Answer:
Expert's answer
let x denote the number of tires.
x~N(36500,50002)
z=\frac{x-\mu}{\sigma}
σ
x−μ
a) P(x>40000)
z=\frac {40000-36500}{5000}
5000
40000−36500
=0.7
we check the value of p(z>0.7) from the z tables.
=0.24196
24.2% of the tires can be expected to last more than 40000 miles.
b) P(z<0.1)
the value of \phi ^{-1}(0.1)ϕ
−1
(0.1) =-1.28
z=\frac{x-\mu}{\sigma}z=
σ
x−μ
-1.28=\frac{x-36500}{5000}−1.28=
5000
x−36500
=30100 miles
BRAIN LESS ANSWER IS CORRECT