Help? Statistics...
The 200-meter race times at a state track meet are normally distributed with a mean of 13.56 seconds and a standard deviation of 2.24 seconds. Using the Standard Normal Probabilities table, what is the approximate probability that a runner chosen at random will have a 200-meter time less than 13.5 seconds?

The mean and standard deviation are given to you. You want to find "less than 13.5 seconds", meaning that 13.5 is the maximum "Upper Limit" and 0 would be the Lower Limit because we can pretend 0 second is the minimum. Don't worry about negatives, in this scenario, values like "-5 seconds" are impossible.

Enter normalcdf(0, 13.5, 13.56, 2.24) in a calculator and the output result is the answer.

mreijepatmat mreijepatmat · Answer 2 · 2017-08-10T03:25:38+02:00

mreijepatmat mreijepatmat

10-08-2017

Mean = μ =13.56; Standard deviation = σ = 2.24.
Let's consider the Z distribution where μ =0 and σ = 1
(with x = 13.50)
Let's calculate the Z score: Z = (x-μ)/σ .

Z = (13.50 - 13.56)/2.24 → Z = - 0.026785

For Z = -0.02 , P(Z) = 0.4920 OR 49.20%
For Z = -0.03 , P(Z) = 0.4880 OR 48.80%
Then Z (Approx) = [(48.2)+(48.8)]/2 = 49% (approx.)

Then 49% of a runner chosen at random will have a 200-meter time less than 13.5 seconds

Answer Link