Question

4 from unit 1, lesson 9 a student has these scores on ten assignments. 0 40 60 70 75 80 85 95 95 100 the teacher is considering dropping a lowest score. what effect does eliminating the lowest value, 0, from the data set have on the mean and median? 5 from unit 1, lesson 9 a. what is the five-number summary for the data? 2 2 4 4 5 5 6 7 9 15 b. when the maximum, 15, is removed from the data set, what is the five-number summary? 6 from unit 1, lesson 4 the box plot summarizes the test scores for 100 students: box plot image which term best describes the shape of the distribution? a bell - shaped b uniform c skewed d symmetric

Explanation:

Problem 4

Step1: Calculate mean with 0

First, find the sum of all scores: \(0 + 40 + 60 + 70 + 75 + 80 + 85 + 95 + 95 + 100 = 600\). There are 10 scores, so the mean is \(\frac{600}{10} = 60\).

To find the median, arrange the scores in order: \(0, 40, 60, 70, 75, 80, 85, 95, 95, 100\). The median is the average of the 5th and 6th terms: \(\frac{75 + 80}{2} = 77.5\).

Step2: Calculate mean without 0

Remove 0, the sum becomes \(600 - 0 = 600\), and the number of scores is 9. So the new mean is \(\frac{600}{9} \approx 66.67\).

For the median, the new ordered list is \(40, 60, 70, 75, 80, 85, 95, 95, 100\). The median is the 5th term, which is 80.

Step3: Analyze the effect

The mean increases (from 60 to ~66.67) because removing a low value (0) that pulled the mean down. The median also increases (from 77.5 to 80) because the middle value shifts when the smallest value is removed.

Step1: Order the data

The data set is \(2, 2, 4, 4, 5, 5, 6, 7, 9, 15\) (already in order).

Step2: Find minimum, Q1, median, Q3, maximum

• Minimum: The smallest value is \(2\).
• Median (Q2): For \(n = 10\) (even), median is the average of the 5th and 6th terms: \(\frac{5 + 5}{2} = 5\).
• Q1: The median of the lower half (\(2, 2, 4, 4, 5\)). For \(n = 5\) (odd), Q1 is the 3rd term: \(4\).
• Q3: The median of the upper half (\(5, 6, 7, 9, 15\)). For \(n = 5\) (odd), Q3 is the 3rd term: \(7\).
• Maximum: The largest value is \(15\).

Step1: Remove 15 and re - order

The new data set (after removing 15) is \(2, 2, 4, 4, 5, 5, 6, 7, 9\) (ordered).

Step2: Find minimum, Q1, median, Q3, maximum

• Minimum: \(2\) (smallest value).
• Median (Q2): For \(n = 9\) (odd), median is the 5th term: \(5\).
• Q1: The median of the lower half (\(2, 2, 4, 4\)). For \(n = 4\) (even), Q1 is the average of the 2nd and 3rd terms: \(\frac{2 + 4}{2}=3\).
• Q3: The median of the upper half (\(5, 6, 7, 9\)). For \(n = 4\) (even), Q3 is the average of the 2nd and 3rd terms: \(\frac{6 + 7}{2}=6.5\).
• Maximum: \(9\) (largest value).

Answer:

Eliminating the lowest value (0) increases both the mean and the median. The mean increases from 60 to approximately 66.67, and the median increases from 77.5 to 80.

Problem 5a