Question

difficulty scores in women’s gymnastics (all-around final) 5, 9, 4, 10, 4, 8, 9, 6, 8 ascending order: 3,4,5,6,7,8,8,9,9,10 lower extreme: __ lower quartile: median: upper quartile: upper extreme: __

Explanation:

First, we need to sort the data set. The given data points are 5, 9, 4, 10, 4, 8, 9, 6, 8. Let's sort them in ascending order:

Step 1: Sort the data

First, list out all the numbers: 4, 4, 5, 6, 8, 8, 9, 9, 10. Wait, wait, let's check again. Wait the original data (from the image) seems to have numbers: 5, 9, 4, 10, 4, 8, 9, 6, 8? Wait no, the handwritten ascending order is 3,4,5,6,7,8,8,9,10? Wait maybe I misread. Wait the image shows "Ascending order: 3,4,5,6,7,8,8,9,10"? Wait no, the first part: "Difficulty Scores in Women’s Gymnastics (All - Around Final)" with numbers: 5, 9, 4, 10, 4, 8, 9, 6, 8? Wait the handwritten ascending order is boxed as 3,4,5,6,7,8,8,9,10? Wait maybe a typo, but let's assume the sorted data is \( 3, 4, 5, 6, 7, 8, 8, 9, 10 \) (maybe the original data had a 3). Let's confirm the number of data points. Let's count: 3,4,5,6,7,8,8,9,10 – that's 9 data points.

Step 2: Find the Lower Extreme (Minimum)

The minimum value in the sorted data is the first element. So Lower Extreme \( = 3 \).

Step 3: Find the Lower Quartile (Q1)

For a data set with \( n = 9 \) (odd number of observations), the position of the median is \( \frac{n + 1}{2}=\frac{9+1}{2}=5 \)-th term. So the median is the 5th term. Then, the lower half of the data (excluding the median) is the first 4 terms: \( 3, 4, 5, 6 \). The lower quartile (Q1) is the median of the lower half. Since there are 4 terms (even), the median of the lower half is the average of the 2nd and 3rd terms? Wait no, for \( n = 9 \), the lower half is the first \( \frac{n - 1}{2}=4 \) terms? Wait, actually, the formula for quartiles: for a data set with \( n \) observations, sorted in ascending order.

- If \( n \) is odd:
- Median (Q2) is at position \( \frac{n + 1}{2} \)
- Q1 is the median of the first \( \frac{n - 1}{2} \) observations
- Q3 is the median of the last \( \frac{n - 1}{2} \) observations

So here, \( n = 9 \), so \( \frac{n - 1}{2}=4 \). So first 4 observations: \( 3, 4, 5, 6 \). The median of these 4 (even number) is the average of the 2nd and 3rd terms: \( \frac{4 + 5}{2}=4.5 \)? Wait no, wait the sorted data: let's list them properly. Wait maybe the original data is 4,4,5,6,8,8,9,9,10 (if the 3 was a mistake). Let's check the handwritten ascending order: "3,4,5,6,7,8,8,9,10" – so 9 numbers. Let's proceed with that.

So sorted data: \( 3, 4, 5, 6, 7, 8, 8, 9, 10 \)

- Number of observations \( n = 9 \)
- Median (Q2) position: \( \frac{9 + 1}{2}=5 \)-th term. So median is the 5th term: \( 7 \)
- Lower half: first 4 terms: \( 3, 4, 5, 6 \)
- Q1: median of lower half. Since 4 terms, average of 2nd and 3rd: \( \frac{4 + 5}{2}=4.5 \)? Wait no, in some methods, for odd \( n \), Q1 is the median of the first \( \frac{n - 1}{2} \) terms, which is 4 terms here. So median of 4 terms is the average of the 2nd and 3rd. So \( (4 + 5)/2 = 4.5 \)
- Upper half: last 4 terms: \( 8, 8, 9, 10 \)
- Q3: median of upper half. Average of 2nd and 3rd terms: \( (8 + 9)/2 = 8.5 \)
- Upper Extreme (Maximum): last term, \( 10 \)

Wait but maybe the original data was 4,4,5,6,8,8,9,9,10 (without the 3 and 7). Let's check the initial numbers: "5,9,4,10,4,8,9,6,8" – let's sort these: 4,4,5,6,8,8,9,9,10 (9 numbers). Then:

- Median (Q2): position 5, term is 8
- Lower half: first 4 terms: 4,4,5,6. Q1: median of these 4: (4 + 5)/2 = 4.5
- Upper half: last 4 terms: 8,9,9,10. Q3: (9 + 9)/2 = 9
- Lower Extreme: 4
- Upper Extreme: 10

But the handwritten ascending order is "3,4,5,6,7,8,8,9,10", so maybe the original data had 3 and 7. Let's go with the handwritten ascending order: \( 3, 4, 5, 6, 7, 8, 8, 9, 10 \)

So:

- Lower Extreme (Minimum): \( 3 \)
- Lower Quartile (Q1): median of first 4 terms (3,4,5,6). Since 4 terms, Q1 = (4 + 5)/2 = 4.5
- Median (Q2): 5th term, \( 7 \)
- Upper Quartile (Q3): median of last 4 terms (8,8,9,10). (8 + 9)/2 = 8.5
- Upper Extreme (Maximum): \( 10 \)

But maybe the problem expects using the data as per the handwritten ascending order. Let's confirm:

Sorted data: 3,4,5,6,7,8,8,9,10 (n=9)

1. Lower Extreme: minimum value = 3
2. Lower Quartile (Q1): The lower half is the first 4 numbers (3,4,5,6). The median of these 4 is (4 + 5)/2 = 4.5
3. Median: The middle number (5th term) is 7
4. Upper Quartile (Q3): The upper half is the last 4 numbers (8,8,9,10). The median of these 4 is (8 + 9)/2 = 8.5
5. Upper Extreme: maximum value = 10

Answer:

• Lower Extreme: \( 3 \)
• Lower Quartile: \( 4.5 \)
• Median: \( 7 \)
• Upper Quartile: \( 8.5 \)
• Upper Extreme: \( 10 \)

(Note: If the original data was 4,4,5,6,8,8,9,9,10, the answers would be: Lower Extreme: 4, Lower Quartile: 4.5, Median: 8, Upper Quartile: 9, Upper Extreme: 10. But based on the handwritten ascending order, we used 3,4,5,6,7,8,8,9,10.)