Question

A lifetime of certain brand of tires is normally distributed with the mean of 80, 000 kilometers and a standard deviation of 6000 kilometers. a. Probability that a randomly selected tire will have lifetime 70,000 kilometers or more is __________ b. The percent of tires with lifetimes between 65, 000 kilometers and 85, 000 kilometers is ______________ c. The manufacturer of tires wants to establish the warranty on the tires. He wants to replace for free only 4% of the tires with the smallest lifetimes. The length of the warranty period (in kilometers) is__________________ kilometers

Answer 1

Solution :

Given that ,

mean = = 80000

standard deviation = = 6000

P(x > 70000 ) = 1 - P( x < 70000)

= 1- P[(x - ) / < ( 70000 - 80000) / 6000]

=1- P(z < -1.67 )

Using z table,

= 1 - 0.0475

= 0.9525

Probability = 0.9525

b.

P( 65000 < x < 85000 )

= P[( 65000 - 80000) / 6000) < (x - ) /  < ( 85000 - 80000) / 6000 ) ]

= P( -2.5 < z < 0.83 )

= P(z < 0.83) - P(z < -2.5 )

Using z table,

= 0.7967 - 0.0062

= 0.7905

= 79.05%

Answer : 79.05%

c.

The z - distribution of the 4% is,

P(Z < z) = 4%

= P(Z < z ) = 0.04

= P(Z < -1.751 ) = 0.04

z = -1.751

Using z-score formula,

x = z * +

x = -1.751 * 6000 + 80000

x = 69494

Answer : x = 69494 kilometers