Question

1. for the following set of scores:

6 3 7 6 1 2 7 3 6 4
1 2 4 7 4 4 5 5 6 6
a. construct a frequency distribution table to organize the scores. include cumulative frequency and cumulative percent.
b. what is the percentile rank of the upper real limit of x = 5?
c. what is the upper real limit of the score that corresponds to the 50th percentile?

Explanation:

Part (a)

Step 1: Identify unique scores

First, we list out the unique scores from the data set. The scores are: 1, 2, 3, 4, 5, 6, 7.

Step 2: Calculate frequency

Frequency (\(f\)) is the number of times each score appears.

• For \(X = 1\): Appears 2 times.
• For \(X = 2\): Appears 2 times.
• For \(X = 3\): Appears 2 times.
• For \(X = 4\): Appears 3 times.
• For \(X = 5\): Appears 2 times.
• For \(X = 6\): Appears 5 times.
• For \(X = 7\): Appears 3 times.

Step 3: Calculate cumulative frequency (\(cf\))

Cumulative frequency is the sum of frequencies up to and including the current score.

• For \(X = 1\): \(cf = 2\)
• For \(X = 2\): \(cf = 2 + 2 = 4\)
• For \(X = 3\): \(cf = 4 + 2 = 6\)
• For \(X = 4\): \(cf = 6 + 3 = 9\)
• For \(X = 5\): \(cf = 9 + 2 = 11\)
• For \(X = 6\): \(cf = 11 + 5 = 16\)
• For \(X = 7\): \(cf = 16 + 3 = 19\) (Wait, let's check the total number of scores. The first row has 10 scores, the second row has 10 scores, total \(n = 20\). Wait, I must have miscalculated the frequencies. Let's recount the scores:

First row: 6, 3, 7, 6, 1, 2, 7, 3, 6, 4 (10 scores)

Second row: 1, 2, 4, 7, 4, 4, 5, 5, 6, 6 (10 scores)

Now, count each \(X\):

• \(X = 1\): 2 (first row:1, second row:1)
• \(X = 2\): 2 (first row:2, second row:2)
• \(X = 3\): 2 (first row:3,3)
• \(X = 4\): 3 (first row:4, second row:4,4,4) Wait, first row has 1 four, second row has 3 fours (4,4,4). So total 1 + 3 = 4? Wait first row: 6,3,7,6,1,2,7,3,6,4 → 4 appears once. Second row:1,2,4,7,4,4,5,5,6,6 → 4 appears 3 times. So total \(f(4)=1 + 3 = 4\).

• \(X = 5\): 2 (second row:5,5)
• \(X = 6\): First row:6,6,6 (three times), second row:6,6 (two times). So \(f(6)=3 + 2 = 5\)
• \(X = 7\): First row:7,7 (two times), second row:7 (one time). So \(f(7)=2 + 1 = 3\)

Now total \(n = 2 + 2 + 2 + 4 + 2 + 5 + 3 = 20\), which is correct (10 + 10).

Now recalculate cumulative frequency:

• \(X = 1\): \(cf = 2\)
• \(X = 2\): \(cf = 2 + 2 = 4\)
• \(X = 3\): \(cf = 4 + 2 = 6\)
• \(X = 4\): \(cf = 6 + 4 = 10\)
• \(X = 5\): \(cf = 10 + 2 = 12\)
• \(X = 6\): \(cf = 12 + 5 = 17\)
• \(X = 7\): \(cf = 17 + 3 = 20\)

Step 4: Calculate cumulative percent (\(c\%\))

Cumulative percent is calculated as \(\frac{cf}{n} \times 100\), where \(n = 20\).

• For \(X = 1\): \(c\%=\frac{2}{20}\times 100 = 10\%\)
• For \(X = 2\): \(c\%=\frac{4}{20}\times 100 = 20\%\)
• For \(X = 3\): \(c\%=\frac{6}{20}\times 100 = 30\%\)
• For \(X = 4\): \(c\%=\frac{10}{20}\times 100 = 50\%\)
• For \(X = 5\): \(c\%=\frac{12}{20}\times 100 = 60\%\)
• For \(X = 6\): \(c\%=\frac{17}{20}\times 100 = 85\%\)
• For \(X = 7\): \(c\%=\frac{20}{20}\times 100 = 100\%\)

Now we can construct the frequency distribution table:

\(X\)	\(f\)	\(cf\)	\(c\%\)
2	2	4	20%
3	2	6	30%
4	4	10	50%
5	2	12	60%
6	5	17	85%
7	3	20	100%

Part (b)

Step 1: Recall the formula for percentile rank

The percentile rank of a score \(X\) is the percentage of scores that are less than or equal to \(X\). For the upper real limit of \(X = 5\), we need to consider the cumulative frequency up to the upper real limit of 5.

The real limits for a discrete score \(X\) (assuming these are discrete scores, so the real limits are \(X - 0.5\) to \(X + 0.5\)). So the upper real limit of \(X = 5\) is \(5.5\).

We need to find the cumulative frequency of all scores less than or equal to \(5.5\). From the frequency distribution table, the cumulative frequency for \(X = 5\) is \(12\) (since all scores up to \(X = 5\) are less than \(5.5\)).

Step 2: Calculate percentile rank

Percentile rank (\(PR\)) is given by \(\frac{cf}{n} \times 100\), where \(cf\) is the cumulative frequency of scores less than or equal to the upper real limit, and \(n = 20\).

Here, \(cf = 12\) (cumulative frequency for \(X = 5\), since scores up to \(X = 5\) are within the upper real limit of \(5.5\)), \(n = 20\).

So \(PR=\frac{12}{20}\times 100 = 60\%\)

Part (c)

Step 1: Recall the concept of 50th percentile

The 50th percentile (median) is the score where 50% of the scores are less than or equal to it. From the cumulative percent column in the frequency distribution table, we look for the score where the cumulative percent is at least 50% and the previous cumulative percent is less than 50%.

Looking at the table:

• For \(X = 4\), cumulative percent is \(50\%\).

Step 2: Find the upper real limit of \(X = 4\)

For a discrete score \(X\), the upper real limit is \(X + 0.5\). So for \(X = 4\), the upper real limit is \(4 + 0.5 = 4.5\).

We can verify: The cumulative frequency for \(X = 4\) is \(10\), which is \(50\%\) of \(20\) (since \(\frac{10}{20}\times 100 = 50\%\)). So the score corresponding to the 50th percentile is \(X = 4\), and its upper real limit is \(4.5\).

Final Answers

Part (a)

\(X\)	\(f\)	\(cf\)	\(c\%\)
2	2	4	20%
3	2	6	30%
4	4	10	50%
5	2	12	60%
6	5	17	85%
7	3	20	100%

Part (b)

Answer:

\(60\%\)

Part (c)