Question

name handwriting, date ???, period

1. for a science project, a student plants seeds of two different varieties of corn plants. the table displays the heights, in inches, of each of the corn plants after 8 weeks.

table: variety a: 46, 46, 47, 48, 49, 49, 50, 46, 48, 49; variety b: 48, 46, 45, 47, 46, 48, 49, 47, 48, 48
a. create a dot plot for both varieties of corn.
b. create a box plot for both varieties of corn.
c. find the iqr for both varieties.
variety a iqr: variety b iqr:
d. is the median the same for both varieties?
e. is the mean the same for both varieties?

Explanation:

Part (a): Dot Plot Creation

To create a dot plot, we first list the unique values (heights) and count the frequency of each value for both varieties.

Variety A Heights: \( 46, 46, 47, 48, 49, 49, 50, 46, 48, 49 \)

• Frequency of \( 46 \): \( 3 \)
• Frequency of \( 47 \): \( 1 \)
• Frequency of \( 48 \): \( 2 \)
• Frequency of \( 49 \): \( 3 \)
• Frequency of \( 50 \): \( 1 \)

Variety B Heights: \( 48, 46, 45, 47, 46, 48, 49, 47, 48, 48 \)

• Frequency of \( 45 \): \( 1 \)
• Frequency of \( 46 \): \( 2 \)
• Frequency of \( 47 \): \( 2 \)
• Frequency of \( 48 \): \( 4 \)
• Frequency of \( 49 \): \( 1 \)

For the dot plot, we draw a number line with values from \( 45 \) to \( 50 \). Above each value, we place dots equal to the frequency. For example, above \( 46 \) in Variety A, we place 3 dots, above \( 47 \) we place 1 dot, etc. For Variety B, above \( 45 \) we place 1 dot, above \( 46 \) we place 2 dots, etc.

Part (b): Box Plot Creation

To create a box plot, we need the minimum, first quartile (\( Q_1 \)), median (\( Q_2 \)), third quartile (\( Q_3 \)), and maximum for each variety.

Variety A:

1. Order the data: \( 46, 46, 46, 47, 48, 48, 49, 49, 49, 50 \) (n = 10, even)
2. Minimum: \( 46 \)
3. Median (\( Q_2 \)): Average of the 5th and 6th values: \( \frac{48 + 48}{2} = 48 \)
4. Lower half (first 5 values): \( 46, 46, 46, 47, 48 \)

• \( Q_1 \): Median of lower half (3rd value): \( 46 \)

1. Upper half (last 5 values): \( 48, 49, 49, 49, 50 \)

• \( Q_3 \): Median of upper half (3rd value): \( 49 \)

1. Maximum: \( 50 \)

Variety B:

1. Order the data: \( 45, 46, 46, 47, 47, 48, 48, 48, 48, 49 \) (n = 10, even)
2. Minimum: \( 45 \)
3. Median (\( Q_2 \)): Average of the 5th and 6th values: \( \frac{47 + 48}{2} = 47.5 \)
4. Lower half (first 5 values): \( 45, 46, 46, 47, 47 \)

• \( Q_1 \): Median of lower half (3rd value): \( 46 \)

1. Upper half (last 5 values): \( 48, 48, 48, 48, 49 \)

• \( Q_3 \): Median of upper half (3rd value): \( 48 \)

1. Maximum: \( 49 \)

For the box plot, we draw a number line. For each variety, we draw a box from \( Q_1 \) to \( Q_3 \), a line inside the box at the median, and whiskers from the minimum to \( Q_1 \) and from \( Q_3 \) to the maximum.

Part (c): IQR Calculation

IQR is calculated as \( Q_3 - Q_1 \).

Variety A:

\( IQR = Q_3 - Q_1 = 49 - 46 = 3 \)

Variety B:

\( IQR = Q_3 - Q_1 = 48 - 46 = 2 \)

Part (d): Median Comparison

• Variety A median: \( 48 \)
• Variety B median: \( 47.5 \)

Since \( 48
eq 47.5 \), the medians are not the same.

Part (e): Mean Calculation

Mean is calculated as \( \frac{\text{Sum of values}}{\text{Number of values}} \).

Variety A:

Sum = \( 46 + 46 + 47 + 48 + 49 + 49 + 50 + 46 + 48 + 49 \)
= \( 46 \times 3 + 47 + 48 \times 2 + 49 \times 3 + 50 \)
= \( 138 + 47 + 96 + 147 + 50 = 478 \)
Mean = \( \frac{478}{10} = 47.8 \)

Variety B:

Sum = \( 48 + 46 + 45 + 47 + 46 + 48 + 49 + 47 + 48 + 48 \)
= \( 45 + 46 \times 2 + 47 \times 2 + 48 \times 4 + 49 \)
= \( 45 + 92 + 94 + 192 + 49 = 472 \)
Mean = \( \frac{472}{10} = 47.2 \)

Since \( 47.8
eq 47.2 \), the means are not the same.

Final Answers:

(c)

• Variety A IQR: \( \boldsymbol{3} \)
• Variety B IQR: \( \boldsymbol{2} \)

(d)

No, the medians are not the same.

(e)

No, the means are not the same.

Answer:

Part (a): Dot Plot Creation

To create a dot plot, we first list the unique values (heights) and count the frequency of each value for both varieties.

Variety A Heights: \( 46, 46, 47, 48, 49, 49, 50, 46, 48, 49 \)

• Frequency of \( 46 \): \( 3 \)
• Frequency of \( 47 \): \( 1 \)
• Frequency of \( 48 \): \( 2 \)
• Frequency of \( 49 \): \( 3 \)
• Frequency of \( 50 \): \( 1 \)

Variety B Heights: \( 48, 46, 45, 47, 46, 48, 49, 47, 48, 48 \)

• Frequency of \( 45 \): \( 1 \)
• Frequency of \( 46 \): \( 2 \)
• Frequency of \( 47 \): \( 2 \)
• Frequency of \( 48 \): \( 4 \)
• Frequency of \( 49 \): \( 1 \)

For the dot plot, we draw a number line with values from \( 45 \) to \( 50 \). Above each value, we place dots equal to the frequency. For example, above \( 46 \) in Variety A, we place 3 dots, above \( 47 \) we place 1 dot, etc. For Variety B, above \( 45 \) we place 1 dot, above \( 46 \) we place 2 dots, etc.

Part (b): Box Plot Creation

To create a box plot, we need the minimum, first quartile (\( Q_1 \)), median (\( Q_2 \)), third quartile (\( Q_3 \)), and maximum for each variety.

Variety A:

1. Order the data: \( 46, 46, 46, 47, 48, 48, 49, 49, 49, 50 \) (n = 10, even)
2. Minimum: \( 46 \)
3. Median (\( Q_2 \)): Average of the 5th and 6th values: \( \frac{48 + 48}{2} = 48 \)
4. Lower half (first 5 values): \( 46, 46, 46, 47, 48 \)

• \( Q_1 \): Median of lower half (3rd value): \( 46 \)

1. Upper half (last 5 values): \( 48, 49, 49, 49, 50 \)

• \( Q_3 \): Median of upper half (3rd value): \( 49 \)

1. Maximum: \( 50 \)

Variety B:

1. Order the data: \( 45, 46, 46, 47, 47, 48, 48, 48, 48, 49 \) (n = 10, even)
2. Minimum: \( 45 \)
3. Median (\( Q_2 \)): Average of the 5th and 6th values: \( \frac{47 + 48}{2} = 47.5 \)
4. Lower half (first 5 values): \( 45, 46, 46, 47, 47 \)

• \( Q_1 \): Median of lower half (3rd value): \( 46 \)

1. Upper half (last 5 values): \( 48, 48, 48, 48, 49 \)

• \( Q_3 \): Median of upper half (3rd value): \( 48 \)

1. Maximum: \( 49 \)

For the box plot, we draw a number line. For each variety, we draw a box from \( Q_1 \) to \( Q_3 \), a line inside the box at the median, and whiskers from the minimum to \( Q_1 \) and from \( Q_3 \) to the maximum.

Part (c): IQR Calculation

IQR is calculated as \( Q_3 - Q_1 \).

Variety A:

\( IQR = Q_3 - Q_1 = 49 - 46 = 3 \)

Variety B:

\( IQR = Q_3 - Q_1 = 48 - 46 = 2 \)

Part (d): Median Comparison

• Variety A median: \( 48 \)
• Variety B median: \( 47.5 \)

Since \( 48
eq 47.5 \), the medians are not the same.

Part (e): Mean Calculation

Mean is calculated as \( \frac{\text{Sum of values}}{\text{Number of values}} \).

Variety A:

Sum = \( 46 + 46 + 47 + 48 + 49 + 49 + 50 + 46 + 48 + 49 \)
= \( 46 \times 3 + 47 + 48 \times 2 + 49 \times 3 + 50 \)
= \( 138 + 47 + 96 + 147 + 50 = 478 \)
Mean = \( \frac{478}{10} = 47.8 \)

Variety B:

Sum = \( 48 + 46 + 45 + 47 + 46 + 48 + 49 + 47 + 48 + 48 \)
= \( 45 + 46 \times 2 + 47 \times 2 + 48 \times 4 + 49 \)
= \( 45 + 92 + 94 + 192 + 49 = 472 \)
Mean = \( \frac{472}{10} = 47.2 \)

Since \( 47.8
eq 47.2 \), the means are not the same.

Final Answers:

(c)

• Variety A IQR: \( \boldsymbol{3} \)
• Variety B IQR: \( \boldsymbol{2} \)

(d)

No, the medians are not the same.

(e)

No, the means are not the same.