Question

which set of population data is the least dispersed from its mean?
○ 2, 3, 2, 9
○ 4, 0, 4, 0
○ 6, 2, 2, 2
○ 9, 3, 5, 3

Explanation:

Step1: Recall the formula for population standard deviation (a measure of dispersion): $\sigma = \sqrt{\frac{\sum_{i = 1}^{N}(x_i - \mu)^2}{N}}$, where $\mu$ is the mean and $N$ is the number of data points. A smaller standard deviation means less dispersion. We can also compare the sum of squared deviations from the mean (since dividing by $N$ and taking the square root is a monotonic transformation, the set with the smallest sum of squared deviations will have the smallest standard deviation).

Step2: Calculate the mean and sum of squared deviations for each set.

For set 1: 2, 3, 2, 9

- Mean $\mu_1=\frac{2 + 3 + 2 + 9}{4}=\frac{16}{4} = 4$
- Sum of squared deviations: $(2 - 4)^2+(3 - 4)^2+(2 - 4)^2+(9 - 4)^2=(-2)^2+(-1)^2+(-2)^2+(5)^2 = 4 + 1 + 4 + 25 = 34$

For set 2: 4, 0, 4, 0

- Mean $\mu_2=\frac{4 + 0 + 4 + 0}{4}=\frac{8}{4}=2$
- Sum of squared deviations: $(4 - 2)^2+(0 - 2)^2+(4 - 2)^2+(0 - 2)^2=(2)^2+(-2)^2+(2)^2+(-2)^2 = 4 + 4 + 4 + 4 = 16$

For set 3: 6, 2, 2, 2

- Mean $\mu_3=\frac{6 + 2 + 2 + 2}{4}=\frac{12}{4}=3$
- Sum of squared deviations: $(6 - 3)^2+(2 - 3)^2+(2 - 3)^2+(2 - 3)^2=(3)^2+(-1)^2+(-1)^2+(-1)^2 = 9 + 1 + 1 + 1 = 12$ Wait, no, wait: 6-3=3, squared is 9; 2-3=-1, squared is 1, three times. So 9 + 1+1+1=12? Wait, no, 6,2,2,2: four data points. So (6-3)^2 = 9, (2-3)^2=1, (2-3)^2=1, (2-3)^2=1. So 9 + 1 + 1 + 1 = 12.

For set 4: 9, 3, 5, 3

- Mean $\mu_4=\frac{9 + 3 + 5 + 3}{4}=\frac{20}{4}=5$
- Sum of squared deviations: $(9 - 5)^2+(3 - 5)^2+(5 - 5)^2+(3 - 5)^2=(4)^2+(-2)^2+(0)^2+(-2)^2 = 16 + 4 + 0 + 4 = 24$

Wait, I made a mistake in set 3. Wait, 6,2,2,2: sum is 6+2+2+2=12, mean is 3. Then (6-3)^2=9, (2-3)^2=1, three times. So 9 + 1+1+1=12. But wait, let's recalculate set 2: 4,0,4,0. Mean is (4+0+4+0)/4=8/4=2. Then (4-2)^2=4, (0-2)^2=4, (4-2)^2=4, (0-2)^2=4. So 4*4=16. Set 3: 6,2,2,2. Mean 3. (6-3)^2=9, (2-3)^2=1, three times. So 9 + 1+1+1=12. Set 4: 9,3,5,3. Mean 5. (9-5)=4, squared 16; (3-5)=-2, squared 4; (5-5)=0; (3-5)=-2, squared 4. So 16+4+0+4=24. Set 1: 2,3,2,9. Mean 4. (2-4)=-2, squared 4; (3-4)=-1, squared 1; (2-4)=-2, squared 4; (9-4)=5, squared 25. So 4+1+4+25=34. Wait, but now set 3 has sum of squared deviations 12, which is less than set 2's 16? But that contradicts my initial thought. Wait, no, let's check again. Wait, 6,2,2,2: data points are 6,2,2,2. Mean is 3. The deviations are 3, -1, -1, -1. Squared deviations: 9,1,1,1. Sum is 12. For set 2: 4,0,4,0. Mean 2. Deviations: 2, -2, 2, -2. Squared deviations: 4,4,4,4. Sum 16. For set 4: 9,3,5,3. Mean 5. Deviations: 4, -2, 0, -2. Squared: 16,4,0,4. Sum 24. For set 1: 2,3,2,9. Mean 4. Deviations: -2, -1, -2, 5. Squared: 4,1,4,25. Sum 34. Wait, but then set 3 has sum 12, which is less than set 2's 16. But that can't be right. Wait, maybe I miscalculated set 3. Wait, 6,2,2,2: the data points are 6,2,2,2. So the mean is (6+2+2+2)/4 = 12/4=3. Correct. Then (6-3)=3, squared 9. (2-3)=-1, squared 1, three times. So 9 + 1+1+1=12. So sum of squared deviations is 12. For set 2: 4,0,4,0. Sum of squared deviations is 16. So set 3 has a smaller sum. But wait, let's check the standard deviation. Standard deviation is square root of (sum of squared deviations / N). So for set 3: sqrt(12/4)=sqrt(3)≈1.732. For set 2: sqrt(16/4)=sqrt(4)=2. For set 4: sqrt(24/4)=sqrt(6)≈2.449. For set 1: sqrt(34/4)=sqrt(8.5)≈2.915. Wait, so set 3 has a smaller standard deviation than set 2? But that seems odd. Wait, let's check the data again. Set 3: 6,2,2,2. The data has one large value (6) and three small values (2). The mean is 3. The deviation of 6 is 3, and the deviations of the 2s are -1. So the squared deviations are 9,1,1,1. Sum 12. Set 2: 4,0,4,0. Deviations are 2, -2, 2, -2. Squared deviations 4,4,4,4. Sum 16. So yes, set 3 has a smaller sum of squared deviations, hence smaller standard deviation. But wait, the options are:

A. 2, 3, 2, 9

B. 4, 0, 4, 0

C. 6, 2, 2, 2

D. 9, 3, 5, 3

Wait, but according to this calculation, set C (6,2,2,2) has a smaller standard deviation than set B (4,0,4,0). But maybe I made a mistake. Wait, let's recalculate set C's mean. 6+2+2+2=12, 12/4=3. Correct. (6-3)=3, squared 9. (2-3)=-1, squared 1, three times. So 9+1+1+1=12. Sum of squared deviations is 12. N=4. So variance is 12/4=3, standard deviation sqrt(3)≈1.732. Set B: sum of squared deviations 16, variance 4, standard deviation 2. So set C has a smaller standard deviation. But wait, maybe I misread the options. Let me check again. The options are:

1. 2, 3, 2, 9

2. 4, 0, 4, 0

3. 6, 2, 2, 2

4. 9, 3, 5, 3

Wait, but maybe I made a mistake in set C. Wait, 6,2,2,2: the data points are 6,2,2,2. So the range is 6-2=4. Set B: 4,0,4,0. Range is 4-0=4. But standard deviation is a better measure. Wait, but let's check the mean absolute deviation, which is another measure of dispersion. For set C: mean is 3. Absolute deviations: |6-3|=3, |2-3|=1, |2-3|=1, |2-3|=1. Sum: 3+1+1+1=6. Mean absolute deviation: 6/4=1.5. For set B: mean is 2. Absolute deviations: |4-2|=2, |0-2|=2, |4-2|=2, |0-2|=2. Sum: 8. Mean absolute deviation: 8/4=2. So set C has a smaller mean absolute deviation as well. So set C is less dispersed? But that contradicts my initial intuition. Wait, maybe the question is about population data, so we use N, not N-1. So the calculations are correct. So the least dispersed is set C: 6,2,2,2? Wait, but let's check set D: 9,3,5,3. Mean is 5. Absolute deviations: |9-5|=4, |3-5|=2, |5-5|=0, |3-5|=2. Sum: 4+2+0+2=8. Mean absolute deviation: 8/4=2. So set C has mean absolute deviation 1.5, set B has 2, set D has 2, set A has |2-4|=2, |3-4|=1, |2-4|=2, |9-4|=5. Sum: 2+1+2+5=10. Mean absolute deviation: 10/4=2.5. So yes, set C has the smallest mean absolute deviation and smallest standard deviation. So the answer should be C. 6, 2, 2, 2. But wait, maybe I made a mistake in the sum of squared deviations for set C. Let me check again: (6-3)^2 = 9, (2-3)^2 = 1, (2-3)^2 = 1, (2-3)^2 = 1. 9 + 1 + 1 + 1 = 12. Correct. For set B: (4-2)^2=4, (0-2)^2=4, (4-2)^2=4, (0-2)^2=4. 4*4=16. Correct. So set C has a smaller sum of squared deviations, hence less dispersion. So the answer is C. 6, 2, 2, 2.

Answer:

C. 6, 2, 2, 2