Question

unit 1 practice #2
name: jahleil haff

1. use the box plot to

answer the following:
box plot with scale 2.8, 3.0, 3.2, 3.4, 3.6, 3.8, 4.0, 4.2, 4.4, 4.6, 4.8, 5.0, 5.2, 5...
calculate the 5 number summary of the given box plot.
table with columns: minimum, q1, median, q3, maximum
2.
the number of writing instruments in some teachers’ desks is displayed in
the dot plot. which is greater, the mean or the median? explain your
reasoning using the shape of the distribution.
dot plot with number of writing instruments: 4, 5, 6, 7, 8, 9, 10, 11, 12, 13
3.
here are two box plots:
box plot a (scale 95, 100, 105, 110, 115)
box plot b (scale 95, 100, 105, 110...)
a. which box plot has a greater median?
b. which box plot has a greater measure of variability?

Explanation:

Question 1: Calculate the 5 - number summary of the given box plot.

• Step 1: Identify the Minimum

The minimum value is the left - most point of the whisker. From the box - plot scale, the left - most whisker starts at 2.8. So, Minimum = 2.8.

• Step 2: Identify Q1 (First Quartile)

Q1 is the left - hand side of the box. Looking at the box, the left side of the box is at 3.0. So, Q1 = 3.0.

• Step 3: Identify the Median

The median is the line inside the box. From the box - plot, the line inside the box is at 3.6. So, Median = 3.6.

• Step 4: Identify Q3 (Third Quartile)

Q3 is the right - hand side of the box. The right side of the box is at 4.0. So, Q3 = 4.0.

• Step 5: Identify the Maximum

The maximum value is the right - most point of the whisker. From the box - plot scale, the right - most whisker ends at 5.2. So, Maximum = 5.2.

1. First, we analyze the shape of the distribution. The dot - plot has data points from 5 to 12. Let's count the number of dots:

• For 5: 1 dot; 6: 1 dot; 7: 1 dot; 8: 1 dot; 9: 3 dots; 10: 4 dots; 11: 2 dots; 12: 1 dot.
• The total number of data points \(n=1 + 1+1 + 1+3 + 4+2 + 1=14\) (even number). The median will be the average of the 7th and 8th values when ordered.
• Ordering the data: 5, 6, 7, 8, 9, 9, 9, 10, 10, 10, 10, 11, 11, 12. The 7th value is 9 and the 8th value is 10. So, median \(=\frac{9 + 10}{2}=9.5\).
• Now, let's calculate the mean. The sum of the data:
• \(5\times1+6\times1 + 7\times1+8\times1+9\times3+10\times4+11\times2+12\times1\)
• \(=5 + 6+7 + 8+27+40+22+12\)
• \(=(5 + 6)+(7 + 8)+27+(40 + 22)+12\)
• \(=11+15 + 27+62+12\)
• \(=11 + 15=26;26+27 = 53;53+62 = 115;115+12 = 127\)
• Mean \(=\frac{127}{14}\approx9.07\) (wait, this is wrong. Wait, let's recalculate the sum:
• 5 (1 time): 5
• 6 (1 time): 6 (total so far: 5 + 6 = 11)
• 7 (1 time): 7 (total: 11+7 = 18)
• 8 (1 time): 8 (total: 18 + 8 = 26)
• 9 (3 times): \(9\times3=27\) (total: 26+27 = 53)
• 10 (4 times): \(10\times4 = 40\) (total: 53+40 = 93)
• 11 (2 times): \(11\times2=22\) (total: 93+22 = 115)
• 12 (1 time): 12 (total: 115 + 12=127). Mean \(=\frac{127}{14}\approx9.07\)? Wait, no, I made a mistake in counting the number of dots. Let's re - count the dots:
• At 5: 1 dot
• At 6: 1 dot
• At 7: 1 dot
• At 8: 1 dot
• At 9: 3 dots (so 3)
• At 10: 4 dots (so 4)
• At 11: 2 dots (so 2)
• At 12: 1 dot (so 1)
• Total \(n=1 + 1+1 + 1+3 + 4+2 + 1 = 14\). Wait, but when we order the data: 5, 6, 7, 8, 9, 9, 9, 10, 10, 10, 10, 11, 11, 12. The 7th term is 9 and the 8th term is 10. Median \(=\frac{9 + 10}{2}=9.5\).
• Wait, maybe my calculation of the mean is wrong. Let's use the concept of skewness. The distribution is symmetric? Wait, no. Wait, the number of dots on the left of the peak (at 10) and right:
• Left of 10: 5,6,7,8,9. Number of dots: \(1 + 1+1 + 1+3=7\)
• Right of 10: 11,12. Number of dots: \(2 + 1 = 3\)
• Wait, the peak is at 10. The data is slightly skewed to the right? Wait, no, the left side (from 5 to 9) has 7 dots and the right side (11,12) has 3 dots. Wait, actually, the distribution is symmetric around the middle. Wait, maybe I made a mistake. Let's calculate the mean again:
• Sum \(=5\times1+6\times1+7\times1+8\times1+9\times3+10\times4+11\times2+12\times1\)
• \(=5 + 6+7 + 8+27+40+22+12\)
• \(=5+6 = 11;11 + 7=18;18+8 = 26;26+27 = 53;53+40 = 93;93+22 = 115;115+12 = 127\)
• Mean \(=\frac{127}{14}\approx9.07\). But median is 9.5. Wait, that can't be. Wait, maybe I miscounted the dots. Let's look at the dot - plot again:
• At 5: 1 dot
• At 6: 1 dot
• At 7: 1 dot
• At 8: 1 dot
• At 9: 3 dots (so 3)
• At 10: 4 dots (so 4)
• At 11: 2 dots (so 2)
• At 12: 1 dot (so 1)
• Total \(n = 1+1 + 1+1+3+4+2+1=14\). The median is the average of the 7th and 8th terms. The first 7 terms: 5,6,7,8,9,9,9. The 8th term is 10. So median \(=\frac{9 + 10}{2}=9.5\). The mean: \(\frac{5+6+7+8+9\times3+10\times4+11\times2+12}{14}=\frac{5 + 6+7+8+27+40+22+12}{14}=\frac{127}{14}\approx9.07\). Wait, this is a contradiction. Wait, maybe the dot - plot is symmetric. Wait, maybe I mad…

• The median in a box - plot is the line inside the box.
• For Box Plot A, the median line is at 105 (since the box is between 100 and 110, and the middle line is at 105).
• For Box Plot B, the median line is at 100 (since the box is between 100 and 110, and the middle line is at 100).
• So, Box Plot A has a greater median.

Answer:

Minimum: 2.8, Q1: 3.0, Median: 3.6, Q3: 4.0, Maximum: 5.2

Question 2: Compare mean and median using the dot - plot