Question

a certain brand of automobile tire has a mean life span of 38,000 miles and a standard deviation of 2,450 miles. (assume the life spans of the tires have a bell - shaped distribution.)
for the life span of 37,000 miles, z - score is - 0.41.
(round to the nearest hundredth as needed.)
for the life span of 31,000 miles, z - score is - 2.86.
(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
no
(b) the life spans of three randomly selected tires are 35,550 miles, 40,450 miles, and 38,000 miles. using the empirical rule, find the percentile that corresponds to each life span.
the life span 35,550 miles corresponds to a percentile of
the life span 40,450 miles corresponds to a percentile of
the life span 38,000 miles corresponds to a percentile of
(round to the nearest whole number as needed.)

Explanation:

Step1: Recall the empirical rule (68 - 95 - 99.7 rule) for a bell - shaped distribution.

The empirical rule states that for a normal (bell - shaped) distribution:

• Approximately 68% of the data lies within $\mu\pm\sigma$ (1 standard deviation of the mean).
• Approximately 95% of the data lies within $\mu\pm2\sigma$ (2 standard deviations of the mean).
• Approximately 99.7% of the data lies within $\mu\pm3\sigma$ (3 standard deviations of the mean).

The mean $\mu = 38000$ miles and the standard deviation $\sigma=2450$ miles.

For 35,550 miles:

First, find how many standard deviations away from the mean this value is.
The formula for the number of standard deviations (z - score) is $z=\frac{x - \mu}{\sigma}$.
Substitute $x = 35550$, $\mu=38000$, and $\sigma = 2450$:
$z=\frac{35550 - 38000}{2450}=\frac{- 2450}{2450}=- 1$
So, $x=\mu-\sigma$.
According to the empirical rule, the percentage of data less than $\mu-\sigma$:
The total percentage of data less than $\mu-\sigma$ is $\frac{1 - 0.68}{2}=0.16$ or 16% (since 68% is within $\mu\pm\sigma$, the remaining 32% is outside, 16% on each side). So the percentile is $16$ (because percentile represents the percentage of data less than or equal to a given value. The data less than $\mu-\sigma$ is 16% of the total data, so the percentile is 16).

For 40,450 miles:

Calculate the z - score: $z=\frac{40450 - 38000}{2450}=\frac{2450}{2450}=1$
So, $x=\mu+\sigma$.
The percentage of data less than $\mu+\sigma$ is $100-\frac{1 - 0.68}{2}=100 - 0.16 = 84$ (because 16% of the data is greater than $\mu+\sigma$, so 84% is less than or equal to $\mu+\sigma$). So the percentile is $84$.

For 38,000 miles:

The z - score is $z=\frac{38000 - 38000}{2450}=0$
This is the mean. In a normal distribution, 50% of the data is less than the mean and 50% is greater than the mean. So the percentile is $50$.

Answer:

The life span 35,550 miles corresponds to a percentile of $\boldsymbol{16}$.
The life span 40,450 miles corresponds to a percentile of $\boldsymbol{84}$.
The life span 38,000 miles corresponds to a percentile of $\boldsymbol{50}$.