Question

the times of all 15 year olds who run a certain race are approximately normally distributed with a given mean μ = sec and standard deviation σ = 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 \(z=\frac{x-\mu}{\sigma}\). Given \(x = 14.4\), \(\mu=18\) (assuming the mean \(\mu = 18\) as it is missing in the problem statement but for a normal distribution \(x-\mu=- 3.6\) and \(\sigma = 1.2\)), then \(z=\frac{14.4 - 18}{1.2}=\frac{-3.6}{1.2}=-3\)

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

The empirical rule for a normal distribution states that:

• 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 to the left of \(z=-3\) is \(\frac{100 - 99.7}{2}=0.15\%\)

Answer:

\(0.15\%\)