Question

a certain brand of automobile tire has a mean life span of 34,000 miles and a standard deviation of 2,400 miles. (ass (a) the life spans of three randomly selected tires are 33,000 miles, 38,000 miles, and 31,000 miles. find the z - score for the life span of 33,000 miles, z - score is - 0.42. (round to the nearest hundredth as needed.) for the life span of 38,000 miles, z - score is 1.67. (round to the nearest hundredth as needed.) for the life span of 31,000 miles, z - score is - 1.25. (round to the nearest hundredth as needed.) according to the z - scores, would the life spans of any of these tires be considered unusual? no yes (b) the life spans of three randomly selected tires are 29,200 miles, 38,800 miles, and 34,000 miles. using the empirical the life span 29,200 miles corresponds to a percentile of

Explanation:

Step1: Recall z - score formula

The z - score formula is $z=\frac{x-\mu}{\sigma}$, where $x$ is the value, $\mu$ is the mean, and $\sigma$ is the standard deviation. Given $\mu = 34000$ and $\sigma=2400$.

Step2: Calculate z - score for $x = 29200$

$z=\frac{29200 - 34000}{2400}=\frac{- 4800}{2400}=-2$

Step3: Use the empirical rule for normal distribution

The empirical rule states that for a normal distribution: about 68% of the data lies within $z=- 1$ and $z = 1$, about 95% lies within $z=-2$ and $z = 2$, and about 99.7% lies within $z=-3$ and $z = 3$. The area to the left of $z=-2$ is $\frac{100 - 95}{2}=2.5\%$. So the percentile corresponding to $z=-2$ is 2.5.

Answer:

2.5