Question

Question: If you randomly select a beginner runner from the population have them run a 5k, what are the chances that they would finish it between 25 and 27 minutes? That is, how many times, out of 100, would you randomly select someone that could run the 5k between 25 and 27 minutes? (Hint: use Appendix B.1 to answer this question. Start by calculating the z score for 25 and 27). Remember, you'll need to know the mean and standard deviation that you calculated for your population of runners. (Write a paragraph explaining, No Plagiarism!)

Time to complete 5k (in minutes):

21
21
22
22
23
23
23
24
24
24
24
24
25
25
25
26
26
27
27

Answer: 38.97% Your friend's z score is -0.28 (z = 23.4-24/1.795), the area between the mean and z is 11.03, 50-11.03 = 38.97% (38.97% have times that are lower than this person's score).

Mean: 24

Standard Deviation: 1.795

Answer 1

The data as given in the question: 21 21 22 22 23 23 23 24 24 24 24 24 25 25 25 26 26 27 27 The mean is calculated by: 21+21+ ......+27 +27 19 456 19 = 24 The sample standard deviation is calculated by: (21 – 24)' +(21 – 24)? + ..... + (27 – 24)? 19-1 S=

19+9+..... +9+9 18 58 18 V3.2222 = 1.795 The Z-score for a runner who completes a 5K run in 25 minutes: X-1 = II 25-24 1.795 1.795 = 0.557 The 2-score for a runner who completes a 5K run in 27 minutes: X-M1 -= = 27 - 24 1.795 3 1.795 = 1.67 The probability that a beginner runner would finish the 5k run between 25 and 27 minutes: = P(- =1.67)- P(= = 0.557) = 0.9525 – 0.7112 (By using the standard normal table.) = 0.2413