For a recent 10k run, the finishers are normally distributed with mean 63 minute

Question

For a recent 10k run, the finishers are normally distributed with mean 63 minutes and standard deviation 10 minutes. Complete parts (a) through (d) below aher oview.pagelofthestandardnormaldistributiontable. Click here to view page 2 of the standard normal distribution table. a. Determine the percentage of finishers with times between 45 and 75 minutes. Approximately 1-1% of finishers had times between 45 and 75 minutes. Round to two decimal places as needed.) b. Detormine the percentage of finishers with times less than 80 minutes. Approximately [1% of firishers had times less than 80 minutes. Round to two decimal places as needed.) c. Obtain and interpret the 40th percentile for the finishing times. The 40th percentile is minutes. Round to two decimal places as needed.) Interpret the 40th percentile. Select the correct choice below and fill in the answer box(es) to complete your choice. (Type integers or decimals rounded to two decimal places as needed.) A. This percentile is the value that is 1,1% less than the average fnishing time. B. This percentile means that% of the finishing times are less than . C. This percentile is the value that is 1% greater than the smallest finishing time. D. This percent1e means that % of the finishina times are equal to (11. lick to select your answerfs)

Explanation / Answer

Ans:

Given that

mean=63

standard deviation=10

a)

z(45)=(45-63)/10=-1.8

z(75)=(75-63)/10=1.2

P(-1.8<z<1.2)=P(z<1.2)-P(z<-1.8)=0.8849-0.0359=0.8490

84.90%

b)

z(80)=(80-63)/10=1.7

P(z<1.7)=0.9554

95.54%

c)40th percentile means that

P(Z<=z)=0.4

z=-0.2533

x=63-0.2533*10=60.47

Option B is correct.(this percentile means that 40% of the finishing times are less than 60.47)

d)8th decile means that

P(Z<=z)=0.8

z=0.8416

x=63+0.8416*10=71.42

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