Given that the bearings are normally distributed with mean of 4 cm and standard deviation of 0.2 cm, then to evaluate the proportion of of the bearings that have the diameter of less than 3.7 cm we proceed as follows:
P(x<3.7)=P(z<Z)
Z=(x-μ)/σ
where:
μ=4 cm
σ=0.2 cm
thus
Z=(4-3.7)/0.2
Z=0.15
Hence:
P(z<0.15)=0.5596
hence the answer is 55.96%
