Question

a certain brand of automobile tire has a mean life span of 34,000 miles and a standard deviation of 2,400 miles. (assume the life span of the tires)
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 rule, find the percentile that corresponds to
the life span 29,200 miles corresponds to a percentile of

Explanation:

Step1: Recall the 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, $\mu = 34000$ and $\sigma=2400$.

Step2: Recall the range of usual z - scores

Usual z - scores are in the range $- 2\leq z\leq2$. Given $z_{1}=-0.42$, $z_{2}=1.67$, $z_{3}=-1.25$. Since all these z - scores are within the range $[-2,2]$, none of the life - spans are unusual.

Step3: Find the z - score for $x = 29200$

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

Step4: Use the empirical rule for the percentile of $z=-2$

The empirical rule states that for a normal distribution, about 95% of the data lies within $z=-2$ and $z = 2$. The area to the left of $z=-2$ is $\frac{100 - 95}{2}=2.5\%$. So the life - span of 29200 miles corresponds to the 2.5th percentile.

Answer:

No
2.5