Question

the times of all 15 year olds who run a certain race are approximately normally distributed with a given mean \\( \mu = 18 \\) sec and standard deviation \\( \sigma = 1.2 \\) sec. what percentage of the runners have times less than 14.4 sec?\
\\( \bigcirc \\) 0.15%\
\\( \bigcirc \\) 0.30%\
\\( \bigcirc \\) 0.60%\
\\( \bigcirc \\) 2.50%

Explanation:

Step1: Calculate the z - score

The formula for the z - score is $z=\frac{x - \mu}{\sigma}$, where $x = 14.4$, $\mu=18$, and $\sigma = 1.2$.
Substitute the values into the formula: $z=\frac{14.4 - 18}{1.2}=\frac{- 3.6}{1.2}=- 3$.

Step2: Use the standard normal distribution table

For a standard normal distribution $N(0,1)$, we know the properties of the normal curve. The percentage of data less than $z=-3$ can be found from the standard normal table or the empirical rule. The empirical rule states that about 99.7% of the data lies within $z = - 3$ and $z = 3$ of the mean. So the percentage of data less than $z=-3$ is $\frac{1 - 0.997}{2}=0.0015$ or 0.15%.

Answer:

0.15%