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 (b) the life spans of three randomly selected tires are 31,600 miles, 40,400 miles, and 36,000 miles. using the empirical rule, find the percentile that corresponds to each life span. the life span 31,600 miles corresponds to a percentile of the life span 40,400 miles corresponds to a percentile of the life span 36,000 miles corresponds to a percentile of (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 $\mu = 36000$ 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 unusual values

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

Step6: Recall the empirical rule for normal distribution

The empirical rule states that about 68% of the data lies within 1 standard deviation of the mean ($\mu\pm\sigma$), about 95% lies within 2 standard deviations ($\mu\pm2\sigma$), and about 99.7% lies within 3 standard deviations ($\mu\pm3\sigma$).

Step7: Calculate z - score for $x = 31600$

$z=\frac{31600 - 36000}{2200}=\frac{-4400}{2200}=-2$
Since about 95% of the data lies within $z=- 2$ and $z = 2$, the percentage of data less than $z=-2$ is $\frac{100 - 95}{2}=2.5\%$, so the percentile is 2.5% or 3% (rounded to the nearest whole number).

Step8: Calculate z - score for $x = 40400$

$z=\frac{40400 - 36000}{2200}=\frac{4400}{2200}=2$
The percentage of data less than $z = 2$ is $95\%+\frac{100 - 95}{2}=97.5\%$ or 98% (rounded to the nearest whole number).

Step9: Calculate z - score for $x = 36000$

$z=\frac{36000 - 36000}{2200}=0$
The mean corresponds to the 50th percentile.

Answer:

The life span 31600 miles corresponds to a percentile of 3
The life span 40400 miles corresponds to a percentile of 98
The life span 36000 miles corresponds to a percentile of 50