Question

1. The amount of time a tire lasts is normally distributed with a mean of 30,000 km and standard
deviation of 2000 km.
(a) 68% of all tires will last between ____________ km and _____________ km.
(b) 95% of all tires will last between ____________ km and _____________ km.
(c) What percent of the tires will have a life that exceeds 26,000 km? _________
(d) What percent of the tires will have a life that lasts between 28,000 and 34,000 km? ______
(e) If a company purchases 2000 tires, how many tires are expected to last
 more than 28,000 km? ___________
 between 24,000 and 32,000 km? ____________

Answer 1

a)
we need to convert 68% into two-sided z value
so, we have


With the help of formula for z, we can say that

68% of all tires will last between 28011 km and 31989 km

b)
we need to convert 95% into two-sided z value
so, we have


With the help of formula for z, we can say that

95% of all tires will last between 26080 km and 33920 km

This is a normal distribution question with

c)
P(x > 26000.0)=?
The z-score at x = 26000.0 is,

This implies that


d)
P(28000.0 < x < 34000.0)=?

This implies that

PS: you have to refer z score table to find the final probabilities.

e)
In the last part first we will find out the probability for a single tire
and multiply that probability with 2000 to gegt final answer.

(i)
P(x > 28000.0)=?
The z-score at x = 28000.0 is,

This implies that

Number of tires expected to last more than 28,000 km
= 0.8413*2000 = 1682.6 tires

(ii)
P(24000.0 < x < 32000.0)=?

This implies that

Number of tires expected to last in between 24,000 km & 32,000 km
​​​​​​​= 0.84*2000 = 1680 tires
PS: you have to refer z score table to find the final probabilities.