Answer: 0.0668073
Step-by-step 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 : [tex]\dfrac{x-\mu}{\sigma}[/tex]
For x= 62,000
[tex]z=\dfrac{62000-50000}{8000}=1.5[/tex]
By using the standard z-value table , the probability for a randomly selected tire to last for at least 62,000 miles will be :_
[tex]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[/tex]
Hence, the probability for a randomly selected tire to last for at least 62,000 miles = 0.0668073
