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graph 1:0:2 points, 1: 4 points, 2: 5 points, 3:3 points, 4:2 points, 5: 2 points

graph 2 4: no points, 5: 1 point, 6; 2 points, 7: 3 points, 8: 5 points, 9: 5 points, 10: 3 points, 9: 1 point, 12 has no points
which graph has a higher standard deviation

1 answer

To determine which graph has a higher standard deviation, we first need to calculate the standard deviation for both sets of points.

Graph 1

Points:

  • \(0: 2\)
  • \(1: 4\)
  • \(2: 5\)
  • \(3: 3\)
  • \(4: 2\)
  • \(5: 2\)

The data points can be expressed as:

  • \(0, 0, 1, 1, 1, 1, 2, 2, 2, 2, 3, 4, 5\)

To find the mean (\(\mu\)): \[ \mu = \frac{\text{Sum of all points}}{\text{Number of points}} = \frac{0 \cdot 2 + 1 \cdot 4 + 2 \cdot 5 + 3 \cdot 3 + 4 \cdot 2 + 5 \cdot 2}{2 + 4 + 5 + 3 + 2 + 2} = \frac{0 + 4 + 10 + 9 + 8 + 10}{18} = \frac{41}{18} \approx 2.28 \]

To find the squared deviations from the mean:

  1. \((0 - 2.28)^2 \cdot 2\)
  2. \((1 - 2.28)^2 \cdot 4\)
  3. \((2 - 2.28)^2 \cdot 5\)
  4. \((3 - 2.28)^2 \cdot 3\)
  5. \((4 - 2.28)^2 \cdot 2\)
  6. \((5 - 2.28)^2 \cdot 2\)

Calculating these:

  • \(2 \cdot (0 - 2.28)^2 \approx 2 \cdot 5.1984 \approx 10.3968\)
  • \(4 \cdot (1 - 2.28)^2 \approx 4 \cdot 1.6124 \approx 6.4496\)
  • \(5 \cdot (2 - 2.28)^2 \approx 5 \cdot 0.0784 \approx 0.392\)
  • \(3 \cdot (3 - 2.28)^2 \approx 3 \cdot 0.5184 \approx 1.5552\)
  • \(2 \cdot (4 - 2.28)^2 \approx 2 \cdot 2.9536 \approx 5.9072\)
  • \(2 \cdot (5 - 2.28)^2 \approx 2 \cdot 7.4084 \approx 14.8168\)

Now, add these up: \[ 10.3968 + 6.4496 + 0.392 + 1.5552 + 5.9072 + 14.8168 \approx 39.5176 \]

Divide by the number of data points (N = 18): \[ \sigma^2 = \frac{39.5176}{18} \approx 2.1954 \]

Now taking the square root to find the standard deviation: \[ \sigma \approx \sqrt{2.1954} \approx 1.48 \]

Graph 2

Points:

  • \(4: 0\)
  • \(5: 1\)
  • \(6: 2\)
  • \(7: 3\)
  • \(8: 5\)
  • \(9: 5\)
  • \(10: 3\)
  • \(11: 1\)
  • \(12: 0\)

The data points can be expressed as:

  • \(5, 6, 7, 7, 8, 9, 9, 10\)

To find the mean (\(\mu\)): \[ \mu = \frac{(0 \cdot 4 + 1 \cdot 5 + 2 \cdot 6 + 3 \cdot 7 + 5 \cdot 8 + 5 \cdot 9 + 3 \cdot 10 + 1 \cdot 11)}{4 + 5 + 2 + 3 + 5 + 5 + 3 + 1} = \frac{0 + 5 + 12 + 21 + 40 + 45 + 30 + 11}{30} \]

Calculating: \[ \mu = \frac{164}{30} \approx 5.47 \]

Calculating squared deviations from the mean:

  1. \((0 - 5.47) \cdot 4\)
  2. \((5 - 5.47) \cdot 5\)
  3. \((6 - 5.47) \cdot 2\)
  4. \((7 - 5.47) \cdot 3\)
  5. \((8 - 5.47) \cdot 5\)
  6. \((9 - 5.47) \cdot 5\)
  7. \((10 - 5.47) \cdot 3\)
  8. \((11 - 5.47) \cdot 1\)

Calculating these:

  • \(4 \cdot (0 - 5.47)^2 = 4 \cdot 29.8009 = 119.2036\)
  • \(5 \cdot (5 - 5.47)^2 = 5 \cdot 0.2209 = 1.1045\)
  • \(2 \cdot (6 - 5.47)^2 = 2 \cdot .2809 = 0.5618\)
  • \(3 \cdot (7 - 5.47)^2 = 3 \cdot 2.3716 = 7.1148\)
  • \(5 \cdot (8 - 5.47)^2 = 5 \cdot 6.4396 = 32.198\)
  • \(5 \cdot (9 - 5.47)^2 = 5 \cdot 12.0609 = 60.3045\)
  • \(3 \cdot (10 - 5.47)^2 = 3 \cdot 20.6409 = 61.9227\)
  • \(1 \cdot (11 - 5.47)^2 = 1 \cdot 30.6409 = 30.6409\)

Summing these values gives: \[ 119.2036 + 1.1045 + 0.5618 + 7.1148 + 32.198 + 60.3045 + 61.9227 + 30.6409 = 312.0815 \]

Finally, divide by the number of data points (N = 30): \[ \sigma^2 = \frac{312.0815}{30} \approx 10.4027 \]

Taking the square root: \[ \sigma \approx \sqrt{10.4027} \approx 3.22 \]

Conclusion

  • Graph 1 Standard Deviation: ≈ 1.48
  • Graph 2 Standard Deviation: ≈ 3.22

Graph 2 has a higher standard deviation than Graph 1.