Question

below are two representations of data.
a: 2,5,5,6,6,6,7,8,9
box plot for b with number line 3,4,5,6,7,8,9,10,11
which statement about a and b is true?

1. median of a > median of b
2. range of a < range of b
3. upper quartile of a < upper quartile of b
4. lower quartile of a > lower quartile of b

which statistic can not be determined from a box plot representing the scores on a math test in mrs. deridder’s algebra class?

1. the lowest score
2. the median score
3. the highest score
4. the score that occurs most frequently

the dot plot shown below represents the number of pets owned by students in a class.
dot plot with x - axis 0,1,2,3,4,5,6
which statement about the data is not true?

1. the median is 3.
2. the interquartile range is 2.
3. the mean is 3.
4. the data contain no outliers.

Explanation:

First Question (About A and B)

Step 1: Analyze Data A

Data A: \(2, 5, 5, 6, 6, 6, 7, 8, 9\) (assuming the first number is 2, maybe a typo like 2 instead of 3? Wait, the box plot for B has a minimum at 3? Wait, let's re - check. Wait, the data A is written as "2,5,5,6,6,6,7,8,9" (let's count the numbers: 9 numbers). Median of A: for 9 numbers, the median is the 5th term. The 5th term is 6. Range of A: \(9 - 2=7\). Lower quartile (Q1) of A: the first 4 numbers (\(2,5,5,6\)), Q1 is the median of these 4, which is \(\frac{5 + 5}{2}=5\). Upper quartile (Q3) of A: the last 4 numbers (\(6,7,8,9\)), Q3 is \(\frac{7+8}{2}=7.5\).

Step 2: Analyze Data B (Box - plot)

From the box - plot, the minimum (min) is 3, the lower quartile (Q1) is 4, the median (Q2) is 5, the upper quartile (Q3) is 9, and the maximum (max) is 10. Range of B: \(10 - 3 = 7\).

Step 3: Check Each Option

1. Median of A: 6, Median of B: 5. So median of A>median of B? Wait, no, 6>5? Wait, but let's re - check. Wait, maybe I misread data A. Wait, the first number in A: if it's a typo and should be 3? Wait, the box - plot for B has min at 3. Wait, the original data A: "2,5,5,6,6,6,7,8,9" – 9 numbers. Median is the 5th term: 6. Box - plot B: median (line in box) is at 5. So median of A (6)>median of B (5)? But let's check other options.
2. Range of A: \(9 - 2 = 7\), Range of B: \(10 - 3=7\). So range of A < range of B is false.
3. Upper quartile of A: 7.5, Upper quartile of B: 9. So upper quartile of A < upper quartile of B is true? Wait, no, 7.5 < 9 is true? Wait, but let's check option 4. Lower quartile of A: 5, Lower quartile of B: 4. So lower quartile of A>lower quartile of B (5 > 4) is true? Wait, there is a contradiction here. Wait, maybe I made a mistake in data A. Let's re - count data A: the numbers are 2,5,5,6,6,6,7,8,9. Number of terms: 9. Q1: for 9 terms, the first quartile is the median of the first 4 terms (positions 1 - 4: 2,5,5,6). Median of these 4: (5 + 5)/2 = 5. Q3: median of last 4 terms (positions 6 - 9: 6,7,8,9). Median of these 4: (7 + 8)/2 = 7.5. Box - plot B: from the plot, the left whisker starts at 3, Q1 is at 4, median at 5, Q3 at 9, right whisker at 10. So:

• Option 1: Median A = 6, Median B = 5. 6>5, so option 1 is true? But let's check option 4: Lower quartile A = 5, Lower quartile B = 4. 5>4, so option 4 is also true? Wait, this can't be. Maybe the data A is written wrong. Wait, the first number in A: maybe it's 3 instead of 2? Let's assume data A is 3,5,5,6,6,6,7,8,9. Then number of terms is 9. Median is 6. Range: 9 - 3 = 6. Q1: median of 3,5,5,6: (5 + 5)/2 = 5. Q3: median of 6,7,8,9: (7 + 8)/2 = 7.5. Box - plot B: range 10 - 3 = 7. Then range of A (6) < range of B (7), so option 2 is true. But this is based on a typo. Alternatively, maybe the original data A is 2,5,5,6,6,6,7,8,9 (9 terms). Then:
• Option 1: Median A = 6, Median B = 5 (from box - plot, the line in the box is at 5). So 6>5, option 1 is true.
• Option 2: Range A = 9 - 2 = 7, Range B = 10 - 3 = 7. So range A < range B is false.
• Option 3: Upper quartile A = 7.5, Upper quartile B = 9. 7.5 < 9, so option 3 is true.
• Option 4: Lower quartile A = 5, Lower quartile B = 4. 5>4, so option 4 is true.

This is a problem. Maybe the data A is 2,5,5,6,6,6,7,8,9 (9 terms) and the box - plot B has min = 3, Q1 = 4, median = 5, Q3 = 9, max = 10. Then let's re - evaluate:

• Option 1: Median A = 6, Median B = 5. 6>5, so option 1 is true.
• Option 2: Range A = 7, Range B = 7. So option 2 is false.
• Option 3: Upper quartile A = 7.5, Upper quartile B = 9. 7.5 < 9, option 3 is true.…

Step 1: Recall Box - plot Components

A box - plot shows the minimum value (lowest score), the median (median score), the maximum value (highest score), and the quartiles. But it does not show the frequency of individual scores.

Step 2: Analyze Each Option

1. The lowest score: can be determined from the left - most point of the whisker.
2. The median score: can be determined from the line inside the box.
3. The highest score: can be determined from the right - most point of the whisker.
4. The score that occurs most frequently (mode): a box - plot does not provide information about the frequency of individual scores, so it can not be determined.

Third Question (Dot - plot of Pets)

First, we need to count the number of dots:

• For 0: 2 dots
• For 1: 2 dots
• For 2: 5 dots
• For 3: 5 dots
• For 4: 3 dots
• For 5: 2 dots
• For 6: 1 dot

Total number of data points: \(2 + 2+5 + 5+3 + 2+1=20\)

Step 1: Find the Median

Since there are 20 data points, the median is the average of the 10th and 11th terms. Let's order the data:

• 0 (2 times), 1 (2 times), 2 (5 times) – cumulative: \(2 + 2+5 = 9\)
• Then 3 (5 times) – the 10th and 11th terms are 3. So median is 3. Option 1 is true.

Step 2: Find the Interquartile Range (IQR)

Q1: median of the first 10 terms. The first 10 terms: 0,0,1,1,2,2,2,2,2,3. The median of these 10 terms (average of 5th and 6th terms) is \(\frac{2 + 2}{2}=2\).
Q3: median of the last 10 terms. The last 10 terms: 3,3,3,3,3,4,4,4,5,5,6 (wait, no, total 20 terms, last 10 terms: positions 11 - 20. The data: after the first 9 terms (0,0,1,1,2,2,2,2,2), the next 5 are 3,3,3,3,3 (positions 10 - 14), then 4,4,4 (positions 15 - 17), 5,5 (positions 18 - 19), 6 (position 20). So the last 10 terms: 3,3,3,3,3,4,4,4,5,5. The median of these 10 terms (average of 5th and 6th terms) is \(\frac{3 + 4}{2}=3.5\)? Wait, no, my counting is wrong. Wait, total number of data points: 2 (0s) + 2 (1s)+5 (2s)+5 (3s)+3 (4s)+2 (5s)+1 (6s) = 20.
Q1: the median of the first 10 data points. The first 10 data points: 0,0,1,1,2,2,2,2,2,3. The 5th term is 2, the 6th term is 2. So Q1 = 2.
Q3: the median of the last 10 data points. The last 10 data points: 3,3,3,3,3,4,4,4,5,5. The 5th term is 3, the 6th term is 4. So Q3=\(\frac{3 + 4}{2}=3.5\). Then IQR = Q3 - Q1=3.5 - 2 = 1.5? Wait, this contradicts option 2. Wait, maybe my counting is wrong. Let's list all data points:
0,0,
1,1,
2,2,2,2,2,
3,3,3,3,3,
4,4,4,
5,5,
6

Now, number the positions from 1 to 20:
1:0, 2:0,
3:1, 4:1,
5:2, 6:2, 7:2, 8:2, 9:2,
10:3, 11:3, 12:3, 13:3, 14:3,
15:4, 16:4, 17:4,
18:5, 19:5,
20:6

Q1: median of first 10 (positions 1 - 10): (position 5 + position 6)/2=(2 + 2)/2 = 2.
Q3: median of last 10 (positions 11 - 20): (position 15 + position 16)/2=(4 + 4)/2 = 4? Wait, no, position 11:3, 12:3, 13:3, 14:3, 15:4, 16:4, 17:4, 18:5, 19:5, 20:6. The median of positions 11 - 20 is the average of position 15 and 16: (4 + 4)/2 = 4. Then IQR = 4 - 2 = 2. So option 2 is true.

Step 3: Find the Mean

Mean=\(\frac{(0\times2)+(1\times2)+(2\times5)+(3\times5)+(4\times3)+(5\times2)+(6\times1)}{20}\)
=\(\frac{0 + 2+10 + 15+12 + 10+6}{20}=\frac{55}{20}=2.75
eq3\). So option 3 is not true.

Step 4: Check for Outliers

Using the IQR method: IQR = 2, Q1 = 2, Q3 = 4. Lower fence = Q1 - 1.5×IQR=2 - 3=-1. Upper fence = Q3+1.5×IQR=4 + 3 = 7. All data points (0,1,2,3,4,5,6) are within [-1,7], so no outliers. Option 4 is true.

Answer:

s:

1. (Assuming the correct analysis) For the first question, if we consider the lower quartile: Lower quartile of A (5) > Lower quartile of B (4), so the answer is 4) lower quartile of \(A>\) lower quartile of \(B\)
2. For the second question, the answer is 4) the score that occurs most frequently
3. For the third question, the answer is 3) The mean is 3.