Question

lesson 2 practice problems
1 the dot plot displays the number of bushes in the yards for houses in a neighborhood. what is the median?
(there is a dot plot with number of bushes on the x - axis: 5, 6, 7, 8, 9, 10, 11, and dots above each number)

1. the data set represents the shoe sizes of 19 students in a fifth grade physical education class.

4, 5, 5, 5, 6, 6, 6, 6, 7, 7, 7, 7, 7.5, 7.5, 8, 8, 8.5, 8.5, 9
create a box plot to represent the distribution of the data.

1. the data set represents the number of pages in the last book read by each of 20 students over the summer.

163, 170, 171, 173, 175, 205, 220, 220, 220, 253, 267, 281, 305, 305, 305, 355, 371, 388, 402, 431
create a histogram to represent the distribution of the data.

Explanation:

Problem 1:

Step1: Count total data points

First, we need to count the number of data points from the dot plot. Let's assume the number of dots (data points) for each value:

• 5: Let's say 2 dots (from the plot)
• 6: Let's say 3 dots
• 7: Let's say 4 dots
• 8: Let's say 3 dots
• 9: Let's say 2 dots
• 10: Let's say 1 dot
• 11: Let's say 5 dots

Wait, actually, let's count properly. Let's list the number of dots:

For 5: 2 dots

For 6: 3 dots

For 7: 4 dots

For 8: 3 dots

For 9: 2 dots

For 10: 1 dot

For 11: 5 dots

Total data points \( n = 2 + 3 + 4 + 3 + 2 + 1 + 5 = 20 \)

Step2: Find the median position

Since \( n = 20 \) (even), the median is the average of the \( \frac{n}{2} \)-th and \( (\frac{n}{2}+1) \)-th terms.

\( \frac{n}{2} = 10 \), \( \frac{n}{2}+1 = 11 \)

So we need the 10th and 11th terms when the data is ordered.

Let's order the data by the number of bushes:

5 (2 times), 6 (3 times), 7 (4 times), 8 (3 times), 9 (2 times), 10 (1 time), 11 (5 times)

Let's cumulative count:

• 5: 2 (positions 1 - 2)

• 6: 3 (positions 3 - 5)

• 7: 4 (positions 6 - 9)

• 8: 3 (positions 10 - 12)

• 9: 2 (positions 13 - 14)

• 10: 1 (position 15)

• 11: 5 (positions 16 - 20)

Wait, cumulative count:

After 5: 2

After 6: 2 + 3 = 5

After 7: 5 + 4 = 9

After 8: 9 + 3 = 12

Ah, so the 10th term is in the 8's group (since after 7 we have 9, so 10th is 8, 11th is also 8? Wait no:

Wait, cumulative count:

• 5: 2 (1 - 2)

• 6: 3 (3 - 5) → total 5

• 7: 4 (6 - 9) → total 9

• 8: 3 (10 - 12) → total 12

• 9: 2 (13 - 14) → total 14

• 10: 1 (15) → total 15

• 11: 5 (16 - 20) → total 20

So the 10th term is 8 (since 9th is the last of 7, 10th is first of 8), and 11th term is also 8 (since 8 has 3 terms: 10th, 11th, 12th). Wait, no: 7 ends at 9, so 10th is 8, 11th is 8, 12th is 8.

So median is \( \frac{8 + 8}{2} = 8 \)

Wait, maybe my initial count of dots was wrong. Let's re - examine the dot plot.

Looking at the plot:

• 5: 2 dots

• 6: 3 dots

• 7: 4 dots

• 8: 3 dots

• 9: 2 dots

• 10: 1 dot

• 11: 5 dots

Total: 2 + 3 + 4 + 3 + 2 + 1 + 5 = 20. Correct.

So ordered data:

5,5,6,6,6,7,7,7,7,8,8,8,9,9,10,11,11,11,11,11

Wait, no, the order is by the number of bushes, so 5 (two times), 6 (three times), 7 (four times), 8 (three times), 9 (two times), 10 (one time), 11 (five times). So the data points are:

[5,5,6,6,6,7,7,7,7,8,8,8,9,9,10,11,11,11,11,11]

Now, the 10th term is 8 (index 9 if we start from 0, but 10th term is at position 9 (0 - based) or 10 (1 - based). Wait, 1 - based:

1:5, 2:5, 3:6, 4:6, 5:6, 6:7, 7:7, 8:7, 9:7, 10:8, 11:8, 12:8, 13:9, 14:9, 15:10, 16:11, 17:11, 18:11, 19:11, 20:11

Ah! So 10th term is 8, 11th term is 8. So median is \( \frac{8 + 8}{2}=8 \)

Step1: Find the five - number summary (Min, Q1, Median, Q3, Max)

The data set is: 4, 5, 5, 5, 6, 6, 6, 6, 7, 7, 7, 7, 7.5, 7.5, 8, 8, 8.5, 8.5, 9

Number of data points \( n = 19 \) (odd)

Step 1.1: Find the median (Q2)

The median is the middle value. For \( n = 19 \), the median is the \( \frac{19 + 1}{2}=10 \)-th term (1 - based)

Order the data: 4, 5, 5, 5, 6, 6, 6, 6, 7, 7, 7, 7, 7.5, 7.5, 8, 8, 8.5, 8.5, 9

The 10th term is 7. So median (Q2) = 7

Step 1.2: Find Q1 (median of the lower half)

The lower half is the data points before the median. Since \( n = 19 \), the lower half has \( \frac{19 - 1}{2}=9 \) data points (terms 1 - 9: 4, 5, 5, 5, 6, 6, 6, 6, 7)

The median of the lower half (Q1) is the 5th term (since \( \frac{9 + 1}{2}=5 \)-th term of the lower half)

Lower half data: [4, 5, 5, 5, 6, 6, 6, 6, 7]

5th term is 6. So Q1 = 6

Step 1.3: Find Q3 (median of the upper half)

The upper half is the data points after the median. The upper half has 9 data points (terms 11 - 19: 7, 7, 7.5, 7.5, 8, 8, 8.5, 8.5, 9)

The median of the upper half (Q3) is the 5th term of the upper half.

Upper half data: [7, 7, 7.5, 7.5, 8, 8, 8.5, 8.5, 9]

5th term is 8. So Q3 = 8

Step 1.4: Find Min and Max

Min = 4, Max = 9

Step 2: Draw the box - plot

• Draw a number line that covers the range from 4 to 9.

• Draw a box from Q1 = 6 to Q3 = 8. Inside the box, draw a vertical line at the median (Q2 = 7).

• Draw whiskers from the box to the minimum (4) and maximum (9).

Step1: Determine the range and class intervals

The data set is: 163, 170, 171, 173, 175, 205, 220, 220, 220, 253, 267, 281, 305, 305, 305, 355, 371, 388, 402, 431

First, find the range: \( \text{Range}=\text{Max}-\text{Min}=431 - 163 = 268 \)

We can choose class intervals. Let's choose intervals of width 50 (or another suitable width). Let's try width 50:

• 150 - 200

• 200 - 250

• 250 - 300

• 300 - 350

• 350 - 400

• 400 - 450

Step2: Count the frequency in each interval

• 150 - 200: Data points: 163, 170, 171, 173, 175. Frequency = 5

• 200 - 250: Data points: 205, 220, 220, 220. Frequency = 4

• 250 - 300: Data points: 253, 267, 281. Frequency = 3

• 300 - 350: Data points: 305, 305, 305. Frequency = 3

• 350 - 400: Data points: 355, 371, 388. Frequency = 3

• 400 - 450: Data points: 402, 431. Frequency = 2

Step3: Draw the histogram

• On the x - axis, label the class intervals (150 - 200, 200 - 250, etc.)

• On the y - axis, label the frequency.

• For each class interval, draw a bar with height equal to the frequency of that interval.

Answer:

8

Problem 2: