To find the standard deviation of the masses of the packets taken in the sample from each machine, we can use the formula:

Standard Deviation = √(Σ(xᵢ - ȳ)² / n)

Where xᵢ is the mass of each packet, ȳ is the mean mass, and n is the number of packets in the sample.

For Machine A:
Mean (ȳ) = (196 + 198 + 198 + 199 + 200 + 200 + 201 + 201 + 202 + 205) / 10 = 200.0 g

Calculate the squared differences from the mean for each packet:
(196 - 200)² = 16
(198 - 200)² = 4
(198 - 200)² = 4
(199 - 200)² = 1
(200 - 200)² = 0
(200 - 200)² = 0
(201 - 200)² = 1
(201 - 200)² = 1
(202 - 200)² = 4
(205 - 200)² = 25

Add up the squared differences: 16 + 4 + 4 + 1 + 0 + 0 + 1 + 1 + 4 + 25 = 56

Standard Deviation = √(56 / 10) = √5.6 ≈ 2.37 g

For Machine B:
Mean (ȳ) = (192 + 194 + 195 + 198 + 200 + 201 + 203 + 204 + 206 + 207) / 10 = 200.0 g

Calculate the squared differences from the mean for each packet:
(192 - 200)² = 64
(194 - 200)² = 36
(195 - 200)² = 25
(198 - 200)² = 4
(200 - 200)² = 0
(201 - 200)² = 1
(203 - 200)² = 9
(204 - 200)² = 16
(206 - 200)² = 36
(207 - 200)² = 49

Add up the squared differences: 64 + 36 + 25 + 4 + 0 + 1 + 9 + 16 + 36 + 49 = 240

Standard Deviation = √(240 / 10) = √24 ≈ 4.90 g

Therefore, the standard deviation of the masses of the packets taken from Machine A is approximately 2.37 g, while the standard deviation for Machine B is approximately 4.90 g.

Since the standard deviation measures the spread or variation of the data, a smaller standard deviation indicates that the data is more consistent and reliable. Therefore, Machine A, with a smaller standard deviation, is more reliable compared to Machine B. This means that the masses of packets from Machine A are closer to the mean mass and have less variability overall.

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