The probability that the mean length of the sample rods would differ from the population mean by greater than 0.65 mm is 0.1698 or 16.98%.
Mean length of the population (μ) = 176 mm.
Variance of the population (σ²) = 100.
Therefore, the standard deviation of the population (σ) = √100 = 10.
The sample size (n) = 446.
The mean of the sample = the mean of the population = μ = 176 mm.
The standard deviation of the sample (s) = σ/√n = 0.4735137.
We are asked to find the probability that the sample rods would differ the population mean by greater than 0.65 mm, that is,
either the length of the sample rods is less than (176 - 0.65) =175.35
or the length of the sample rods is more than (176 + 0.65) = 176.65.
This can be shown as P(X < 175.35 or X > 176.65) = 1 - P(175.35 < X < 176.65) = 1 - Normalcdf(175.35,176.65,176,0.4735137) = 1 - 0.8302 = 0.1698 or 16.98%.
Thus, the probability that the mean length of the sample rods would differ from the population mean by greater than 0.65 mm is 0.1698 or 16.98%.
Learn more about the probability of sampling distributions at
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