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A chemical company sells a specialized industrial lubricant in 20 liter containers. A
random sample of the latest production lot shows the contents of 10 containers to be
20.72, 20.6, 20.71, 20.92, 19.79, 20.93, 20.09, 20.27, 20.49, and 20.2 liters. Find the
standard deviation of the sampled containers' contents.
A. o = 0.55
B. o = 0.53
C. o = 0.36
D. o = 0.42

Answer :

Eduard22slygo

The standard deviation of the sampled containers' contents is 0.36

How to determine the mean

  • Data = 20.72, 20.6, 20.71, 20.92, 19.79, 20.93, 20.09, 20.27, 20.49, 20.2
  • Number of data (n) = 10
  • Summation of data = 20.72 + 20.6 + 20.71 + 20.92 + 19.79 + 20.93 + 20.09 + 20.27 + 20.49 + 20.2 = 204.72
  • Mean (μ) =?

Mean = summation of data / number

μ = 204.72 / 10

μ = 20.472

How to determine the standard deviation

  • Data = 20.72, 20.6, 20.71, 20.92, 19.79, 20.93, 20.09, 20.27, 20.49, 20.2
  • Number of data (n) = 10
  • Mean (μ) = 20.472
  • Standard deviation (σ) =?

σ = √[[(x₁ - μ)² + (x₂ - μ)² + (x₃ - μ)² + (x₄ - μ)² + (x₅ - μ)² + (x₆ - μ)² + (x₇ - μ)² + (x₈ - μ)² + (x₉ - μ)² + (x₁₀ - μ)²] / n]

σ = √[[(20.72 - 20.472)² + (20.6 - 20.472)² + (20.71 - 20.472)² + (20.92 - 20.472)² + (19.79 - 20.472)² + (20.93 - 20.472)² + (20.09 - 20.472)² + (20.27 - 20.472)² + (20.49 - 20.472)² + (20.2 - 20.472)²] / 10]

σ = 0.36

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