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The times of all 15-year-olds who run a certain race are approximately normally distributed with a given mean [tex]\mu = 18 \, \text{sec}[/tex] and standard deviation [tex]\sigma = 1.2 \, \text{sec}[/tex]. What percentage of the runners have times less than 14.4 seconds?

A. 0.15%
B. 0.30%
C. 0.60%
D. 2.50%

Sagot :

Sure, let's solve this step-by-step.

1. Understand the Problem:
- We have a normally distributed variable, which in this case is the time taken by 15-year-olds to run a certain race.
- The mean (average time) is given as [tex]\(\mu = 18\)[/tex] seconds.
- The standard deviation is [tex]\(\sigma = 1.2\)[/tex] seconds.
- We want to find the percentage of runners who complete the race in less than [tex]\(14.4\)[/tex] seconds.

2. Calculate the Z-Score:
- The Z-score tells us how many standard deviations away a data point is from the mean.
- The formula for the Z-score [tex]\(z\)[/tex] is:
[tex]\[ z = \frac{x - \mu}{\sigma} \][/tex]
where [tex]\(x\)[/tex] is the value we are interested in (14.4 seconds in this case).

Plugging in the values:
[tex]\[ z = \frac{14.4 - 18}{1.2} = \frac{-3.6}{1.2} = -3.0 \][/tex]

3. Find the Probability:
- The Z-score of [tex]\(-3.0\)[/tex] tells us how far 14.4 seconds is from the mean in terms of standard deviations.
- We use the cumulative distribution function (CDF) for a standard normal distribution to find the probability that a value is less than a given Z-score.
- The CDF value for a Z-score of [tex]\(-3.0\)[/tex] is approximately [tex]\(0.0013498980316300933\)[/tex].

This value represents the probability that a runner completes the race in less than [tex]\(14.4\)[/tex] seconds.

4. Convert the Probability to Percentage:
- To convert the probability to a percentage, we multiply by 100:
[tex]\[ \text{Percentage} = 0.0013498980316300933 \times 100 \approx 0.13498980316300932 \% \][/tex]

5. Select the Appropriate Answer:
- Given the options:
[tex]\[ \begin{array}{c} 0.15 \% \\ 0.30 \% \\ 0.60 \% \\ 2.50 \% \\ \end{array} \][/tex]
- The closest and correct answer is [tex]\(0.15\%\)[/tex].

Therefore, the percentage of runners who have times less than [tex]\(14.4\)[/tex] seconds is approximately [tex]\(0.15\%\)[/tex].
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