Question

I have got everything correct except for D. If you could be detailed on how you calculate D with the numbers, that would be great!!

For a recent 10k run, the finishers are normally distributed with mean 59 minutes and standard deviation 12 minutes. Complete parts (a) through (d) below. Click here to view page 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 45 and 70 minutes. Approximately 70.02 % of finishers had times between 45 and 70 minutes. (Round to two decimal places as needed.) b. Determine the percentage of finishers with times less than 75 minutes. Approximately 90.82% of finishers had times less than 75 minutes. (Round to two decimal places as needed.) c. Obtain and interpret the 35th percentile for the finishing times. The 35th percentile is 54.32 minutes. (Round to two decimal places as needed.) Interpret the 35th 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.) CA. This percentile means that 35 % of the finishing times are less than 54.32. OB. This percentile means that % of the finishing times are equal to OC. This percentile is the value that is % greater than the smallest finishing time. D. This percentile is the value that is % less than the average finishing time. d. Find the seventh decile for the finishing times. The seventh decile is 68.44 minutes. (Round to two decimal places as needed.)

Answer 1

Given: el = 59 6 = 12 a) with times between Percentage of and 70 finishers minutes is, 45 P( 45 < x <70) P(x< 70) - P(XL45) P(x-4 < 70-2) - P(x-4 (45-21) P(Z < 70-59-412<4575) P(Z < 0.9167) - P(Z < -1.1667) 0.217 0.8204 PC 45 < «<70) 0.6987 In percent, PC45 < x < 70) - 69.87 % finishers . Thie Percentage of 45 and To minutes with times 69.87 % between is with Hmes less b) Percentage of finishers than 75 minutes is P( * < 75) P ( x < 7524) PC 75-59 z < 12 P(Z < 1.3333) 0.90 88 PCXC75) CSScanned with CamScanner

In percent. ( x (15) = 90.88 % :. The with times percentage of finishers less than 75 minutes is go.88 % 35th percentile for the finishing times is, X-u 35 % 35 al 6 x- c 100 0.35 roll 6 using Excel, NORM SINV (0.35) =-0.3853 X- 59 -0.3853 Il 12 12 X(-0.3853) +59 54.38 for the 35 th finishing & The percentile 54.38, times is & that ob means 35% The percentile 54.38 times are less than the finishing CSScanned with CamScanner

d) Seventh decile for the finishing times, is, 7 11 a-u 6 10 X-4 e-4 6 0-7 By using encel, NORMS INV (0.7) = 0.5244 X-59 12 0.5244 re 12* 0.5244 + 59 TX = 65.29 for the finishing : The decile serenth times is CSScanned with CamScanner 65.29.