 101 has \(Q_3 = 7\) minutes and a maximum recorded delay of \(15\) minutes. Route 102 has \(Q_3 = 6\) minutes and a maximum recorded delay of \(8\) minutes. 2. Route 101 has the smaller median: \(4\) minutes compared with \(5\) minutes for Route 102. At least half of the recorded Route 101 trips were delayed by no more than \(4\) minutes. 3. Route 102 had a much smaller maximum recorded delay, \(8\) minutes instead of \(15\). For a time-sensitive connection, that smaller observed worst delay may make Route 102 preferable, although the sample does not guarantee future delays.

a) Route 101: \(Q_3 = 7\) minutes, maximum \(15\) minutes Route 102: \(Q_3 = 6\) minutes, maximum \(8\) minutes b) Route 101 has the smaller median, \(4\) minutes. c) Route 102 had the smaller maximum recorded delay, which may reduce the risk of missing a connection.

- Identify the five values represented by the endpoints, box edges, and median line. - The range uses the maximum and minimum; the IQR uses the two quartiles. - Think about what fraction of the data lies at or above \(Q_1\). - Convert that percent to a number of students out of \(24\).

1. The box plot shows a minimum of \(\$15\), \(Q_1 = \$25\), a median of \(\$35\), \(Q_3 = \$45\), and a maximum of \(\$65\). 2. The range is \(\$65 - \$15 = \$50\). 3. The IQR is \(\$45 - \$25 = \$20\). 4. Since \(Q_1 = \$25\), at least \(75\%\) of the students receive \(\$25\) or more. The minimum number is \(0.75 \times 24 = 18\) students.

a) Minimum: \(\$15\); first quartile: \(\$25\); median: \(\$35\); third quartile: \(\$45\); maximum: \(\$65\) b) Range: \(\$50\); IQR: \(\$20\) c) At least \(75\%\), or at least \(18\) students

- Identify the five values shown by each box plot. - The IQR is the length of the box. - For race times, a smaller median is faster. - Recall what percent of the data lies at or below \(Q_3\).

1. Class A has range \(21 - 13 = 8\) seconds. Class B has range \(20 - 12 = 8\) seconds. 2. Class A has IQR \(18 - 15 = 3\) seconds. Class B has IQR \(19 - 14 = 5\) seconds, so Class B has the requested IQR. 3. The median times are \(16\) seconds for Class A and \(17\) seconds for Class B. A lower sprint time is faster, so Class A has the faster median. 4. Class A has \(Q_3 = 18\) seconds. At least \(75\%\) of its recorded times are at or below \(18\) seconds, so the statement is true.

a) Class A: \(8\) seconds; Class B: \(8\) seconds b) Class B c) Class A, with a median of \(16\) seconds d) True, because \(Q_3 = 18\) seconds for Class A.

- Read the values at the left and right edges of Type B's box. - Which quartile separates the lowest \(25\%\) from the highest \(75\%\)? - The maximum is at the far end of the right whisker.

1. For Type B, the left edge of the box is \(Q_1 = 8500\,\text{h}\), and the right edge is \(Q_3 = 10{,}000\,\text{h}\). 2. For Type B, \(Q_1 = 8500\,\text{h}\). Therefore, at least \(75\%\) of its bulbs lasted \(8500\,\text{h}\) or more, so Type B clearly meets the \(8000\,\text{h}\) criterion. 3. For Type A, \(Q_1 = 7500\,\text{h}\) and the median is \(8500\,\text{h}\). The box plot does not show whether at least \(75\%\) of the bulbs lasted at least \(8000\,\text{h}\), so Type A is not guaranteed to meet the criterion. 4. The maximum for Type A is \(11{,}500\,\text{h}\), compared with \(10{,}500\,\text{h}\) for Type B. Type A produced the longest-lasting individual bulb.

a) \(Q_1 = 8500\,\text{h}\); \(Q_3 = 10{,}000\,\text{h}\) b) Type B clearly meets the criterion. The box plot does not guarantee that Type A does. c) Type A, with a maximum of \(11{,}500\,\text{h}\), compared with \(10{,}500\,\text{h}\) for Type B.

- The median is the line inside each box. - The range is the maximum minus the minimum. - The box represents the middle \(50\%\) of the data. - The IQR is \(Q_3 - Q_1\).

1. The median for Class A is \(24\) points, and the median for Class B is \(26\) points. Class B has the higher median. 2. Class A's range is \(40 - 12 = 28\) points. Class B's range is \(36 - 14 = 22\) points. Class A has the greater range. 3. Class A's IQR is \(30 - 18 = 12\) points. Class B's IQR is \(28 - 22 = 6\) points. Because Class B has the smaller IQR, its middle \(50\%\) of scores are closer together.

a) Class B, with a median of \(26\) points b) Class A, with a range of \(28\) points c) Class B. Its IQR is \(6\) points, compared with \(12\) points for Class A.

- The two plots share some landmarks, so compute Felix's quartiles rather than matching only the minimum, maximum, or median. - For each half of five values, use its middle value as the quartile. - Compare box lengths using \(Q_3-Q_1\).

1. Felix's median is \((8750+9250)\div2=9000\) steps. 2. The lower five values have median \(8000\), so \(Q_1=8000\). The upper five have median \(11{,}000\), so \(Q_3=11{,}000\). 3. Plot A has \(Q_1=8000\), median \(9000\), and \(Q_3=11{,}000\), so Plot A is Felix's. 4. Felix's IQR is \(11{,}000-8000=3000\) steps. 5. Plot B has \(Q_1=8500\) and \(Q_3=10{,}000\), so Lara's IQR is \(1500\) steps. 6. Lara's IQR is \(3000-1500=1500\) steps smaller.

a) Felix: \(Q_1=8000\), median \(9000\), \(Q_3=11{,}000\); Plot A. b) Lara has the smaller IQR by \(1500\) steps.

- The median is the line inside each box. - Shorter boxes and whiskers indicate less variability. - Recall what percentage of the data lies at or above \(Q_3\). - Find the range by subtracting the minimum from the maximum.

1. Both Model X and Model Y have a median of \(8\,\text{h}\), so neither has a higher median. 2. Model Y is more consistent. Its range is \(9 - 7 = 2\,\text{h}\), compared with \(12 - 4 = 8\,\text{h}\) for Model X. Its IQR is also smaller: \(8.5 - 7.5 = 1\,\text{h}\), compared with \(10 - 6 = 4\,\text{h}\) for Model X. 3. For Model X, \(Q_3 = 10\,\text{h}\). At least the highest \(25\%\) of the values are therefore \(10\,\text{h}\) or greater. 4. Model Y's range is \(9 - 7 = 2\,\text{h}\).

a) Neither; both medians are \(8\,\text{h}\). b) Model Y, because its range and IQR are smaller. c) At least \(25\%\) d) \(2\,\text{h}\)

- Compare the endpoints of the whiskers for the extrema and ranges. - The line inside each box marks the median. - The box contains the middle \(50\%\) of the data. - The range is the maximum minus the minimum.

1. Statement 1 is true. Class B's maximum is \(35\), while Class A's maximum is \(32\). 2. Statement 2 is false. Class A's median is \(23.5\), and Class B's median is \(25.5\). 3. Statement 3 is true. Class B's box extends from \(Q_1 = 23\) to \(Q_3 = 27.5\), and the box represents the middle \(50\%\) of the data. 4. Statement 4 is true. Class A's range is \(32 - 15 = 17\), while Class B's range is \(35 - 10 = 25\).

Statements \(1\), \(3\), and \(4\) are true.

- For race times, the smallest value represents the fastest runner. - The range is the maximum minus the minimum. - The median divides the ordered data into two halves.

1. The fastest runner has the minimum time. Group A's minimum is \(20\) minutes, and Group B's minimum is \(23\) minutes, so Group A has the fastest runner. 2. Group A's range is \(32 - 20 = 12\) minutes. Group B's range is \(29 - 23 = 6\) minutes. Group B has more consistent times because its range is smaller. 3. Group A's median is \(25\) minutes, so at least \(50\%\) of its runners finished in \(25\) minutes or less. Group B's median is \(26\) minutes, so at least \(50\%\) finished in \(26\) minutes or less. The condition holds for both groups.

a) Group A, with a minimum time of \(20\) minutes b) Group B, because its range is \(6\) minutes rather than \(12\) minutes c) Both groups

- The lower and upper edges of a box represent \(Q_1\) and \(Q_3\). - The box contains the middle \(50\%\) of the data. - Find each range by subtracting the minimum from the maximum.

1. For Class A, the lower and upper edges of the box show \(Q_1 = 2\) and \(Q_3 = 8\). 2. For Class B, the box extends from \(4\) to \(7\). Therefore, the middle \(50\%\) read from \(4\) to \(7\) books. 3. Class A has range \(20 - 0 = 20\) books. Class B has range \(12 - 1 = 11\) books. 4. Class A has the greater range, and the difference is \(20 - 11 = 9\) books.

a) \(Q_1 = 2\) and \(Q_3 = 8\) b) At least \(4\) books and at most \(7\) books c) Class A; the difference between the ranges is \(9\) books.

- The lower whisker and the left edge of the box describe the lowest quarter of the data. - Compare the teams' minima and first quartiles. - Higher point totals indicate stronger scoring performance.

1. Team Alpha's lowest \(25\%\) of scores extend from \(40\) points to \(Q_1 = 50\) points. 2. Team Beta's lowest \(25\%\) extend from \(50\) points to \(Q_1 = 65\) points. 3. Team Beta's minimum of \(50\) points equals Team Alpha's first quartile, and Team Beta's first quartile of \(65\) points is greater than Team Alpha's median of \(60\) points. 4. Team Beta appears stronger based on its lower-scoring games because even its lower quarter is shifted toward higher scores.

Team Beta appears stronger. Its minimum is \(50\) points, equal to Team Alpha's \(Q_1\), and its \(Q_1\) of \(65\) points is greater than Team Alpha's median of \(60\) points.

- Compare the full lengths from minimum to maximum. - The line inside each box marks the median. - The interval from \(Q_1\) to \(Q_3\) contains the middle \(50\%\) of the data.

1. Service A's range is \(48 - 12 = 36\) hours. Service B's range is \(30 - 18 = 12\) hours. Service C's range is \(60 - 24 = 36\) hours. Service B is the most consistent because it has the smallest range. 2. Services A and B each have a median of \(24\) hours. Service C has a median of \(36\) hours. Services A and B are fastest according to the median. 3. The middle \(50\%\) of Service A's times are from \(20\) to \(28\) hours. The middle \(50\%\) of Service B's times are from \(22\) to \(26\) hours.

a) Service B, with a range of \(12\) hours b) Services A and B, each with a median of \(24\) hours c) Service A: \(20\) to \(28\) hours; Service B: \(22\) to \(26\) hours

- The whisker endpoints show the minimum and maximum. - The IQR is the width of the box, \(Q_3 - Q_1\). - At least \(25\%\) of the values are at or above \(Q_3\).

1. For the South sample, the minimum is \(130\,\text{g}\), and the maximum is \(210\,\text{g}\). 2. The North orchard's IQR is \(175 - 145 = 30\,\text{g}\). The South orchard's IQR is \(180 - 150 = 30\,\text{g}\). 3. The middle \(50\%\) vary equally in the two samples because the IQRs are equal. 4. For the South orchard, \(Q_3 = 180\,\text{g}\), so at least \(25\%\) of the sample values are \(180\,\text{g}\) or more. The North orchard has \(Q_3 = 175\,\text{g}\), so the same conclusion cannot be guaranteed there.

a) Lightest: \(130\,\text{g}\); heaviest: \(210\,\text{g}\) b) Both IQRs are \(30\,\text{g}\). c) Neither; the middle \(50\%\) vary equally. d) The South orchard

- The median is the line inside each box. - For reaction time, a smaller value means a faster response. - The range is the maximum minus the minimum. - Compare the box widths to compare the middle \(50\%\).

1. The regular players have a median of \(215\,\text{ms}\), and the infrequent players have a median of \(260\,\text{ms}\). Because a lower reaction time is faster, the regular players reacted faster typically in these samples. 2. The regular players' range is \(280 - 180 = 100\,\text{ms}\). The infrequent players' range is \(380 - 210 = 170\,\text{ms}\). The infrequent players have the greater range. 3. The regular players' IQR is \(230 - 200 = 30\,\text{ms}\). The infrequent players' IQR is \(290 - 240 = 50\,\text{ms}\). The middle \(50\%\) are closer together for the regular players.

a) Regular players: \(215\,\text{ms}\); infrequent players: \(260\,\text{ms}\). The regular-player sample has the faster typical reaction time. b) The infrequent-player group, with a range of \(170\,\text{ms}\) c) The regular-player group, with an IQR of \(30\,\text{ms}\)

- Identify the whisker endpoints, box edges, and line inside the box. - Find each IQR using \(Q_3 - Q_1\). - A smaller IQR indicates more consistent values in the middle half of the data.

1. For Fertilizer A, the minimum is \(15\,\text{cm}\), \(Q_1 = 22\,\text{cm}\), the median is \(28\,\text{cm}\), \(Q_3 = 35\,\text{cm}\), and the maximum is \(45\,\text{cm}\). 2. Fertilizer A's IQR is \(35 - 22 = 13\,\text{cm}\). 3. Fertilizer B has \(Q_1 = 25\,\text{cm}\) and \(Q_3 = 33\,\text{cm}\), so its IQR is \(33 - 25 = 8\,\text{cm}\). 4. The box plots support the claim for the middle \(50\%\) of the data because Fertilizer B has the smaller IQR. Its total range is also smaller: \(38 - 20 = 18\,\text{cm}\), compared with \(45 - 15 = 30\,\text{cm}\) for Fertilizer A.

a) Minimum \(15\,\text{cm}\), \(Q_1 = 22\,\text{cm}\), median \(28\,\text{cm}\), \(Q_3 = 35\,\text{cm}\), maximum \(45\,\text{cm}\) b) Yes. Fertilizer B has an IQR of \(8\,\text{cm}\), compared with \(13\,\text{cm}\) for Fertilizer A.

- The two edges of each box represent \(Q_1\) and \(Q_3\). - Think about what percent of the data lies at or above \(Q_1\).

1. For Class C, \(Q_1 = 20\) and \(Q_3 = 40\). 2. For Class D, \(Q_1 = 30\) and \(Q_3 = 42\). 3. At least \(75\%\) of the values are at or above the first quartile. 4. Since Class D has \(Q_1 = 30\), at least \(75\%\) of its students scored \(30\) points or more. Class C has \(Q_1 = 20\), so the box plot does not guarantee that result for Class C.

a) Class C: \(Q_1 = 20\), \(Q_3 = 40\) Class D: \(Q_1 = 30\), \(Q_3 = 42\) b) Class D, because its first quartile is \(30\) points.

- The minimum, median, and maximum are deliberately the same for both teams, so they cannot determine the match. - Use the stated quartile convention on the four values below and above each median. - Compare your two quartiles with the left and right edges of each displayed box.

1. Team Blue has minimum \(12\), \(Q_1=(14+18)\div2=16\), median \(24\), \(Q_3=(29+31)\div2=30\), and maximum \(36\). 2. Team Red has minimum \(12\), \(Q_1=(21+23)\div2=22\), median \(24\), \(Q_3=(27+29)\div2=28\), and maximum \(36\). 3. Since both teams have the same minimum, median, and maximum, those three values cannot distinguish the plots. 4. Box plot (1) has quartiles \(16\) and \(30\), so it represents Team Blue. Box plot (2) has quartiles \(22\) and \(28\), so it represents Team Red.

Team Blue: five-number summary \(12,16,24,30,36\); box plot (1). Team Red: five-number summary \(12,22,24,28,36\); box plot (2).

- Locate the middle value of the complete data set. - Divide the remaining values into two equal halves. - Use only the minimum and maximum to find the range.

1. The data are already ordered. The sixth value is the median, \(12\) minutes. 2. The lower half is \(3, 5, 7, 8, 9\), so \(Q_1 = 7\) minutes. 3. The upper half is \(13, 15, 18, 21, 24\), so \(Q_3 = 18\) minutes. 4. The range is \(24 - 3 = 21\) minutes.

\(Q_1 = 7\) minutes, median \(12\) minutes, \(Q_3 = 18\) minutes, and range \(21\) minutes.

- Decide which endpoints belong in each calculation. - An outlier can lie beyond a whisker. - Do not confuse the width of the box with the range.

1. The range between the whisker endpoints is \(80 - 20 = 60\) pages. 2. For all displayed values, the outlier at \(100\) is the maximum. 3. The range of all displayed values is \(100 - 20 = 80\) pages.

a) \(60\) pages b) \(80\) pages

- The choices deliberately agree on the median and range, so compute both quartiles. - Each half contains four values and therefore has two middle values. - Compare your two quartiles with all three choices.

1. All three choices use the same median and range, so those values cannot distinguish the choices. 2. The lower half is \(2,4,5,7\), so \(Q_1=(4+5)\div2=4.5\). 3. The upper half is \(10,12,13,16\), so \(Q_3=(12+13)\div2=12.5\). 4. Therefore Summary A is correct.

Summary A is correct.

- Compare the \(\times\) with the open circle. - The box and whiskers do not include every displayed value. - Consider how one very large value affects a mean.

1. The median is \(\$40\), and the mean is \(\$45\). 2. A high outlier is shown at \(\$120\). 3. This unusually large value pulls the mean to the right of the median. 4. Therefore, the positions are possible and are not a drawing error.

No. The outlier at \(\$120\) pulls the mean to \(\$45\), which is to the right of the median at \(\$40\).

- The bar heights are frequencies. - Identify the relevant positions in the ordered data. - Use cumulative frequencies instead of writing every value.

1. The frequencies total \(2 + 4 + 8 + 6 + 4 = 24\) values. 2. For the lower half, positions \(6\) and \(7\) are \(1\) and \(2\), so \(Q_1 = (1 + 2) \div 2 = 1.5\). 3. Positions \(12\) and \(13\) are both \(2\), so the median is \(2\). 4. For the upper half, positions \(18\) and \(19\) are both \(3\), so \(Q_3 = 3\).

\(Q_1 = 1.5\), median \(2\), and \(Q_3 = 3\).

- Check each statement separately. - Distinguish the whisker endpoints from all displayed values. - Distinguish the \(\times\) marking the mean from the median line.

1. The box length is \(70 - 30 = 40\), so statement 1 is true. 2. All displayed values extend from the outlier at \(5\) to \(90\), so the range is \(90 - 5 = 85\). Statement 2 is false. 3. The median is \(50\), and the mean is \(45\), so statement 3 is true. 4. The open circle at \(5\) is an outlier, not a whisker endpoint. Statement 4 is false.

Statements 1 and 3 are true.

- Divide the even number of values into two equal halves. - Each half also has two middle values. - Use the minimum and maximum for the range.

1. The median is \((11 + 12) \div 2 = 11.5\). 2. The lower half is \(4, 6, 7, 8, 9, 11\), so \(Q_1 = (7 + 8) \div 2 = 7.5\). 3. The upper half is \(12, 14, 15, 18, 20, 24\), so \(Q_3 = (15 + 18) \div 2 = 16.5\). 4. The range is \(24 - 4 = 20\).

\(Q_1 = 7.5\), median \(11.5\), \(Q_3 = 16.5\), and range \(20\).

- Look for values on the side toward which the mean is shifted. - Compare the median with the interval represented by the box. - A measure of a usual value should not be controlled by a few extremes.

1. Two low outliers are shown at \(2.5\) and \(5\) points. 2. These low values pull the mean downward. 3. The middle \(50\%\) of the scores lie from \(15\) to \(25\) points. 4. The median of \(20\) points better describes a usual score.

The low outliers pull the mean downward. The median of \(20\) points better describes a usual score.

- Compare each statement with the box, median, mean, and outliers. - The mean does not always show where most values lie. - Choose the statement that accounts for the entire displayed distribution.

1. The median is \(10\) minutes, and the box extends from \(8\) to \(12\) minutes. 2. The high outliers at \(30\) and \(40\) minutes pull the mean to \(14\) minutes. 3. Statement A mistakes the shifted mean for the location of most values. 4. Statement C is false because the outliers increase the mean. 5. Statement B agrees with the box and the median.

Statement B is best supported. The median of \(10\) minutes is at the center of the box, while high outliers raise the mean.

- Read the median, box edges, mean markers, and outlier separately. - Use the specific measure requested in each part. - Consider how one unusually large value affects the mean.

1. Studio North has a median of \(4\) minutes, and Studio South has a median of \(5\) minutes. Studio North is typically faster according to the median. 2. Studio North's box has length \(6 - 3 = 3\) minutes. Studio South's box also has length \(6 - 3 = 3\) minutes. Their middle \(50\%\) have equal variability according to the IQR. 3. Studio North has a high outlier at \(14\) minutes. This value raises North's mean to \(5\) minutes, which is greater than South's mean of \(4.5\) minutes.

a) Studio North is typically faster: median \(4\) minutes versus \(5\) minutes for Studio South. b) Both IQRs are \(3\) minutes. c) North's \(14\)-minute outlier raises its mean to \(5\) minutes, above South's mean of \(4.5\) minutes.

- Read all five marked locations before changing units. - Use the relationship between seconds and minutes consistently for every value. - Find each spread from the converted summary, not from the picture’s physical length.

1. Read the five-number summary in seconds: \(90, 120, 150, 180, 270\). 2. Divide each value by \(60\) to convert seconds to minutes. The summary is \(1.5, 2, 2.5, 3, 4.5\) minutes. 3. The interquartile range is \(3-2=1\) minute. 4. The range is \(4.5-1.5=3\) minutes.

a) Minimum \(1.5\) minutes, \(Q_1=2\) minutes, median \(2.5\) minutes, \(Q_3=3\) minutes, maximum \(4.5\) minutes b) Interquartile range \(1\) minute; range \(3\) minutes

- Apply the same calibration change to every marked statistic. - Compare the distances between box-plot landmarks before and after a common shift. - Recheck the two measures of spread using the corrected values.

1. The original five-number summary is \(40,45,50,55,60\), in degrees Fahrenheit. 2. Increasing every value by \(4\) gives the corrected summary \(44,49,54,59,64\). 3. The corrected range is \(64-44=20\,^{\circ}\text{F}\). 4. The corrected interquartile range is \(59-49=10\,^{\circ}\text{F}\). 5. Every marked location shifts \(4\) units to the right, while the box length, whisker lengths, range, and interquartile range stay unchanged.

a) Minimum \(44\,^{\circ}\text{F}\), \(Q_1=49\,^{\circ}\text{F}\), median \(54\,^{\circ}\text{F}\), \(Q_3=59\,^{\circ}\text{F}\), maximum \(64\,^{\circ}\text{F}\) b) Range \(20\,^{\circ}\text{F}\); interquartile range \(10\,^{\circ}\text{F}\) c) All five positions shift by \(4\), while the box length, whisker lengths, range, and interquartile range stay the same.

- Move from left to right, adding one section length at a time. - Keep the five box-plot landmarks in their required order. - Use the box endpoints for one spread and the outer endpoints for the other.

1. \(Q_1=5+3=8\). 2. The median is \(8+4=12\). 3. \(Q_3=12+6=18\). 4. The maximum is \(18+2=20\). 5. The interquartile range is \(18-8=10\), and the range is \(20-5=15\).

Five-number summary: \(5, 8, 12, 18, 20\) Interquartile range: \(10\) Range: \(15\)

- Use the middle three entries of each summary to compare the boxes. - Use the two outer entries for the full spread. - Separate information about the middle half from information about all values.

1. Both medians are \(8\). 2. Both interquartile ranges are \(11-5=6\), so their boxes have the same endpoints and length. 3. Plot A has range \(15-2=13\). 4. Plot B has range \(17-1=16\). 5. Identical boxes show the same middle-half center and spread, but different whisker endpoints can produce different overall ranges.

Both medians are \(8\), and both IQRs are \(6\). Plot A’s range is \(13\); Plot B’s range is \(16\).

- Interpret the quartile section as a portion of ordered positions. - Consider what happens when several observations equal a quartile value. - Separate “at or above” from “strictly greater than.”

1. The interval from \(Q_3\) to the maximum represents the upper quarter of the ordered data, about \(25\%\) of the observations. 2. For \(24\) values, one quarter corresponds to about \(6\) positions. 3. A box plot does not show how many observations equal \(Q_3\). 4. If repeated values occur at \(Q_3\), the number strictly greater than \(Q_3\) may be fewer than \(6\). 5. Therefore, the statement of exactly \(6\) strictly greater values is not guaranteed.

The statement is not guaranteed. The upper quarter contains about \(6\) ordered positions from \(Q_3\) toward the maximum, but the exact number strictly greater than \(Q_3\) is hidden.

- Apply the transformation to each box-plot landmark. - Check whether the rule keeps the order increasing. - Recompute both spreads from the transformed endpoints.

1. The rule is increasing, so it preserves the order of all five landmarks. 2. Applying \(2x-1\) gives \(5,9,15,19,27\). 3. The new median is \(15\). 4. The new interquartile range is \(19-9=10\). 5. The new range is \(27-5=22\).

New five-number summary: \(5,9,15,19,27\) Median: \(15\) Interquartile range: \(10\) Range: \(22\)

- Use the outer landmarks for one spread and the quartiles for the other. - Interpret a section length of zero as equal endpoint values. - Separate numerical width from the number of ordered observations.

1. The range is \(12-2=10\). 2. The interquartile range is \(9-5=4\). 3. Because \(Q_1\) and the median are both \(5\), the second quarter of ordered positions has no numerical spread. 4. It does not mean that the quarter contains no observations; repeated values at \(5\) can occupy those positions.

Range \(10\), interquartile range \(4\). The zero-length section means that quarter’s values are concentrated at \(5\), not that the quarter is empty.

- Relate the box from \(Q_1\) to \(Q_3\) to a fraction of the whole data set. - Work backward from the count represented by one-half. - Divide the full count into four equal positional groups.

1. The middle \(50\%\) is one-half of all observation positions. 2. If one-half is \(18\), the total is \(18\times2=36\). 3. Each quarter represents about \(\frac{36}{4}=9\) observation positions.

Total observations: \(36\) Observation positions per quarter: about \(9\)

- Read the five-number summary from the box plot. - Determine where the median and quartiles occur in an ordered list of eight values. - Place the known measurements in order and identify where the missing value must lie. - Use the third quartile to write an equation for the missing value.

1. The box plot shows a minimum of \(3\), \(Q_1 = 7\), a median of \(11\), \(Q_3 = 17\), and a maximum of \(21\). 2. With eight ordered values, the median is the mean of the fourth and fifth values. The known values \(10\) and \(12\) must remain in those positions because \((10 + 12) \div 2 = 11\). Thus, the missing value lies between \(12\) and \(19\). 3. The upper half is then \(12, x, 19, 21\), so \(Q_3 = (x + 19) \div 2\). 4. Set \((x + 19) \div 2 = 17\). Then \(x + 19 = 34\), so \(x = 15\). 5. This value fits between \(12\) and \(19\), and the complete data set has the five-number summary shown.

Jordan's sunflower seedling was \(15\,\text{cm}\) tall.

- Read the five-number summary from the box plot first. - Order each data set before finding its median and quartiles. - Check whether the minimum and maximum eliminate any choices. - Compare the remaining quartiles and median one at a time.

1. The box plot shows the five-number summary \(2, 5, 9, 14, 18\). 2. Data set (1), in order, is \(2, 3, 4, 6, 8, 9, 9, 11, 13, 15, 16, 18\). Its minimum is \(2\), \(Q_1 = (4 + 6) \div 2 = 5\), median is \((9 + 9) \div 2 = 9\), \(Q_3 = (13 + 15) \div 2 = 14\), and maximum is \(18\). These values match the box plot. 3. Data set (2) has median \((10 + 12) \div 2 = 11\), so it does not match. 4. Data set (3) has \(Q_1 = (6 + 8) \div 2 = 7\), so it does not match. 5. Therefore, only data set (1) matches the box plot.

Data set (1) matches the box plot. Its five-number summary is \(2, 5, 9, 14, 18\).

- Work only with the lower half of the ordered data set. - The lower half has an even number of values. - Check that your result lies between its neighboring values.

1. The overall median, \(10\), is excluded when the halves are formed. 2. The lower half is \(2, 4, x, 8\). 3. Its median is \((4 + x) \div 2\). Set \((4 + x) \div 2 = 5\). 4. Then \(4 + x = 10\), so \(x = 6\). 5. The value \(6\) lies between \(4\) and \(8\), so the list remains ordered.

\(x = 6\)

- Fix the positions of the minimum, median, and maximum first. - Each quartile can be the mean of two neighboring values. - More than one data set can satisfy the conditions.

1. Place the minimum, median, and maximum in positions \(1\), \(5\), and \(9\): \(1\), \(10\), and \(20\). 2. For \(Q_1 = 5\), the second and third values must have a mean of \(5\). Choose \(4\) and \(6\). 3. For \(Q_3 = 15\), the seventh and eighth values must have a mean of \(15\). Choose \(14\) and \(16\). 4. Choose \(8\) and \(12\) for the remaining positions. One valid data set is \(1, 4, 6, 8, 10, 12, 14, 16, 20\).

One possible data set is \(1, 4, 6, 8, 10, 12, 14, 16, 20\).

- Find the four given statistics for each data set. - A box plot does not show every individual data value. - More than one data set can have the same five-number summary.

1. Data set A has \(Q_1 = (3 + 5) \div 2 = 4\) and \(Q_3 = (13 + 15) \div 2 = 14\), so it does not match. 2. Data set B has \(Q_1 = 5\), median \(9\), \(Q_3 = 13\), and range \(17 - 1 = 16\). 3. Data set C also has \(Q_1 = 5\), median \(9\), \(Q_3 = 13\), and range \(16\). 4. Therefore, both B and C could produce the given box plot.

Data sets B and C both match. A box plot does not determine every value in the original data set.

- Identify which two values are still missing from a five-number summary. - Choose different valid positions for those two values while keeping the required order. - Compare the resulting box lengths.

1. One valid summary is \(2, 4, 7, 10, 14\). Its interquartile range is \(10-4=6\). 2. Another valid summary is \(2, 6, 7, 8, 14\). Its interquartile range is \(8-6=2\). 3. Both summaries have the required minimum, median, and maximum, but their boxes have different endpoints and lengths. 4. Therefore, \(Q_1\) and \(Q_3\) are also needed to determine the box plot.

For example, \(2, 4, 7, 10, 14\) has interquartile range \(6\), while \(2, 6, 7, 8, 14\) has interquartile range \(2\). The minimum, median, and maximum alone do not determine a box plot.

- Find the opposite of each of the five original statistics. - Put the reflected values back in increasing order. - Treat range and interquartile range as distances on the number line.

1. The opposites of the five original statistics are \(-2,-5,-8,-11,-15\). 2. In increasing order, the reflected five-number summary is \(-15,-11,-8,-5,-2\). 3. The distance from \(-15\) to \(-2\) is \(13\), so the range is \(13\). 4. The distance from \(-11\) to \(-5\) is \(6\), so the interquartile range is \(6\). 5. Reflection across \(0\) reverses left-right order, so the old maximum becomes the new minimum and the old \(Q_3\) becomes the new \(Q_1\).

a) \(-15,-11,-8,-5,-2\) b) Range \(13\); interquartile range \(6\) c) Reflection across \(0\) reverses the order of the values.

- Verify the five marked statistics for each ordered list. - Count repeated values separately from finding quartiles. - Compare what a box plot records with what it leaves hidden.

1. For set A, the median is \(4\), \(Q_1=(2+2)\div2=2\), and \(Q_3=(6+6)\div2=6\). Its modes are \(2\) and \(6\). 2. For set B, the median is \(4\), \(Q_1=(1+3)\div2=2\), and \(Q_3=(5+7)\div2=6\). Its modes are \(3\) and \(5\). 3. Both sets have minimum \(0\), maximum \(8\), and the same quartiles and median. 4. They produce the same box plot but have different modes, so the mode is not determined by a box plot.

Both data sets have five-number summary \(0, 2, 4, 6, 8\), but set A has modes \(2\) and \(6\), while set B has modes \(3\) and \(5\). The box plot does not determine the mode.

- Place the required minimum, middle value, and maximum first. - Choose lower-half values whose two middle entries give the first quartile. - Check the upper half separately, then consider which interior entries could vary.

1. One valid ordered data set is \(2, 4, 4, 5, 6, 8, 9, 9, 12\). 2. The middle value is \(6\), so the median is \(6\). 3. In the lower half \(2, 4, 4, 5\), the middle pair averages to \((4+4)\div2=4\), so \(Q_1=4\). 4. In the upper half \(8, 9, 9, 12\), the middle pair averages to \((9+9)\div2=9\), so \(Q_3=9\). 5. The minimum is \(2\) and the maximum is \(12\). 6. Other interior values can be changed while preserving these five landmarks, so the box plot does not determine a unique data set.

One possible data set is \(2, 4, 4, 5, 6, 8, 9, 9, 12\). It is not unique because a box plot records only the five-number summary, not every observation.

- Use the quartile convention stated in the problem. - Compute each mean from all nine values. - Identify which information a five-number summary leaves hidden.

1. In both ordered sets, the minimum is \(0\), the median is the fifth value \(4\), and the maximum is \(8\). 2. In each lower half, the middle two values are \(2,2\), so \(Q_1=2\). 3. In each upper half, the middle two values are \(6,6\), so \(Q_3=6\). 4. Data Set A has total \(36\), so its mean is \(36\div9=4\). 5. Data Set B has total \(35\), so its mean is \(\frac{35}{9}\approx3.89\). 6. A box plot does not show all interior values, so symmetry of the five-number summary does not force the mean to equal the median.

Both sets have five-number summary \(0,2,4,6,8\). Set A has mean \(4\), while Set B has mean \(\frac{35}{9}\approx3.89\). The claim is false.

- Separate information about extreme values from information about ordered middle positions. - Ask what the old box plot does not reveal about the original data set. - Consider whether adding one value could shift the positions used for the center and quartiles.

1. The old minimum remains \(14\,\text{g}\), and the new maximum is \(50\,\text{g}\), because the added value is larger than every displayed value. 2. The new range is \(50-14=36\,\text{g}\). 3. The old box plot does not show the sample size or the individual interior values. Adding one observation can change which ordered positions determine the median and quartiles. 4. Therefore, only the new minimum, maximum, and range are certain; the new \(Q_1\), median, and \(Q_3\) cannot be found from the box plot alone.

a) The new minimum is \(14\,\text{g}\), and the new maximum is \(50\,\text{g}\). The quartiles and median are not determined. b) \(36\,\text{g}\) c) The box plot does not provide the number or positions of all individual data values, which are needed after adding a new observation.

- Identify the median position in a sorted list of \(25\) values. - Identify the two positions that determine the median of the upper half. - Compare values in later positions with the overall median. - Decide when the mean of two values that are each at least \(\$20\) can equal \(\$20\).

1. In a sorted list of \(25\) values, the median is the value in position \(13\). Therefore, the value in position \(13\) is \(\$20\). 2. The upper half consists of positions \(14\) through \(25\), so \(Q_3\) is the mean of the values in positions \(19\) and \(20\). 3. Because the list is sorted, the values in positions \(19\) and \(20\) are each at least \(\$20\). Their mean cannot be \(\$15\), so the claim is impossible. 4. For their mean to equal \(\$20\), both values must equal \(\$20\). The sorted order then requires every value from position \(13\) through position \(20\) to equal \(\$20\).

a) The claim is impossible. The values in positions \(19\) and \(20\), which determine \(Q_3\), cannot be less than the median value of \(\$20\). b) Every value from position \(13\) through position \(20\) must be \(\$20\). Then the median is \(\$20\) and \(Q_3 = (\$20 + \$20) \div 2 = \$20\).