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This week, a very large running race (5K) occured in Denver. The times were normally distributed,...

This week, a very large running race (5K) occured in Denver. The times were normally distributed, with a mean of 20.91 minutes and a standard deviation of 2.17 minutes.
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a. What percent of runners took 16.5 minutes or less to complete the race?  %


b. What time in minutes is the cutoff for the fastest 10.65 %?  Minutes


c. What percent of runners took more than 15 minutes to complete the race? %This week, a very large running race (5K) occured in Denver. The times were normally distributed, with a mean of 20.91 minutes and a standard deviation of 2.17 minutes.
Report your answers accurate to 2 decimals


a. What percent of runners took 16.5 minutes or less to complete the race?  %


b. What time in minutes is the cutoff for the fastest 10.65 %?  Minutes


c. What percent of runners took more than 15 minutes to complete the race? %

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Answer #1

Define random variable X: The time taken by runner to complete the race

X follows normal distribution with mean = \mu = 20.91 and standard deviation = \sigma = 2.17

a)

Here we have to find P(X\leq 16.5)

P(X\leq 16.5)=P(\frac{X-\mu}{\sigma}\leq \frac{16.5-\mu}{\sigma})

= P(Z < 16.5 – 20.91 2.17 where z is standard normal variable

=P(z\leq \frac{-4.41}{2.17})

  =P(z\leq -2.03) (Round to 2 decimal)

= 0.0212 (From statistical table of z values)

2.12% of runners took 16.5 minutes or less to complete the race.

b)
We have to find x such that P(X > x) = 0.1065

1 - P(X < x) = 0.1065

P(X < x) = 0.8935

z for area = 0.8935 is

z = 1.25 (From statistical table of z values)

x=\mu+z*\sigma

x=20.91+1.25*2.17

x = 20.91 + 2.702428

x = 23.61 (Round to 2 decimal)

23.61 minutes is the cutoff for the fastest 10.65%

c)

Here we have to find P(X> 15)

P(X>15)=P(\frac{X-\mu}{\sigma}> \frac{15-\mu}{\sigma})

=P(z> \frac{15-20.91}{2.17}) where z is standard normal variable

=P(z> \frac{-5.91}{2.17})

  =P(z> -2.72) (Round to 2 decimal)

= 1 - P(z < -2.72)

= 1 - 0.0033 (From statistical table of negative z values)

= 0.9967

99.67% of runners took more than 15 minutes to complete the race.

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