The Chemco Company manufactures car tires thar last distance

The Chemco Company manufactures car tires thar last distances. Assuming such distances are normally distributed with a mean of 35,600 miles and a standard deviation of 4,280 miles.

(a). If a tire is randomly selected, what is the probability it last less than 30,000 mi?

(b). If 36 tires are randomly selected, what is the probability they last distances that have a mean less than 30,000 mi?

(c). If the manufacturers want to guarantee the tires so that only 3% will be replaced because of failure before before be guarantee number of miles, for how many miles should the tire be guaranteed?

Solution

Mean ( u ) =35600
Standard Deviation ( sd )=4280
Normal Distribution = Z= X- u / sd ~ N(0,1)                  
a)
P(X < 30000) = (30000-35600)/4280
= -5600/4280= -1.3084
= P ( Z <-1.3084) From Standard Normal Table
= 0.0954                  
b)
P(X < 30000) = (30000-35600)/4280/ Sqrt ( 36 )
= -5600/713.3333= -7.8505
= P ( Z <-7.8505) From Standard NOrmal Table
= 0                  
c)
P ( Z < x ) = 0.03
Value of z to the cumulative probability of 0.03 from normal table is -1.881
P( x-u/s.d < x - 35600/4280 ) = 0.03
That is, ( x - 35600/4280 ) = -1.88
--> x = -1.88 * 4280 + 35600 = 27549.32