Question

The lifetime of a particular brand of tire is modeled with a normal distribution with mean μ = 75,000 miles and standard deviation σ = 5,000 miles.

a) What is the probability that a randomly selected tire lasts less than 67,000 miles?

b) If a random sample of 35 tires is taken, what is the probability that the sample mean is greater than 70,000 miles?

Answer 1

Solution :

Given that ,

mean = = 75000 miles

standard deviation = = 5000 miles     

P(X< 67000) = P[(X- ) / < (67000 - 75000) / 5000]

= P(z <-1.6 )

Using z table

probability = 0.0548   

b). If a random sample of 35 tires is taken

so,

Given that ,

mean = = 75000

standard deviation = = 5000

n=35

sampling distribution of sample mean = 75000

standard error = / n = 5000 / 35 = 845.15

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

= 1 - P[( - ) / < (70000-75000) /845.15 ]

= 1 - P(z <-5.92 )

Using z table

= 1 - 0

probability= 1.000