Question

compute the mean, median, and mode for the set of scores shown in the following frequency distribution table.

x f
7 1
6 2
5 3
4 4
3 1
2 0
1 1

the mean is select
the mode is select
the median is select

Explanation:

Step1: Calculate the total number of scores (N)

First, we sum up the frequencies (f). The frequencies are 1, 2, 3, 4, 1, 0, 1. So:
\( N = 1 + 2 + 3 + 4 + 1 + 0 + 1 = 12 \)

Step2: Calculate the sum of \( X \times f \) (for mean)

We multiply each \( X \) by its corresponding \( f \) and sum them up:

• For \( X = 7 \), \( 7 \times 1 = 7 \)
• For \( X = 6 \), \( 6 \times 2 = 12 \)
• For \( X = 5 \), \( 5 \times 3 = 15 \)
• For \( X = 4 \), \( 4 \times 4 = 16 \)
• For \( X = 3 \), \( 3 \times 1 = 3 \)
• For \( X = 2 \), \( 2 \times 0 = 0 \)
• For \( X = 1 \), \( 1 \times 1 = 1 \)

Now sum these products: \( 7 + 12 + 15 + 16 + 3 + 0 + 1 = 54 \)

Step3: Calculate the mean

The mean (\( \bar{X} \)) is given by \( \bar{X} = \frac{\sum (X \times f)}{N} \). Substituting the values:
\( \bar{X} = \frac{54}{12} = 4.5 \)

Step4: Find the mode

The mode is the value of \( X \) with the highest frequency (f). Looking at the frequencies: 1 (for 7), 2 (for 6), 3 (for 5), 4 (for 4), 1 (for 3), 0 (for 2), 1 (for 1). The highest frequency is 4, which corresponds to \( X = 4 \). So the mode is 4.

Step5: Find the median

First, we need to find the position of the median. For \( N = 12 \) (even), the median is the average of the \( \frac{N}{2} \)-th and \( (\frac{N}{2} + 1) \)-th values. So \( \frac{12}{2} = 6 \)-th and \( 7 \)-th values.

Now we list the scores in order (using the frequency distribution):

• 1 (appears 1 time)
• 3 (appears 1 time)
• 4 (appears 4 times)
• 5 (appears 3 times)
• 6 (appears 2 times)
• 7 (appears 1 time)

Let's list the cumulative frequencies:

• Up to \( X = 1 \): 1
• Up to \( X = 3 \): 1 + 1 = 2
• Up to \( X = 4 \): 2 + 4 = 6
• Up to \( X = 5 \): 6 + 3 = 9
• Up to \( X = 6 \): 9 + 2 = 11
• Up to \( X = 7 \): 11 + 1 = 12

The 6th value is 4 (since cumulative frequency up to 4 is 6) and the 7th value is 4 (since cumulative frequency up to 4 is 6, and the next values start at 5, so the 7th value is still 4? Wait, no. Wait, let's list the actual scores:

1, 3, 4, 4, 4, 4, 5, 5, 5, 6, 6, 7

Wait, let's count:

1 (1st), 3 (2nd), 4 (3rd), 4 (4th), 4 (5th), 4 (6th), 5 (7th), 5 (8th), 5 (9th), 6 (10th), 6 (11th), 7 (12th)

Ah, so the 6th value is 4 and the 7th value is 5? Wait, no, my mistake earlier. Let's recalculate the cumulative frequencies correctly:

• \( X = 1 \), f = 1: cumulative = 1 (scores: [1])
• \( X = 3 \), f = 1: cumulative = 1 + 1 = 2 (scores: [1, 3])
• \( X = 4 \), f = 4: cumulative = 2 + 4 = 6 (scores: [1, 3, 4, 4, 4, 4])
• \( X = 5 \), f = 3: cumulative = 6 + 3 = 9 (scores: [1, 3, 4, 4, 4, 4, 5, 5, 5])
• \( X = 6 \), f = 2: cumulative = 9 + 2 = 11 (scores: [1, 3, 4, 4, 4, 4, 5, 5, 5, 6, 6])
• \( X = 7 \), f = 1: cumulative = 11 + 1 = 12 (scores: [1, 3, 4, 4, 4, 4, 5, 5, 5, 6, 6, 7])

So the 6th value is 4 (since the first 6 scores are 1, 3, 4, 4, 4, 4) and the 7th value is 5 (the 7th score is 5). Wait, no, the 6th score is the last 4 in the first 6, and the 7th is the first 5. Wait, let's list all 12 scores:

1 (1), 3 (2), 4 (3), 4 (4), 4 (5), 4 (6), 5 (7), 5 (8), 5 (9), 6 (10), 6 (11), 7 (12)

Yes, so the 6th score is 4 and the 7th score is 5. Then the median is the average of 4 and 5: \( \frac{4 + 5}{2} = 4.5 \)

Wait, but let's check again. Wait, the cumulative frequency up to X=4 is 6, so the 6th score is 4. Then the 7th score is the first score in X=5, which is 5. So median is (4 + 5)/2 = 4.5.

Answer:

Mean: \( 4.5 \)
Mode: \( 4 \)
Median: \( 4.5 \)