Question

A factory produces car tires. The number of miles each tire lasts before it completely wares out follows a normal distribution with mean μ = 50,000 miles and standard deviation σ = 8,000 miles. Suppose that the factory promises to the consumers that the tires will last for at least 62,000 miles. What is the probability for a randomly selected tire to last for at least 62,000 miles, i.e., P(x ≥ 62,000)?

Answer 1

Answer: 0.0668073

Explanation:

Given : The number of miles each tire lasts before it completely wares out follows a normal distribution with mean μ = 50,000 miles and standard deviation σ = 8,000 miles.

Let x be the random variable that represents the number of miles each tire lasts.

z-score :
`(x-\mu)/(\sigma)`

For x= 62,000

`z=(62000-50000)/(8000)=1.5`

By using the standard z-value table , the probability for a randomly selected tire to last for at least 62,000 miles will be :_

`P(x\geq62000)=P(z\geq1.5)=1-P(z<1.5)\ \ [\because\ P(Z\geq z)=1-P(Z<z)]\\\\=1-0.9331927\\\\=0.0668073`

Hence, the probability for a randomly selected tire to last for at least 62,000 miles = 0.0668073