Question

this histogram shows the distribution of exam scores for a class of 33 students. which of the following is the most reasonable estimate for the median?
○ 50
○ 75
○ 85

Explanation:

Step1: Determine the median position

For \( n = 33 \) (odd number of observations), the median is the \( \frac{n + 1}{2}=\frac{33+ 1}{2}=17^{\text{th}} \) value when the data is ordered.

Step2: Cumulate the frequencies

- First bar (0 - 20?): Frequency \( f_1 = 1 \) (approx, from the graph)
- Second bar (20 - 40?): Frequency \( f_2 = 1 \) (approx)
- Third bar (40 - 60?): Frequency \( f_3 = 0 \) (no bar)
- Fourth bar (60 - 80?): Wait, let's re - examine the x - axis. Let's assume the intervals: 0 - 20, 20 - 40, 40 - 60, 60 - 80, 80 - 100? Wait, no, looking at the graph, the bars are at 0 - 20 (height ~1), 20 - 40 (height ~1), then a gap, then 50 - 60? Wait, maybe the intervals are 0 - 20, 20 - 40, 40 - 60, 60 - 80, 80 - 100? Wait, the first bar (0 - 20): frequency 1, second (20 - 40): frequency 1, then 40 - 60: 0, then 60 - 80? Wait, no, the next bar starts around 50 - 60 with frequency 3 (approx), then 60 - 80: frequency 6, then 80 - 100? Wait, no, the bars are:

First bar (0 - 20): height 1 (frequency 1)

Second bar (20 - 40): height 1 (frequency 1)

Third bar (50 - 60?): height 3 (frequency 3)

Fourth bar (60 - 70? 60 - 80?): height 6 (frequency 6)

Fifth bar (70 - 80? No, next bar height 8 (frequency 8))

Sixth bar (80 - 90? Height 10 (frequency 10))

Seventh bar (90 - 100? Height 4 (frequency 4))

Wait, let's sum the frequencies: \( 1+1 + 3+6+8+10+4=33 \), which matches the total number of students.

Now, let's cumulate the frequencies:

- After 0 - 20: \( 1 \)
- After 20 - 40: \( 1 + 1=2 \)
- After 50 - 60: \( 2+3 = 5 \)
- After 60 - 70? Wait, maybe the intervals are 0 - 20, 20 - 40, 40 - 60, 60 - 80, 80 - 100? No, the x - axis has marks at 0, 20, 40, 60, 80, 100. Let's assume the intervals are [0,20), [20,40), [40,60), [60,80), [80,100).

Wait, the first bar (0 - 20): frequency 1

Second bar (20 - 40): frequency 1

Third bar (40 - 60): frequency 0 (no bar)

Fourth bar (60 - 80): frequency \( 6 + 8=14 \)? No, looking at the graph, the bars are:

First bar (0 - 20): height 1 (frequency 1)

Second bar (20 - 40): height 1 (frequency 1)

Third bar (50 - 60?): height 3 (frequency 3)

Fourth bar (60 - 70?): height 6 (frequency 6)

Fifth bar (70 - 80?): height 8 (frequency 8)

Sixth bar (80 - 90?): height 10 (frequency 10)

Seventh bar (90 - 100?): height 4 (frequency 4)

Cumulative frequencies:

- 0 - 20: 1
- 20 - 40: \( 1+1 = 2 \)
- 40 - 60: \( 2 + 0=2 \) (wait, no, there is a bar between 40 - 60? No, the graph shows a gap between 20 - 40 and the next bar which starts around 50. Maybe the intervals are 0 - 20, 20 - 40, 40 - 60, 60 - 80, 80 - 100 with frequencies:

0 - 20: 1

20 - 40: 1

40 - 60: 3 (the bar with height 3)

60 - 80: 6 + 8=14? No, the bar with height 6 is between 60 - 70, height 8 between 70 - 80? Wait, maybe the correct way is to find the cumulative frequency until we reach or exceed the 17th value.

Let's list the frequencies:

- Interval 0 - 20: frequency \( f_1 = 1 \), cumulative \( c_1=1 \)
- Interval 20 - 40: frequency \( f_2 = 1 \), cumulative \( c_2=1 + 1=2 \)
- Interval 40 - 60: frequency \( f_3 = 3 \), cumulative \( c_3=2+3 = 5 \)
- Interval 60 - 80: Wait, no, the next bar has height 6 (frequency 6), cumulative \( c_4=5 + 6=11 \)
- Interval 80 - 90: height 10 (frequency 10), cumulative \( c_5=11+10 = 21 \)
- Interval 90 - 100: height 4 (frequency 4), cumulative \( c_6=21 + 4=25 \)? No, this doesn't add up to 33. Wait, I think I misread the frequencies. Let's look again:

The y - axis is frequency, with marks at 0,2,4,6,8,10,12.

First bar (0 - 20): height ~1 (frequency 1)

Second bar (20 - 40): height ~1 (frequency 1)

Third bar (50 - 60?): height ~3 (frequency 3)

Fourth bar (60 - 70?): height ~6 (frequency 6)

Fifth bar (70 - 80?): height ~8 (frequency 8)

Sixth bar (80 - 90?): height ~10 (frequency 10)

Seventh bar (90 - 100?): height ~4 (frequency 4)

Now sum: \( 1+1+3 + 6+8+10+4=33 \), correct.

Now cumulative frequencies:

- After 0 - 20: 1
- After 20 - 40: \( 1 + 1=2 \)
- After 50 - 60: \( 2+3 = 5 \)
- After 60 - 70: \( 5+6 = 11 \)
- After 70 - 80: \( 11+8 = 19 \)

Ah, here we go. After the 70 - 80 interval, the cumulative frequency is 19, which is greater than 17. The previous cumulative frequency (after 60 - 70) is 11, which is less than 17. So the median lies in the 70 - 80 interval? Wait, no, maybe the intervals are different. Wait, the x - axis has 0,20,40,60,80,100. Maybe the intervals are [0,20), [20,40), [40,60), [60,80), [80,100).

Wait, let's re - calculate with intervals [0,20), [20,40), [40,60), [60,80), [80,100):

- [0,20): frequency \( f_1 = 1 \), cumulative \( c_1 = 1 \)
- [20,40): frequency \( f_2 = 1 \), cumulative \( c_2=1 + 1=2 \)
- [40,60): frequency \( f_3 = 3 \), cumulative \( c_3=2+3 = 5 \)
- [60,80): frequency \( f_4=6 + 8=14 \)? No, the bar between 60 - 80: the first part (60 - 70) has height 6, (70 - 80) has height 8. So [60,80) has frequency \( 6 + 8 = 14 \), cumulative \( c_4=5+14 = 19 \)
- [80,100): frequency \( f_5=10 + 4=14 \)? No, [80,100) has frequency \( 10+4 = 14 \), cumulative \( c_5=19+14 = 33 \)

Now, the median is the 17th value. Since \( c_3 = 5<17 \) and \( c_4=19\geq17 \), the median is in the [60,80) interval? Wait, no, the options are 50,75,85. Wait, maybe my interval assumption is wrong. Let's look at the x - axis labels: 0,20,40,60,80,100. The bars:

First bar: 0 - 20, height 1

Second bar: 20 - 40, height 1

Third bar: 40 - 60, height 0 (no bar)

Fourth bar: 60 - 80, height 6 + 8=14? No, the bar with height 6 is at 60 - 70, height 8 at 70 - 80, height 10 at 80 - 90, height 4 at 90 - 100.

Wait, maybe the correct way is to list the cumulative frequencies until we reach the middle. For \( n = 33 \), the median is the 17th term.

Let's add the frequencies step by step:

- First bar (0 - 20): 1 (total 1)
- Second bar (20 - 40): 1 (total 2)
- Third bar (50 - 60): 3 (total 5)
- Fourth bar (60 - 70): 6 (total 11)
- Fifth bar (70 - 80): 8 (total 19)

Ah, when we add the fifth bar (70 - 80) with frequency 8, the total becomes 19, which is greater than 17. So the 17th term is in the 70 - 80 interval. The mid - point of 70 - 80 is 75. So the most reasonable estimate for the median is 75.

Answer:

75