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?
0.15%
0.30%
0.60%
2.50%

Explanation:

Step1: Calculate the z - score

The formula for the z - score is \(z=\frac{x-\mu}{\sigma}\). Here, \(x = 14.4\), \(\mu=18\), and \(\sigma = 1.2\).

$$ z=\frac{14.4 - 18}{1.2}=\frac{- 3.6}{1.2}=-3 $$

Step2: Use the empirical rule (68 - 95 - 99.7 rule)

The empirical rule states that for a normal distribution:

• Approximately \(68\%\) of the data lies within \(z=-1\) and \(z = 1\)
• Approximately \(95\%\) of the data lies within \(z=-2\) and \(z = 2\)
• Approximately \(99.7\%\) of the data lies within \(z=-3\) and \(z = 3\)

The percentage of data within \(z=-3\) and \(z = 3\) is \(99.7\%\). The percentage of data outside of \(z=-3\) and \(z = 3\) is \(100\%-99.7\% = 0.3\%\). Since the normal distribution is symmetric, the percentage of data less than \(z=-3\) is \(\frac{100\% - 99.7\%}{2}=0.15\%\)

Answer:

0.15%