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.)