Question

For a recent 10k run, the finishers are normally distributed with mean 65 minutes and standard deviation 15 minutes. Complete parts (a) through (d) below Click here to view.age 1 of the standard normal distribution table Click here to view page 2 of the standard normal distribution table a. Determine the percentage of finishers with times between 50 and 75 minutes Approximately of finishers had times between 50 and 75 minutes. (Round to two decimal places as needed.) b. Determine the percentage of finishers with times less than 30 minutes, Approximately % of finishers had times less than 80 minutes. (Round to two decimal places as needed.) c. Obtain and interpret the 35th percentile for the finishing times. The 35th percentile is minutes. (Round to two decimal places as needed.) Interpret the 35th percentile. Select the correct choice below and fill in the answer boxes) to complete your choice (Type integers or decimals rounded to two decimal places as needed.) O A. This percentile is the value that is less than the average finishing time. OB. This percentle means that % of the finishing times are equal to OC. This percentile is the value that is % greater than the smallest finishing time. This percentile means that of the finishing times are less than d. Find the seventh decile for the finishing times. The seventh decile is mir minutes (Round to two decimal places as needed.) Click to select your answer(s). Save for Later Type here to search ORI 3 Prisen Home

Answer 1

Page 1

let the finishers be denoted by x. (finishing Given Nonnal distribution (es, o 2) u= 65 O = 15 times between Percentage finishers ers with and 75 minutes 50 le P[50 < x < 75] x 100% Now 15 P [50 <x<75] =-2/ - P[507665 <x - 65 < 75-05) P[ - 0.667] z = x-65 (re 2= Y4) Yanny) ~ N(0,1) -) < 2 C Z where From nound table we have P[-C2C 0.667) = 0.5889 58.89 % - > finishers had times between 50 and 75 minutes.

Page 2

P[ x POJZ? P[x-65 < 80-65 15 15 >> P[2<1] = 0.0413 Approximately 84.13 % of finishers had times less than 80 minutes. © 35 th percentile = ? REZ (Zoois] = -2 P[z < Zooss] = 0.35 = 0.35 P[YTES ( 20:35 ] P[ X < 65+ (20.35*15) 5)] = 0,3 Zo.is =-0.3853 From nound table 65+ (-0.3853 XIS) 35 th percentile is 59-2205 Therefore the required 35th percentile & for the finishing 59.22 minutes of time is

Page 3

Inlerpretation : Option D: time This percentile (35th percentile = 59.22) that the finishing less than 59.22 minutes 35% of means au d Seventh decile = ? to 70 th percentile 7th decide is equivalent P[2 < 20.70J = 0.70 PE X- 65 15 20.70] =0.70 > P[ X < 65+ (70.70+15)) = 0.70 From Nound table 20.70 = 0.524 65+(0.524815) » 7th decile = = 72.86 times are that It means 70% of the finishing 72.86 minutes. less than