Part a, b and c please The average time to run the 5K fun run is 24 minutes and the standard deviation is 2.7 minutes. 16 runners arerandomly selected to run the 5K fun run. Round all answers to 4 decimal places where possible and assume anormal distribution.a. What is the distribution of X? X - N(b. What is the distribution of ? - N(c. What is the distribution of ? a - N(С.d. If one randomly selected runner is timed, find the probability that this runner's time will be between23.1875 and 23.9875 minutes.e. For the 16 runners, find the probability that their average time is between 23.1875 and 23.9875minutes.f. Find the probability that the randomly selected 16 person team will have a total time less than 372.8.For part e) and f), is the assumption of normal necessary? O No Yesg.h. The top 10% of all 16 person team relay races will compete in the championship round. These are the10% lowest times. What is the longest total time that a relay team can have and still make it to thechampionship round?minutesHint:Some Helpful Videos:• Finding the Sampling Distribution [+]• Finding a Probability Using the Central Limit TheoremFinding Value Given a Probability Using the Central Limit Theorem 2 [+]• The Central Limit Theorem For Sums E [+][+]HintRasmusseREhabits-9 gedpdf 7-

Name: Part a, b and c please The average time to run the 5K fun run is 24 mi
Uploaded: 2024-03-05T11:53:49.000Z
Channel: Bartleby
Description: Part a, b and c please The average time to run the 5K fun run is 24 minutes and the standard deviation is 2.7 minutes. 16 runners arerandomly selected to run the 5K fun run. Round all answers to 4 decimal places where possible and assume anormal dist

Given that the average time to run the 5K fun run is 24 minutes and the standard deviation 2.7minutes. 16 runners are randomly selected to run the 5k fun run. a) From the given information, the distribution of X follows normal with mean 24 and standard deviation 2.7. X~N(24,2.7). b) Central Limit Theorem for mean: If a random sample of size n is taken from a population having mean µ and standard deviation σ then, as the sample size increases, the sample mean approaches the normal distribution with mean µ and standard deviation σ/ sqrt(n). The mean and standard deviation are obtained as follows: μx¯=μ=24σx¯=σn=2.716=0.675 x¯~N24,0.675c) The mean and standard deviation is calculated as follows; μ∑x=nμ=16×24=352σ∑x=nσ=16×2.7=10.8 The distribution of ∑x with mean 352 and standard deviation is 10.8. Thus, ∑x~352,10.8