Question

1. final grades of irma in the major subjects in the first quarter

90 89 88 96 92 90

1. the first quarter scores of norman in the mathematics tests

82 82 75 78 96 92 91 96

Explanation:

Based on the image provided, here are the step-by-step calculations for the mean, median, and mode of both datasets.

⚡ Using what you learned: comparing measures of center & spread

Step 1: Analyze Irma's Grades (Problem 6)

Dataset: \( 90, 89, 88, 96, 92, 90 \) (Number of values, \( n = 6 \))

• Mean:

\[
\text{Mean} = \frac{90 + 89 + 88 + 96 + 92 + 90}{6} = \frac{545}{6} \approx 90.83
\]

• Median:

Arrange in ascending order: \( 88, 89, 90, 90, 92, 96 \)
Since \( n \) is even, take the average of the two middle values (3rd and 4th):
\[
\text{Median} = \frac{90 + 90}{2} = 90
\]

• Mode:

The value that appears most frequently is \( 90 \) (appears twice).

---

Step 2: Analyze Norman's Scores (Problem 7)

Dataset: \( 82, 82, 75, 78, 96, 92, 91, 96 \) (Number of values, \( n = 8 \))

• Mean:

\[
\text{Mean} = \frac{82 + 82 + 75 + 78 + 96 + 92 + 91 + 96}{8} = \frac{692}{8} = 86.5
\]

• Median:

Arrange in ascending order: \( 75, 78, 82, 82, 91, 92, 96, 96 \)
Since \( n \) is even, take the average of the two middle values (4th and 5th):
\[
\text{Median} = \frac{82 + 91}{2} = 86.5
\]

• Mode:

The values that appear most frequently are \( 82 \) and \( 96 \) (each appears twice). This dataset is bimodal.

Answer:

6. Irma's Grades:

• Mean: \( 90.83 \)
• Median: \( 90 \)
• Mode: \( 90 \)

7. Norman's Scores:

• Mean: \( 86.5 \)
• Median: \( 86.5 \)
• Mode: \( 82 \) and \( 96 \)