Question

The average time to run the 5K fun run is 22 minutes and the standard deviation is 2.3 minutes. 47 runners are randomly selected to run the 5K fun run. Round all answers to 4 decimal places where possible and assume a normal distribution. a. What is the distribution of X? X-NI b. What is the distribution of æ? - NCI c. What is the distribution of «? *-N d. If one randomly selected runner is timed, find the probability that this runner's time will be between 21.8967 and 22.2967 minutes. e. For the 47 runners, find the probability that their average time is between 21.8967 and 22.2967 minutes. f. Find the probability that the randomly selected 47 person team will have a total time more than 1029.3. g. For parte) and f), is the assumption of normal necessary? O No Yes h. The top 10% of all 47 person team relay races will compete in the championship round. These are the 10% 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

Then average time to run the sk run fun ai is 22 minutes and the standard deviation is 2.3 minutes... 47 gunner's are randomly selected to run the sk Given data, o catutortib Mean: = 22 minutes Standard deviation ir=2.3 minutes Sample Sizeen) :44 dhe (a) what is the distribution of X? x follows normal distribution with Parameters, (x~ N(M= 22, & ² (2.37²) (6) what is the distribution of ň? As per central limit theorem, n230 distribution of SO روز 3 In NC 1222,0? (2.33%) :ہ Mean 2.22) Cfn) Standard deviation - 2.3 = 0.3355 147 INN(4=22, 8?- (0.3355²), CC) What is the distribution of Σα? sampling distribution of Sample Sum is a normal

mean = n ll= (47) (22) = 1034 1.034 Po I standard deviation = n 6=(147) (2.3) = 15.7680 Ex oN(M = 1034, 8² = 45.7680) 2) 9 S.OST POSS

از زیر Cod). Ti one randomly selected dunner is timed Find the probability that this runner's time be between 21.8967 and 22. 2967 minutes. want to find. P(21.89682X<22.2967 )= P(21.8967-22 < X-4 < 22.2967-22 ولی کی 2.3 p 2.3 = P(-0:0449 < ZL 0.129) (Pool - El sol (2<0.129) - PC2-0.0449) o8df.al = 0.55132 -0.51791 = 0.03341 P(21.8967<« <22.2967) - 010334 18.09 The probability that randomly selected runner will be between 21.8967 time and 22.2967 I minutes is abo.0334 tilld day

(e) for the 47 vunners, find the probability that their average time is between 21.8967 io and 22.2907 minutes. we want to find u na P(21.89675 < 22.29.7) = P(21.8867-22 <IMC 22.2957-22 lit i renoor betale 0.3355 al Peso be = P603079 <2<0.8844) - P(Z <0.8844)- P(Z <-0.3079) 19-2 0-81176-0-37908 P(21.896745<22.2967). 0.43268 ñ 0.4327 0.3355 time for Monit The probability that the average the 47 runner's is between 21.8967 and 2011 22.2961 is 0.43271 1 more (F) Find the probability that the mandomly selected 47 person teams will have a total time than to 29.3 → we want to find P( Ex > 4029.3) - P(Z -4 1023.9780034) = P(Z > -0.2981) P(Z < 0.2981) PCEx >10 29.3) 0.61719 70:61.72 The probability that ranelomly selected 47 person team will have a total time than mome

1029.3 is 0.6172 (g) for past (e) and (f), is the assumption of normal necessary? No o yes As per central limit theorem , any distribution Can be approximated to normal it is sufficiently large. +