# A certain brand of automobile tire has a mean life span of 36,000 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 34,000 miles, 38,000 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 0.95 .(Round to the nearest hundredth as​ needed.) For the life span of 38,000 miles, z-score is 0.95 . ​(Round to the nearest hundredth as​ needed.) For the life span of 32,000 ​miles, z-score is −1.90 . ​(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 Your answer is correct. ​(b) The life spans of three randomly selected tires are 31,800 ​miles, 40,200 ​miles, and 36,000 miles. Using the empirical​ rule, find the percentile that corresponds to each life span. The life span 31,800 miles corresponds to the -th percentile. The life span 40,200 miles corresponds to the -th percentile. The life span 36,000 miles corresponds to the - th percentile.

We are given that the population mean (the average lifespan of a certain type of tires in miles) is `mu=36000` with a standard deviation of `sigma=2100` . We are also told that the population distribution is approximately bell-shaped (or normal.)

(a) We are asked to compute the z-score for a set of life expectancies. The z-score gives the number of standard deviations away from the mean; it is positive if the value is above the mean and negative if the value is below the mean. The formula for the z-score is `z=(x-mu)/sigma` where x is the value we are interested in.

(i) For x=34000 we have `z=(34000-36000)/2100~~-0.95` (This means that 34000 miles is just less than 1 standard deviation below the mean.)

(ii) If x=38000 we have `z~~0.95` . This should be expected -- 38000 is 2000 above the mean and 34000 is 200 below the mean, so they should have the same numerical z-score and opposite sign.

(iii) If x=32000 we have `z~~-1.90`

None of these values for x would be particularly unexpected. A value with a z-score greater than 3 or less than 3 would certainly be unexpected.

(b) We can use the z-score to give percentiles for various values of the variable, assuming the underlying population is normal.

The empirical normal rule gives that 68% of the population is within 1 standard deviation of the mean. (34% will be less than 1 standard deviation below the mean and 34% will be less than 1 standard deviation above the mean since the distribution is symmetric.) Also 95% of the population is within two standard deviations of the mean and 99.7% of the population is within 3 standard deviations of the mean.

We end up with:

.0015 of the population is greater than 3 std. dev. below the mean.
.0235 of the population is between 2-3 std. dev. below the mean.
.135 of the population is between 1-2 std. dev. below the mean.
.34 of the population is between 0-1 std. dev. below the mean.
.34 of the population is between 0-1 std. dev. above the mean.
.135 of the population is between 1-2 std. dev. above the mean.
.0235 of the population is between 2-3 std. dev. above the mean.
.0015 of the population is greater than 3 std. dev. above the mean.

(i) For x=31800 we have z=-2. Thus .0015+.0235= .025 of the population is below that value. I.e. 2.5% of the tire lives will be 31800 miles or less. Thus x=31800 is the 2.5th percentile.

(ii) For x=40200 we have z=2. Then .0015+.0235+.135+.34+.34+.135=.975
so 97.5% of the values lie below this value. (By symmetry we see from (i) that 2.5% lie above, so 1-.02.975 is an equally valid approach.)This is the 97.5th percentile.

(iii) For x=36000 we have z=0 which is the 50th percentile. Half of the population lies below the mean (in a normal distribution the mean, median, and mode are the same.)