Question

Suppose I make several time measurements, and compute an average time of 3.4 s, with a standard deviation of 0.8 s and an SDOM of 0.2 s. If I make another time measurement, how often would I expect the result to be smaller than 2.6s\

Answer 1

The result will be smaller than 2.6s, 16% of the time.

Explanation:

This is a normal distribution problem

The standardized score for any value is the value minus the mean then divided by the standard deviation.

z = (x - xbar)/σ

x = the random value = 2.6 s

xbar = mean = the average = 3.4 s

σ = standard deviation = 0.8 s

z = (2.6 - 3.4)/0.8 = - 1

P(x < 2.6) = P(z < - 1) = 1 - P(z ≥ - 1) = 1 - P(z ≤ 1) = 1 - 0.841 = 0.159 = 16%