Question

a company that manufactures tires produces a tire that has an average life span of 65,000 km with a standard deviation of 5,200 km. the distribution of the life spans of the tire is normal. what percent of the tires last at least 60,000 km? 31.5% 81.0% 82.4% 83.2%

Explanation:

Step1: Calculate the z - score

The formula for the z - score is \(z=\frac{x-\mu}{\sigma}\), where \(x = 60000\), \(\mu=65000\), and \(\sigma = 5200\).

$$ z=\frac{60000 - 65000}{5200}=\frac{- 5000}{5200}\approx - 0.96 $$

Step2: Find the probability using the z - table

We want to find \(P(X\geq60000)\), which is equivalent to \(P(Z\geq - 0.96)\).
Since \(P(Z\geq z)=1 - P(Z\lt z)\), and from the standard normal table \(P(Z\lt - 0.96)=0.1685\).

$$ P(Z\geq - 0.96)=1 - 0.1685 = 0.8315\approx83.2\% $$

Answer:

83.2%