1.X 1 (both) = (2.45+2.53+2.48+2.55+2.51)/5 = 2.504 Error = 2.504-2.50 = 0.004.

X2 (average) = (2.36+2.38+2.48+2.50+2.42)/5 = 2.428 Error =2.50-2.428=0.072.

2. The average number of additional periods = (427+439+428+438+436+440+432+435+436+439)/10 = 435.

Distance = 435 *100/22 =1977.3 (m)

3.36000*0.25+ 30000*0.3+20000*0.4+ 15000*0.5=9000+9000+8000+7500=33500

36000+30000+20000+ 15000= 10 1000

n = 33500/ 10 1000 = 0.33