Question

1. for the following set of scores:

7 5 6 4 4 3 8 9 4 7 5 5 6

9 4 7 5 10 6 8 5 6 3 4 8 5

a. construct a frequency distribution table.

b. sketch a histogram showing the distribution.

c. what is the shape of the distribution?

Answer:

Part (a): Construct a Frequency Distribution Table

Step 1: Identify the Scores and Their Range

First, we list out all the scores and find the minimum and maximum values. The scores are: 7, 5, 6, 4, 4, 3, 8, 9, 4, 7, 5, 5, 6, 9, 4, 7, 5, 10, 6, 8, 5, 6, 3, 4, 8, 5. The minimum score is 3 and the maximum score is 10. So the possible scores (classes) are 3, 4, 5, 6, 7, 8, 9, 10.

Step 2: Count the Frequency of Each Score

• For score 3: Let's count how many times 3 appears. Looking at the list, 3 appears at positions: 6 (3), 23 (3). So frequency \( f = 2 \).
• For score 4: Count the number of 4s. Positions: 4 (4), 5 (4), 9 (4), 15 (4), 24 (4). So \( f = 5 \).
• For score 5: Count the number of 5s. Positions: 2 (5), 11 (5), 12 (5), 17 (5), 21 (5), 26 (5). Wait, let's list all 5s: 5, 5, 5, 5, 5, 5? Wait, let's recount:

Original list: 7, 5, 6, 4, 4, 3, 8, 9, 4, 7, 5, 5, 6, 9, 4, 7, 5, 10, 6, 8, 5, 6, 3, 4, 8, 5. Let's count 5s:

1. 5 (position 2)
2. 5 (position 11)
3. 5 (position 12)
4. 5 (position 17)
5. 5 (position 21)
6. 5 (position 26)

Wait, that's 6? Wait, let's count again:
Looking at the numbers:
First row: 7, 5, 6, 4, 4, 3, 8, 9, 4, 7, 5, 5, 6 → 5 appears at 2, 11, 12 → 3 times.
Second row: 9, 4, 7, 5, 10, 6, 8, 5, 6, 3, 4, 8, 5 → 5 appears at 4, 8, 13 → 3 times.
Total 5s: 3 + 3 = 6? Wait, no, the second row has 13 numbers? Wait, the original data: first row: 7, 5, 6, 4, 4, 3, 8, 9, 4, 7, 5, 5, 6 (13 numbers). Second row: 9, 4, 7, 5, 10, 6, 8, 5, 6, 3, 4, 8, 5 (13 numbers). Total 26 numbers.
Let's list all 5s:
First row: 5 (index 2), 5 (index 11), 5 (index 12) → 3.
Second row: 5 (index 4), 5 (index 8), 5 (index 13) → 3. Total 6. Wait, but let's check again:
Wait, the second row is: 9, 4, 7, 5, 10, 6, 8, 5, 6, 3, 4, 8, 5. So positions (0-based or 1-based? Let's use 1-based):
First row (1-13):
1:7, 2:5, 3:6, 4:4, 5:4, 6:3, 7:8, 8:9, 9:4, 10:7, 11:5, 12:5, 13:6.
Second row (14-26):
14:9, 15:4, 16:7, 17:5, 18:10, 19:6, 20:8, 21:5, 22:6, 23:3, 24:4, 25:8, 26:5.
So 5s are at 2, 11, 12, 17, 21, 26 → that's 6 times. So frequency of 5 is 6.

• For score 6: Count the number of 6s. First row: 3 (6), 13 (6) → 2. Second row: 19 (6), 22 (6) → 2. Wait, first row: 6 (position 3), 13 (position 13) → 2. Second row: 6 (position 19), 22 (position 22) → 2. Wait, no, first row: 7, 5, 6, 4, 4, 3, 8, 9, 4, 7, 5, 5, 6 → 6 at 3, 13 → 2. Second row: 9, 4, 7, 5, 10, 6, 8, 5, 6, 3, 4, 8, 5 → 6 at 6, 9 → 2. Wait, 6 (position 19: 6), 22 (position 22: 6). So total 2 + 2 = 4? Wait, no, 3 (6), 13 (6), 19 (6), 22 (6) → 4? Wait, let's list all 6s:

First row: 6 (3), 6 (13) → 2.
Second row: 6 (19), 6 (22) → 2. Total 4? Wait, no, the first row has 7, 5, 6, 4, 4, 3, 8, 9, 4, 7, 5, 5, 6 → two 6s. Second row: 9, 4, 7, 5, 10, 6, 8, 5, 6, 3, 4, 8, 5 → two 6s. So total 4? Wait, but let's check again:
Wait, the numbers: 6 appears at 3, 13, 19, 22 → 4 times? Wait, no, 3 (6), 13 (6), 19 (6), 22 (6) → 4. Wait, but maybe I missed. Let's count all numbers:
List of scores:
7, 5, 6, 4, 4, 3, 8, 9, 4, 7, 5, 5, 6,
9, 4, 7, 5, 10, 6, 8, 5, 6, 3, 4, 8, 5.
Now, let's list each score and count:

• 3: 3, 3 → 2 (positions 6, 23)
• 4: 4, 4, 4, 4, 4 → 5 (positions 4, 5, 9, 15, 24)
• 5: 5, 5, 5, 5, 5, 5 → 6 (positions 2, 11, 12, 17, 21, 26)
• 6: 6, 6, 6, 6 → 4? Wait, 6 (3), 6 (13), 6 (19), 6 (22) → 4? Wait, no, 3 (6), 13 (6), 19 (6), 22 (6) → 4. Wait, but in the second row, 6 is at 19 (6) and 22 (6) → 2. First row: 2. Total 4.
• 7: 7, 7, 7 → 3 (positions 1, 10, 16)
• 8: 8, 8, 8 → 3 (positions 7, 20, 25)
• 9: 9, 9 → 2 (positions 8, 14)
• 10: 10 → 1 (position 18)

Wait, let's verify the total frequency: 2 (3) + 5 (4) + 6 (5) + 4 (6) + 3 (7) + 3 (8) + 2 (9) + 1 (10) = 2+5=7, 7+6=13, 13+4=17, 17+3=20, 20+3=23, 23+2=25, 25+1=26. Which matches the total number of scores (26). Good.

So the frequency distribution table is:

Score (X)	Frequency (f)
4	5
5	6
6	4
7	3
8	3
9	2
10	1

Part (b): Sketch a Histogram

To sketch a histogram, we use the scores (X) on the horizontal axis (x-axis) and the frequency (f) on the vertical axis (y-axis). Each bar has a width corresponding to the score (since the scores are discrete, we can consider each bar to have a width of 1, centered on the score). The height of each bar is equal to the frequency of the corresponding score.

• For X=3: height = 2
• For X=4: height = 5
• For X=5: height = 6
• For X=6: height = 4
• For X=7: height = 3
• For X=8: height = 3
• For X=9: height = 2
• For X=10: height = 1

We draw vertical bars for each score with the above heights, ensuring that the bars are adjacent (since the scores are consecutive integers).

Part (c): Shape of the Distribution

To determine the shape, we look at the frequency distribution:

• The highest frequency is at X=5 (f=6).
• The frequencies increase from X=3 to X=5 (2 → 5 → 6) and then decrease from X=5 to X=10 (6 → 4 → 3 → 3 → 2 → 1).
• The left tail (scores below the peak, X=3,4) and the right tail (scores above the peak, X=6,7,8,9,10) are not symmetric. Wait, actually, let's check the symmetry. The peak is at X=5. Let's see the frequencies on either side:

Left of 5 (X=3,4): frequencies 2, 5 → cumulative 7.

Right of 5 (X=6,7,8,9,10): frequencies 4, 3, 3, 2, 1 → cumulative 4+3+3+2+1=13? Wait, no, 4 (X=6) + 3 (X=7) + 3 (X=8) + 2 (X=9) + 1 (X=10) = 13. Wait, but the peak is at X=5. Wait, actually, the distribution is approximately symmetric? Wait, no, let's check the frequencies around the peak:

• X=4 (f=5) and X=6 (f=4) → close.
• X=3 (f=2) and X=7 (f=3) → close.
• X=9 (f=2) and X=10 (f=1) → but X=10 is the maximum. Wait, maybe it's a slightly skewed distribution, but more likely, it's approximately symmetric (or maybe slightly negatively skewed? Wait, no, negative skew has a long left tail, positive skew has a long right tail. Here, the right tail (X=6 to 10) has frequencies 4,3,3,2,1, and the left tail (X=3,4) has 2,5. Wait, the peak is at X=5. Let's check the mean, median, and mode. Mode is 5 (highest frequency). Median: since there are 26 scores, the median is the average of the 13th and 14th scores. Let's order the scores:

First, sort the scores:

3, 3,

4, 4, 4, 4, 4,

5, 5, 5, 5, 5, 5,

6, 6, 6, 6,

7, 7, 7,

8, 8, 8,

9, 9,

10.

Wait, let's list them in order:

3, 3,

4, 4, 4, 4, 4, (5 numbers)

5, 5, 5, 5, 5, 5, (6 numbers)

6, 6, 6, 6, (4 numbers)

7, 7, 7, (3 numbers)

8, 8, 8, (3 numbers)

9, 9, (2 numbers)

10 (1 number)

Now, the 13th and 14th scores:

Let's count the positions:

1:3, 2:3,

3:4, 4:4, 5:4, 6:4, 7:4,

8:5, 9:5, 10:5, 11:5, 12:5, 13:5,

14:6, 15:6, 16:6, 17:6,

18:7, 19:7, 20:7,

21:8, 22:8, 23:8,

24:9, 25:9,

26:10.

So the 13th score is 5, the 14th score is 6. So median is \( \frac{5 + 6}{2} = 5.5 \).

Mode is 5 (frequency 6).

Mean: Let's calculate the mean. Sum of scores:

For X=3: 3*2 = 6

X=4: 4*5 = 20

X=5: 5*6 = 30

X=6: 6*4 = 24

X=7: 7*3 = 21

X=8: 8*3 = 24

X=9: 9*2 = 18

X=10: 10*1 = 10

Total sum = 6 + 20 + 30 + 24 + 21 + 24 + 18 + 10 = let's calculate:

6 + 20 = 26

26 + 30 = 56

56 + 24 = 80

80 + 21 = 101

101 + 24 = 125

125 + 18 = 143

143 + 10 = 153

Mean = \( \frac{153}{26} \approx 5.88 \)

Now, mode (5) < median (5.5) < mean (≈5.88). Wait, no: mode (5) < median (5.5) < mean (5.88). Wait, that would indicate positive skewness (mean > median > mode), but wait, no: positive skew has a long right tail, and mean > median > mode. But in our case, the right tail (X=6 to 10) has frequencies decreasing, but the mean is higher than the median. Wait, but let's check the histogram. The peak is at X=5, and the frequencies decrease as we move away from 5 in both directions, but the right tail (X=6,7,8,9,10) has more scores (4+3+3+2+1=13) than the left tail (X=3,4: 2+5=7). Wait, no, the left tail is X < 5 (3,4) with frequency 7, right tail is X > 5 (6,7,8,9,10) with frequency 13. So the right tail is longer, which would mean positive skewness. But the mode is 5, median 5.5, mean ~5.88. So mean > median > mode, which is positive skewness. But wait, maybe I made a mistake in the frequency