5 Fintse re finish te was measured 0 ad ecively If the 5 Fif

5- Fintse re finish te was measured 0 ad, ecively If the 5- Fifty three men athletes were qualified for a downhill race competition and participated in the race. The finish time was measured in second and, based on the results, the mean and the standard deviation of the data were 113.02 and 3.24, respectively. If the Normal model is appropriate,: a- What percentage of the finish time is more than 109.78 seconds? b- What percentage of the competitors finished the race between 110.72 and 116.58 second?

Solution

Normal Distribution
Mean ( u ) =113.02
Standard Deviation ( sd )=3.24
Normal Distribution = Z= X- u / sd ~ N(0,1)                  
a)
P(X > 109.78) = (109.78-113.02)/3.24
= -3.24/3.24 = -1
= P ( Z >-1) From Standard Normal Table
= 0.8413                  
b)
To find P(a < = Z < = b) = F(b) - F(a)
P(X < 110.72) = (110.72-113.02)/3.24
= -2.3/3.24 = -0.7099
= P ( Z <-0.7099) From Standard Normal Table
= 0.23889
P(X < 116.58) = (116.58-113.02)/3.24
= 3.56/3.24 = 1.0988
= P ( Z <1.0988) From Standard Normal Table
= 0.86406
P(110.72 < X < 116.58) = 0.86406-0.23889 = 0.6252