Question

The average time to run the 5K fun run is 22 minutes and the standard deviation is 2.2 minutes. 44 runners are randomly selected to run the 5K fun run. Round all answers to 4 decimal places where possible and assume a normal distribution.

- What is the distribution of XX? XX ~ N(,)
- What is the distribution of ¯xx¯? ¯xx¯ ~ N(,)
- What is the distribution of ∑x∑x? ∑x∑x ~ N(,)
- If one randomly selected runner is timed, find the probability that this runner's time will be between 21.9024 and 22.2024 minutes.
- For the 44 runners, find the probability that their average time is between 21.9024 and 22.2024 minutes.
- Find the probability that the randomly selected 44 person team will have a total time less than 985.6.
- For part e) and f), is the assumption of normal necessary? NoYes
- The top 20% of all 44 person team relay races will compete in the championship round. These are the 20% lowest times. What is the longest total time that a relay team can have and still make it to the championship round? minutes

Answer #1

a)X ~ N(22,2.2)

b)

sample size =n= | 44 |

std error=σ_{x̅}=σ/√n= |
0.3317 |

x¯ ~ N(22 , 0.3317)

c)

total expected value=22*44 =968

standard deviation =2.2*sqrt(44)=14.5931

∑x ~ N(968 ,14.5931)

d)

probability
=P(21.9024<X<22.2024)=P((21.9024-22)/2.2)<Z<(22.2024-22)/2.2)=P(-0.04<Z<0.09)=0.5367-0.4823=0.0543 |

e)

probability
=P(21.9024<X<22.2024)=P((21.9024-22)/0.332)<Z<(22.2024-22)/0.332)=P(-0.29<Z<0.61)=0.7292-0.3843=0.3449 |

f)

probability =P(X<985.6)=(Z<985.6-968)/14.593)=P(Z<(1.206)=0.8861 |

g)

No

h)

for 20th percentile critical value of z= | -0.842 | ||

therefore corresponding value=mean+z*std deviation= | 955.7181 |

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