Question

It is estimated that the settlement amount of a building does not exceed 5 cm with a 95% probability. In addition, through the records of similar buildings, it was found that the coefficient of variation (cov=standard deviation/average) for settlement was about 20%. When the settlement amount of this building follows a normal distribution, what is the probability of having a settlement amount of more than 5.5cm?

Answer 1

let mean is µ and standard deviation is σ

for 95 percentile: critical z =1.645

therefore 95th percentile: mean+z*standard deviation=µ+1.645*σ=5 .........(1)

also coefficient of variation =σ/µ =0.20

solving above : µ+1.645*0.2*µ =5

µ=5/(1+1.645*0.2)=3.7622

and σ =0.2*3.7622=0.7524

therefore from normal distribution:

probability of having a settlement amount of more than 5.5cm :

probability =P(X>5.5)=P(Z>(5.5-3.76222723852521)/0.752)=P(Z>2.31)=1-P(Z<2.31)=1-0.9896=0.0104