Please only do that one that is incorrect Or isn't done The average time to run the 5K fun run is 25 minutesand the standard deviation is 2.4 minutes. 6 runnersare randomly selected to run the 5K fun run. Round allanswers to 4 decimal places where possible and assumea normal distribution.a. What is the distribution of X? X - N( 252.4b. What is the distribution of ? - N( 250.9798c. What is the distribution of > æ? > x - N(15014.4x) od. If one randomly selected runner is timed, findthe probability that this runner's time will bebetween 24.9303 and 26.1303 minutes.e. For the 6 runners, find the probability that theiraverage time is between 24.9303 and 26.1303minutes.f. Find the probability that the randomly selected 6person team will have a total time less than 141.g. For part e) and f), is the assumption of normalnecessary?O YesO Noh. The top 20% of all 6 person team relay races willcompete in the championship round. These arethe 20% lowest times. What is the longest totaltime that a relay team can have and still make itto the championship round?minutes

Note: Hey, since multiple subparts are posted, we will answer three subparts according to our policy. If you want any specific subpart to be answered, then please submit that subparts only in your message Given, X=Time to run the 5k fun run Average time to run the 5k fun run=μ=25 minutes Standard deviation =σ=2.4 Sample of runners =n=6 X~N25,2.42 c) Distribution of ∑X: The distribution of ∑X is calculated as follows: By additive property of normal distribution, If X~Nμ,σ2 then ∑X~N∑μ,∑σ2 So, ∑Xi~N(25+25+25+25+25+25,2.42+2.42+2.42+2.42+2.42+2.42)∑Xi~N150,34.56 Thus, the distribution of ∑X is ∑Xi~N150,34.56.d) Probability that the runner's time is between 24.9303 and 26.1303: The probability value is calculated as follows: P24.9303<X<26.1303 Note: Z=X-μσ~N0,1 So, P24.9303<X<26.1303=P24.9303-μσ<X-μσ<26.1303-μσ =P24.9303-252.4<Z<26.1303-252.4 =P-0.029<Z<0.4709 Note:P(a<X<b)=ϕb-ϕa where ϕZ is the cumulative distribution of the variable Z. P24.9303<X<26.1303=P-0.029<Z<0.4709 =ϕ0.4709-ϕ-0.029 =ϕ0.4709-1-ϕ0.029 =ϕ0.4709+ϕ0.029-1 ϕ0.4709 is calculated using the EXCEL formula, "=NORMSDIST(0.4709)". The value obtained is 0.6811. ϕ0.029 is calculated using the EXCEL formula, "=NORMSDIST(0.029)". The value obtained is 0.5115. So, P24.9303<X<26.1303=ϕ0.4709+ϕ0.029-1 =0.6811+0.5115-1 =0.1926 Thus, the probability that the runner's time is between 24.9303 and 26.1303 is 0.1926. e)Probability that the average time is between 24.9303 and 26.1303: The probability value is calculated as follows: P24.9303<X<26.1303 Note: If X~Nμ,σ2 then X~Nμ,σ2n and X-μσn~N0,1 So, P24.9303<X<26.1303=P24.9303-μσn<X-μσn<26.1303-μσn =P24.9303-252.46<Z<26.1303-252.46 =P-0.06970.9797<Z<1.13030.9797 =P-0.0711<Z<1.1537 =ϕ1.1537-ϕ-0.0711 =ϕ1.1537-1-ϕ0.0711 =ϕ1.1537+ϕ0.0711-1 ϕ1.1537 value is calculated using the EXCEL formula, "=NORMSDIST(1.1537)". The value obtained is 0.8756. ϕ0.0711 value is calculated using the EXCEL formula, "=NORMSDIST(0.0711)". The value obtained is 0.5283. P24.9303<X<26.1303=ϕ1.1537+ϕ0.0711-1 =0.8756+0.5283-1 =0.4039 Thus, the probability that the average time is between 24.9303 and 26.1303 is 0.4039.