Question

a certain brand of automobile tire has a mean life - span of 33,000 miles and a standard deviation of 2,100 miles. (assume the life - spans of the tires have a bell - shaped distribution.)
(a) the life - spans of three randomly selected tires are 35,000 miles, 36,000 miles, and 32,000 miles. find the z - score that corresponds to each life - span.
for the life - span of 35,000 miles, the z - score is - 3.20
(round to the nearest hundredth as needed.)
for the life - span of 36,000 miles, the z - score is 1.20
(round to the nearest hundredth as needed.)
for the life - span of 32,000 miles, the z - score is - 0.40
(round to the nearest hundredth as needed.)
according to the z - scores, would the life - spans of any of these tires be considered unusual? yes
(b) the life - spans of three randomly selected tires are 30,900 miles, 35,100 miles, and 33,000 miles. using the empirical rule, find the percentile that corresponds to each life - span.
the life - span 30,900 miles corresponds to a percentile of 14
(round to the nearest whole number as needed.)
the life - span 35,100 miles corresponds to a percentile of 84
(round to the nearest whole number as needed.)
the life - span 33,000 miles corresponds to a percentile of 50
(round to the nearest whole number as needed.)

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. Given $\mu = 33000$ and $\sigma=2100$.

Step2: Calculate z - score for $x = 35000$

$z=\frac{35000 - 33000}{2100}=\frac{2000}{2100}\approx0.95$

Step3: Calculate z - score for $x = 36000$

$z=\frac{36000 - 33000}{2100}=\frac{3000}{2100}\approx1.43$

Step4: Calculate z - score for $x = 32000$

$z=\frac{32000 - 33000}{2100}=\frac{- 1000}{2100}\approx - 0.48$

Step5: Determine unusual values

Unusual values have $|z|>2$. Since none of the calculated z - scores have an absolute value greater than 2, the answer is No.

Step6: Recall empirical rule for percentiles

For a normal distribution: about 68% of the data is within 1 standard - deviation of the mean ($z=-1$ to $z = 1$), about 95% is within 2 standard - deviations ($z=-2$ to $z = 2$), and about 99.7% is within 3 standard - deviations ($z=-3$ to $z = 3$).

Step7: Calculate percentile for $x = 30900$

$z=\frac{30900 - 33000}{2100}=\frac{-2100}{2100}=-1$. The percentile corresponding to $z=-1$ is 16 (since about 16% of the data is below $z=-1$ in a normal distribution).

Step8: Calculate percentile for $x = 35100$

$z=\frac{35100 - 33000}{2100}=\frac{2100}{2100}=1$. The percentile corresponding to $z = 1$ is 84 (since about 84% of the data is below $z = 1$ in a normal distribution).

Step9: Calculate percentile for $x = 33000$

$z=\frac{33000 - 33000}{2100}=0$. The percentile corresponding to $z = 0$ is 50 (since the mean corresponds to the 50th percentile in a normal distribution).

Answer:

For the life - span of 35000 miles, the z - score is 0.95.
For the life - span of 36000 miles, the z - score is 1.43.
For the life - span of 32000 miles, the z - score is - 0.48.
According to the z - scores, would the life - spans of any of these tires be considered unusual? No
The life - span of 30900 miles corresponds to a percentile of 16.
The life - span of 35100 miles corresponds to a percentile of 84.
The life - span of 33000 miles corresponds to a percentile of 50.