I posted this question before and all of them were wrong . Can you please re send correctly. I don't need the part e and g The average time to run the 5K fun run is 20 minutes and the standard deviation is 2.4 minutes. 41 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(С.c. What is the distribution of) x? ) x - Nd. If one randomly selected runner is timed, find the probability that this runner's time will be between19.5378 and 20.0378 minutes.e. For the 41 runners, find the probability that their average time is between 19.5378 and 20.0378minutes.f. Find the probability that the randomly selected 41 person team will have a total time more than 795.4.g. For part e) and f), is the assumption of normal necessary? ONOO Yesh. The top 20% of all 41 person team relay races will compete in the championship round. These are the20% 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 Theorem [+]• Finding Value Given a Probability Using the Central Limit Theorem [+]• The Central Limit Theorem For Sums [+]Hint 2

Name: I posted this question before and all of them were wrong . Can you ple
Uploaded: 2024-03-20T06:38:44.000Z
Channel: Bartleby
Description: I posted this question before and all of them were wrong . Can you please re send correctly. I don't need the part e and g The average time to run the 5K fun run is 20 minutes and the standard deviation is 2.4 minutes. 41 runners arerandomly selected

mean = 20 s = 2.4 n = 41a)The distirbutionof X is X~N(20,2.4) b) The standard deviation of the mean isSE = σn SE = 2.441 = 0.3748 The distribution of the mean is x ~N(20,0.3748) c)The distribution of sum x is ∑x ~ N(20 x41,2.4 x 40)∑x ~ N(820,96) d)The probability is given by P(19.5378 < X < 20.0378) = P(19.5378-uσ < X -uσ< 20.0378-uσ) P(19.5378 < X < 20.0378) = P(19.5378-202.4 <Z< 20.0378-202.4) P(19.5378 < X < 20.0378) = P(-0.1926 <Z< 0.01575) P(19.5378 < X < 20.0378) =0.08265 f) The probability is given by P(X>795.4) = P(X-uσ > 795.4 -uσ)P(X>795.4) = P(Z > 795.4 -82096)P(X>795.4) = P(Z > -0.2563)P(X>795.4) = 0.6011