Question

according to a physical fitness organization, the amount of time, in seconds, it takes high school students to run a mile is normally distributed, with a standard deviation of 56 seconds. the accompanying figure shows this distribution and marks where the mean of the distribution is based on what you know about the empirical rule and what you observe in the accompanying figure, which one of the following statements is most accurate?
about 5% of students take longer than 560 seconds to run a mile.
about 68% of students have run times between 336 seconds and 560 seconds.
about 16% of students take less than 392 seconds to run a mile.
about 34% of students take between 336 seconds and 392 seconds to run a mile.
about 95% of students take between 392 seconds and 504 seconds to run a mile.

Explanation:

Step1: Recall the Empirical Rule

The Empirical Rule states that for a normal distribution:

• Approximately \(68\%\) of the data lies within \(1\) standard deviation (\(\sigma\)) of the mean (\(\mu\)): \(\mu\pm\sigma\)
• Approximately \(95\%\) of the data lies within \(2\) standard deviations of the mean: \(\mu\pm2\sigma\)
• Approximately \(99.7\%\) of the data lies within \(3\) standard deviations of the mean: \(\mu\pm3\sigma\)

Given \(\mu = 448\) seconds and \(\sigma=56\) seconds.

Step2: Calculate the values for \(1\), \(2\), and \(3\) standard deviations

• For \(1\) standard deviation:
• Lower bound: \(\mu-\sigma=448 - 56=392\)
• Upper bound: \(\mu+\sigma=448 + 56 = 504\)
• For \(2\) standard deviations:
• Lower bound: \(\mu - 2\sigma=448-2\times56=448 - 112 = 336\)
• Upper bound: \(\mu+2\sigma=448+2\times56=448 + 112=560\)

Step3: Analyze each option

• Option 1:
• \(P(X>560)\): Since \(560=\mu + 2\sigma\), \(P(X>\mu + 2\sigma)=\frac{1 - 0.95}{2}=0.025 = 2.5\%

eq5\%\)

• Option 2:
• \(P(336<X<560)\): Since \(336=\mu - 2\sigma\) and \(560=\mu + 2\sigma\), \(P(\mu - 2\sigma<X<\mu + 2\sigma)=0.95

eq0.68\)

• Option 3:
• \(P(X < 392)\): Since \(392=\mu-\sigma\), \(P(X<\mu-\sigma)=\frac{1 - 0.68}{2}=0.16 = 16\%\)
• Option 4:
• \(P(336<X<392)\): \(336=\mu - 2\sigma\) and \(392=\mu-\sigma\). The proportion between \(\mu - 2\sigma\) and \(\mu-\sigma\) is \(\frac{0.95 - 0.68}{2}=0.135

eq0.34\)

• Option 5:
• \(P(392<X<504)\): Since \(392=\mu-\sigma\) and \(504=\mu+\sigma\), \(P(\mu-\sigma<X<\mu+\sigma)=0.68

eq0.95\)

Answer:

About \(16\%\) of students take less than \(392\) seconds to run a mile.