Answer:
The probability that a randomly selected tire will have a life of at least 47,500 miles
P( X≥47,500)= 0.0668
Step-by-step explanation:
Given The life expectancy of a particular brand of tire is normally distributed with a mean of 40,000 and a standard deviation of 5,000 miles
mean μ=40,000
standard deviation of 5,000 miles
σ = 5,000
by using normally distributed
[tex]z= \frac{x-mean}{S.D}[/tex]
here X = 47,500
[tex]z= \frac{47500-40000}{5000}[/tex]
z=1.5
Step(ii)
The probability that a randomly selected tire will have a life of at least 47,500 miles
P( X≥47,500) = P(z≥1.5)
= 0.5- A(z₁)
= 0.5 - A(1.5)
= 0.5-0.4332 ( see from normal table)
= 0.0668
Conclusion:-
The probability that a randomly selected tire will have a life of at least 47,500 miles
P( X≥47,500)= 0.0668
