Question

a 10 - item statistics quiz was given to 30 students. the table below gives the scores received along with the corresponding frequencies. what was the median score on the quiz? 7.5 8.5 9 10

Explanation:

Step1: Determine the position of the median

For \( n = 30 \) (even number of observations), the median is the average of the \( \frac{n}{2} \)-th and \( (\frac{n}{2}+ 1) \)-th values. So, \( \frac{30}{2}=15 \)-th and \( 15 + 1=16 \)-th values.

Step2: Calculate cumulative frequencies

- For score 5: Cumulative frequency \( = 1 \)
- For score 6: Cumulative frequency \( = 1+2 = 3 \)
- For score 7: Cumulative frequency \( = 3 + 5=8 \)
- For score 8: Cumulative frequency \( = 8+5 = 13 \)
- For score 9: Cumulative frequency \( = 13+7 = 20 \)
- For score 10: Cumulative frequency \( = 20 + 10=30 \)

Step3: Find the 15th and 16th values

The 13th value is the last value with score 8. The 14th, 15th, 16th, ..., 20th values have score 9 (since cumulative frequency for 9 is 20). So the 15th and 16th values are both 9? Wait, no, wait. Wait, cumulative frequency for 8 is 13, so the 14th value starts at score 9. Wait, let's re - check:

Wait, cumulative frequency:

- Score 5: 1 (1st value)
- Score 6: 3 (1st - 3rd values)
- Score 7: 8 (1st - 8th values)
- Score 8: 13 (1st - 13th values)
- Score 9: 20 (14th - 20th values)
- Score 10: 30 (21st - 30th values)

So the 15th value is in the score 9 group, and the 16th value is also in the score 9 group? Wait, no, the median for even \( n \) is the average of the \( \frac{n}{2} \)-th (\( 15 \)-th) and \( (\frac{n}{2}+1) \)-th (\( 16 \)-th) values. Since both the 15th and 16th values are 9? Wait, no, wait, maybe I made a mistake. Wait, the cumulative frequency for 8 is 13, so the 14th value is the first value of score 9, the 15th is the second value of score 9, the 16th is the third value of score 9? Wait, no, the frequency of score 9 is 7, so the values from 14th to 20th (inclusive) are score 9. So the 15th and 16th values are both 9? Wait, no, the median is the average of the 15th and 16th values. Wait, but let's check again. Wait, maybe I messed up the cumulative frequency.

Wait, let's list the positions:

- Score 5: positions 1
- Score 6: positions 2 - 3 (2 values)
- Score 7: positions 4 - 8 (5 values)
- Score 8: positions 9 - 13 (5 values)
- Score 9: positions 14 - 20 (7 values)
- Score 10: positions 21 - 30 (10 values)

Ah, here is the mistake earlier. The first value of score 5 is position 1, score 6: positions 2 and 3 (2 values), so cumulative frequency for 6 is 1 + 2=3. Score 7: 3+5 = 8, so positions 4 - 8 (5 values). Score 8: 8 + 5=13, so positions 9 - 13 (5 values). Score 9: 13+7 = 20, so positions 14 - 20 (7 values). Score 10: 20+10 = 30, positions 21 - 30 (10 values).

So the 15th value is in position 15, which is within the score 9 group (positions 14 - 20), and the 16th value is also in the score 9 group. Wait, but the median is the average of the 15th and 16th values. But both are 9? Wait, no, that can't be. Wait, maybe the question has a typo? Wait, no, wait, maybe I made a mistake in the cumulative frequency calculation.

Wait, let's recalculate cumulative frequencies:

- Score 5: frequency = 1, cumulative = 1
- Score 6: frequency = 2, cumulative = 1+2 = 3
- Score 7: frequency = 5, cumulative = 3 + 5=8
- Score 8: frequency = 5, cumulative = 8+5 = 13
- Score 9: frequency = 7, cumulative = 13+7 = 20
- Score 10: frequency = 10, cumulative = 20+10 = 30

Now, the median is the average of the 15th and 16th terms.

The 13th term is the last term with score 8. So the 14th term is the first term with score 9, the 15th term is the second term with score 9, the 16th term is the third term with score 9? Wait, no, the frequency of score 9 is 7, so there are 7 terms with score 9 (positions 14 - 20). So the 15th and 16th terms are both 9? Wait, but that would make the median 9. But wait, maybe I made a mistake. Wait, let's check the answer options. The options are 7.5, 8.5, 9, 10.

Wait, maybe I messed up the position. Wait, for \( n = 30 \), the median is the average of the \( \frac{n}{2}=15 \)-th and \( \frac{n}{2}+ 1 = 16 \)-th observations.

Looking at cumulative frequencies:

- Up to score 8: 13 observations. So the 14th observation is the first one with score 9, the 15th is the second one with score 9, the 16th is the third one with score 9? Wait, no, the number of observations with score 9 is 7. So the observations with score 9 are from observation 14 to observation 14 + 7-1=20. So observation 14,15,16,17,18,19,20 have score 9.

So the 15th observation is 9, the 16th observation is 9. Then the median is \( \frac{9 + 9}{2}=9 \)? But that's one of the options. Wait, but maybe I made a mistake. Wait, let's check again.

Wait, cumulative frequency:

Score 5: 1

Score 6: 1+2 = 3

Score 7: 3+5 = 8

Score 8: 8+5 = 13

Score 9: 13+7 = 20

Score 10: 20+10 = 30

So the 13th observation is the last one with score 8. The 14th observation is the first one with score 9, the 15th is the second one with score 9, the 16th is the third one with score 9, ..., the 20th is the seventh one with score 9.

So the 15th and 16th observations are both 9. So the median is \( \frac{9 + 9}{2}=9 \).

Answer:

9