Question

The life of a type of tire is normally distributed with a mean of 30,000 km and a standard deviation of 2,000 km.

(a) [2] 15% of this type of tire will have a life greater than L km, find the value of L.

(b) [2] Find the probability that the average life of a sample of 100 tires is between 29,750 km and 30,450 km.

Answer 1

Part a)

X ~ N ( µ = 30000 , σ = 2000 )
P ( X > x ) = 1 - P ( X < x ) = 1 - 0.15 = 0.85
To find the value of x
Looking for the probability 0.85 in standard normal table to calculate Z score = 1.0364
Z = ( X - µ ) / σ
1.0364 = ( X - 30000 ) / 2000
X = 32072.8
P ( X > 32072.8 ) = 0.15

Part b)

X ~ N ( µ = 30000 , σ = 2000 )
P ( 29750 < X < 30450 )
Standardizing the value
Z = ( X - µ ) / ( σ / √(n))
Z = ( 29750 - 30000 ) / ( 2000 / √(100))
Z = -1.25
Z = ( 30450 - 30000 ) / ( 2000 / √(100))
Z = 2.25
P ( -1.25 < Z < 2.25 )
P ( 29750 < X̅ < 30450 ) = P ( Z < 2.25 ) - P ( Z < -1.25 )
P ( 29750 < X̅ < 30450 ) = 0.9878 - 0.1056
P ( 29750 < X̅ < 30450 ) = 0.8821