Let X= difference in time of arrival with respect to 9.40 am in minutes.
X∼N(μ,σ2). Then Z=σX−μ∼N(0,1)
Given μ=0,σ=20 min
a.
P(X<−40)=P(Z<20−40−0)
=P(Z<−2)≈0.02275 0.02275(250)=6(occasions)
b.
P(X>−20)=1−P(X≤−20)
=1−P(Z≤20−20−0)=1−P(Z≤−1)≈0.84134
0.84134(250)=210(occasions)
c.
P(20<X<40)=P(X<40)−P(X≤20)
=P(Z<2040−0)−P(Z≤2020−0)
=P(Z<2)−P(Z≤1)
≈0.97725−0.84134≈0.13591
0.13591(250)=34(occasions)
Comments