To solve this problem, we use the t statistic. The formula for z score is:
z = (x – u) / s
where x is the sample value = 15.5 seconds or below, u is the sample mean = 14.62 seconds, s is standard dev = 2.13 seconds
z = (15.5 – 14.62) / 2.13
z = 0.41
Using standard distribution tables at z = 0.41, the value of P is:
P = 0.6591 = 65.91%
Hence there is a 65.91% chance the runner will have less than 15.5 seconds of time
answer:
c. 0.6591