A certain brand of automobile tire has a mean life span of 37,000 miles and a standard deviation of 2,050 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, 36,000 ​miles, and 32,000 miles. Find the​ z-score that corresponds to each life span.For the life span of 34,000 ​miles, z-score is -1.46.​(Round to the nearest hundredth as​ needed.) For the life span of 36,000 ​miles, z-score is -0.49.​(Round to the nearest hundredth as​ needed.) For the life span of 32,000 ​miles, z-score is -2.44.​(Round to the nearest hundredth as​ needed.) According to the​ z-scores, would the life spans of any of these tires be considered​ unusual?NoYesYour answer is correct.​(b) The life spans of three randomly selected tires are 34,950 ​miles, 39,050 ​miles, and 37,000 miles. Using the empirical​ rule, find the percentile that corresponds to each life span.The life span 34,950 miles corresponds to the 15th percentile.The life span 39,050 miles corresponds to the nothingth percentile.The life span 37,000 miles corresponds to what percentile?

We are given that the mean lifetime mileage of a certain type of tire is `mu=37000` with a standard deviation of `sigma=2050` and that the distribution of tire lifetimes is approximately bell-shaped (or approximately normal).

(a) To find the `z-"score"` for a given lifetime means to find the number of standard deviations above or below the mean. The formula is `z=(x-mu)/sigma` , where x is the lifetime in question.

(i) If x=34000, then `z=(34000-37000)/2050~~-1.46` (Note that the requirements for rounding are not standard. Some might want 1.463 or even 1.5. Consult your text/instructor's directions for rounding rules). This means that this lifetime is about one and a half standard deviations below the mean.

(ii) If x=36000, then `z~~-0.49` (One reason to limit to 2 decimal places is that many standard normal charts are designed with z-scores reported to 2 decimal places).

(iii) If x=32000, then `z~~-2.44`

Since the empirical normal rule indicates that approximately 95% of all values lie within two standard deviations of the mean, a lifetime of 32000 miles is a little unusual.

(b) We can use the z-scores to place values by percentiles within the population. The empirical normal rule states that approximately 68% of the population is within 1 standard deviation of the mean, 95% within 2 standard deviations of the mean, and 99.7% within 3 standard deviations of the mean.

Since the normal curve is symmetric we have:

.0015 lies below 3 standard deviations of the mean.
.0235 lies within 2-3 std. dev. below the mean.
.135 lies within 1-2 std. dev. below the mean.
.34 lies within 1 std. dev. below the mean.
.34 lies within 1 std dev. above the mean.
.135 lies within 1-2 std. dev. of the mean.
.0235 lies within 2-3 std. dev. of the mean.
.0015 lies above 3 standard deviations of the mean.

(i) If x=34950, then x lies 1 standard deviation below the mean. Then .0015+.0235+.125=.16 or 16% of the population lies below that value. So the percentile is 16. (Another way to see this is that 50% lie above the mean and 34% lie within 1 std. dev. below the mean, so 84% lie above this value and 16% below. (Using technology, a closer approximation is 0.1586552596)

(ii) If x=39050, then x lies 1 standard deviation above the mean. So .5+.34=.84 means that this value is at the 84th percentile (84% of the population lies below this value).

(iii) If x=37000, then z=0, which is the 50th percentile. (One-half of the population falls below this score.)