Question

given the data below, determine the correct statement comparing the mean and the median.
distribution of exam scores
(the bar graph has score on the x - axis with intervals 1 - 10, 11 - 20, 21 - 30, 31 - 40, 41 - 50, 51 - 60, 61 - 70, 71 - 80, 81 - 90, 91 - 100 and frequency on the y - axis from 0 to 18. the frequencies for each interval are: 1 - 10: ~1, 11 - 20: ~1, 21 - 30: ~2, 31 - 40: ~4, 41 - 50: 10, 51 - 60: ~13, 61 - 70: 16, 71 - 80: ~15, 81 - 90: 10, 91 - 100: ~3)
\\(\circ\\) mean = median
\\(\circ\\) mean < median
\\(\circ\\) mean > median

Explanation:

Step1: Analyze the Histogram Shape

The histogram of exam scores is symmetric? Wait, no, let's check the frequencies. Wait, actually, the distribution looks symmetric? Wait, no, let's calculate cumulative frequency. Wait, first, let's list the intervals and frequencies:

1 - 10: ~1 (frequency)
11 - 20: ~1
21 - 30: ~2
31 - 40: ~4
41 - 50: 10
51 - 60: 13
61 - 70: 16
71 - 80: 15
81 - 90: 10
91 - 100: 3

Wait, no, maybe my initial count is wrong. Wait, the y-axis is frequency. Let's count the bars:

1-10: height ~1
11-20: ~1
21-30: ~2
31-40: ~4
41-50: 10
51-60: 13 (since between 12 and 14)
61-70: 16 (top)
71-80: 15 (below 16)
81-90: 10
91-100: ~3

Now, cumulative frequency:

1-10: 1, total so far 1
11-20: 1, total 2
21-30: 2, total 4
31-40: 4, total 8
41-50: 10, total 18
51-60: 13, total 31
61-70: 16, total 47
71-80: 15, total 62
81-90: 10, total 72
91-100: 3, total 75

Total number of data points: 75. Median is the 38th value (since 75/2=37.5, so 38th). Let's find the interval of the median. Cumulative up to 51-60: 31. Then 61-70: 31 +16=47. So 38th is in 61-70.

Now, mean: we need to calculate the midpoint of each interval, multiply by frequency, sum, then divide by 75.

Midpoints:

1-10: 5.5, frequency 1: 5.5*1=5.5
11-20: 15.5, frequency 1: 15.5*1=15.5
21-30: 25.5, frequency 2: 25.5*2=51
31-40: 35.5, frequency 4: 35.5*4=142
41-50: 45.5, frequency 10: 45.5*10=455
51-60: 55.5, frequency 13: 55.5*13=721.5
61-70: 65.5, frequency 16: 65.5*16=1048
71-80: 75.5, frequency 15: 75.5*15=1132.5
81-90: 85.5, frequency 10: 85.5*10=855
91-100: 95.5, frequency 3: 95.5*3=286.5

Now sum all these: 5.5 +15.5=21; +51=72; +142=214; +455=669; +721.5=1390.5; +1048=2438.5; +1132.5=3571; +855=4426; +286.5=4712.5

Mean = 4712.5 /75 ≈ 62.83

Median: 38th value. Cumulative up to 51-60: 31. So we need 38-31=7th value in 61-70. The midpoint of 61-70 is 65.5, but the median position is in 61-70. Wait, but the distribution: let's check symmetry. Wait, the left side (lower scores) has lower frequencies, and the right side (higher scores) after 61-70: 71-80 (15), 81-90 (10), 91-100 (3). Wait, actually, the left tail (lower scores) is longer? Wait no, the left side (1-10,11-20,21-30,31-40) has low frequencies, and the right side after 61-70: 71-80 is 15, 81-90 is 10, 91-100 is 3. Wait, maybe the distribution is symmetric? Wait, 41-50 (10), 51-60 (13), 61-70 (16), 71-80 (15), 81-90 (10), 91-100 (3). Wait, 41-50 and 81-90 both 10; 51-60 (13) and 71-80 (15) close; 61-70 (16) is peak. So maybe symmetric around 61-70? Wait, but the left tail (1-40) has lower frequencies. Wait, but when we calculated mean ≈62.83, median: let's recalculate cumulative frequency correctly.

Wait, maybe my initial frequency count was wrong. Let's look at the bar heights:

1-10: height ~1 (y=1)
11-20: ~1 (y=1)
21-30: ~2 (y=2)
31-40: ~4 (y=4)
41-50: 10 (y=10)
51-60: 13 (y=13, between 12 and 14)
61-70: 16 (y=16)
71-80: 15 (y=15, between 14 and 16)
81-90: 10 (y=10)
91-100: 3 (y=3, between 2 and 4)

So cumulative:

1-10:1, total 1
11-20:1, total 2
21-30:2, total 4
31-40:4, total 8
41-50:10, total 18
51-60:13, total 31
61-70:16, total 47
71-80:15, total 62
81-90:10, total 72
91-100:3, total 75

So median is at 38th term, which is in 61-70 (since 31 <38 ≤47). The median in a histogram is the value where cumulative frequency is n/2. So median interval is 61-70. The median can be approximated, but let's see the mean: ~62.83, and the median is in 61-70. Wait, but 61-70 midpoint is 65.5? No, 61-70 is 61 to 70, so midpoint is 65.5. But the mean is ~62.83, which is less than 65.5? Wait, no, wait my calculation of mean: 4712.5 /75 ≈62.83. Wait, that's in 61-70? 61-70 is 61 to 70, so 62.83 is in that interval. Wait, but the median is also in 61-70. Wait, but maybe I made a mistake in frequency counts. Wait, maybe the bar for 1-10 is 1, 11-20 is 1, 21-30 is 2, 31-40 is 4, 41-50 is 10, 51-60 is 13, 61-70 is 16, 71-80 is 15, 81-90 is 10, 91-100 is 3. That seems correct.

Wait, but the key is: in a symmetric distribution, mean = median. In a left-skewed (tail on left), mean < median. In right-skewed (tail on right), mean > median. Wait, but here, the left tail (lower scores) has lower frequencies, so the tail is on the left? Wait no, left tail is lower scores, so if left tail is longer (more low scores), it's left-skewed, mean < median. But in our case, the left side (1-40) has low frequencies, so the tail is on the left (lower scores), so left-skewed? But our mean was ~62.83, median in 61-70. Wait, maybe my frequency count is wrong. Wait, maybe the bar for 31-40 is 4, 41-50 is 10, 51-60 is 13, 61-70 is 16, 71-80 is 15, 81-90 is 10, 91-100 is 3. The left side (1-30) has 1+1+2=4, and 31-40 is 4, so 1-40: 8. Then 41-100: 10+13+16+15+10+3=67. Wait, no, total should be 8+67=75, correct.

Wait, maybe the distribution is symmetric around 61-70. Let's check the frequencies:

41-50:10, 81-90:10

51-60:13, 71-80:15 (close)

61-70:16 (peak)

Left of peak (1-60): 1+1+2+4+10+13=31

Right of peak (71-100):15+10+3=28

Wait, not symmetric. But the mean is ~62.83, median is 38th term. Let's calculate median position: n=75, median is (75+1)/2=38th term (since median is the middle value, for odd n, it's (n+1)/2 th term). So cumulative frequency:

1-10:1 (1)

11-20:1 (2)

21-30:2 (4)

31-40:4 (8)

41-50:10 (18)

51-60:13 (31)

61-70:16 (47) → 38th term is in 61-70.

Now, to find the median, we use the formula for grouped data:

Median = L + ((n/2 - CF)/f) * w

Where L is lower limit of median class (61), n/2=37.5, CF is cumulative frequency before median class (31), f is frequency of median class (16), w is class width (10).

So median = 61 + ((37.5 -31)/16)*10 = 61 + (6.5/16)*10 ≈61 +4.0625=65.0625

Wait, that's different from my earlier thought. So median ≈65.06, mean ≈62.83. So mean < median? Wait, that would mean left-skewed. But why? Because the left tail (lower scores) has some low scores, which pull the mean down, while the median is in the middle of the data. Wait, but my calculation of mean: let's recalculate the sum.

Midpoints:

1-10:5.5, frequency 1:5.5

11-20:15.5, frequency1:15.5 → total 21

21-30:25.5, frequency2:51 → total 72

31-40:35.5, frequency4:142 → total 214

41-50:45.5, frequency10:455 → total 669

51-60:55.5, frequency13:55.5*13=721.5 → total 1390.5

61-70:65.5, frequency16:65.5*16=1048 → total 2438.5

71-80:75.5, frequency15:75.5*15=1132.5 → total 3571

81-90:85.5, frequency10:855 → total 4426

91-100:95.5, frequency3:286.5 → total 4712.5

Yes, sum is 4712.5, mean=4712.5/75≈62.83

Median≈65.06

So mean (≈62.83) < median (≈65.06), so Mean < Median.

Step2: Confirm Skewness

The distribution has a left tail (lower scores with low frequencies), which means it's left-skewed. In left-skewed distributions, the mean is pulled down by the low scores, so mean < median.

Answer:

Mean < Median