Question

a certain brand of automobile tire has a mean life span of 36,000 miles and a standard deviation of 2,200 miles (assume the life spans of the tires have a bell - shaped distribution.) (a) the life spans of three randomly selected tires are 34,000 miles, 37,000 miles, and 31,000 miles. find the z - score that corresponds to each life span. for the life span of 34,000 miles, z - score is - 0.91 (round to the nearest hundredth as needed.) for the life span of 37,000 miles, z - score is 0.45 (round to the nearest hundredth as needed.) for the life span of 31,000 miles, z - score is - 2.27 (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

Explanation:

Step1: Recall z - score formula

The z - score formula is $z=\frac{x-\mu}{\sigma}$, where $x$ is the value from the data set, $\mu$ is the mean, and $\sigma$ is the standard deviation. Here, $\mu = 36000$ and $\sigma=2200$.

Step2: Calculate z - score for $x = 34000$

$z=\frac{34000 - 36000}{2200}=\frac{- 2000}{2200}\approx - 0.91$

Step3: Calculate z - score for $x = 37000$

$z=\frac{37000 - 36000}{2200}=\frac{1000}{2200}\approx0.45$

Step4: Calculate z - score for $x = 31000$

$z=\frac{31000 - 36000}{2200}=\frac{-5000}{2200}\approx - 2.27$

Step5: Determine if the values are unusual

In a normal distribution, values with $|z|>2$ are considered unusual. Since $| - 2.27|>2$, the life - span of 31000 miles is unusual.

Answer:

Yes