A certain brand of tires of automobile tires has a mean life spanof 35,000 miles

Question

A certain brand of tires of automobile tires has a mean life spanof 35,000 miles and a standard deviation of 2250 miles. (Assume thelife spans of the tires have a bell-shaped distribution.)

(a) The life spans of the three randomly selected tires are 34,000miles, 37,000 miles, and 31,000 miles. Find the z-score thatcorresponds to each life span. According to the z-scores, would thelife span of any of these tires be considered unusual?

(b) The life span of three randomly selected tires are 30,500miles, 37,250 miles, and 35,000 miles. Using the Empirical Rule,find the percentile that corresponds to each life span.

For part A I understand how to find the z-scores by using theequation: Z = X - /

34,000 miles z-score = -.44
37,000 miles z-score = .88
31,000 miles z-score = -1.77

If any of the z-scores are wrong please let me know and I alsounderstand whether or not the scores are considered unusual. Forthis care the scores are not considered unusual because they fallbetween -2 and 2.

I needed help for part B, I don't know where to begin withfinding the percentiles.

Explanation / Answer

Your formula is correct for the z-scores. A) I got .89 for the 2nd one and -1.78. Just a small roundingerror. Looking at a z-table, 1.78 has a p-value of.0375. This indicates that there is only a 3.75% chance ofseeing a tire lifetime of this small or less (quite small). When doing a z-test, you typically look for a p-value of .025chance or less when doing a 2-sided test (or .05 or less when doinga 1-sided test). The other two tires have have somewhathigher p-values and are less noticeable. B) z(30500) = -2 z(37250) = 1 z(35000) = 0 http://en.wikipedia.org/wiki/68-95-99.7_rule -1 to 1 stdev encloses 68% of the outcomes in a normaldistribution. -2 to 2 stdev encloses 95% of the outcomes -3 to 3 stdev encloses 99.7% of the outcomes Thus, a z score of -2 implies that 5% (1-95%) lies outside of the-2 to 2 range. This is divided evenly due to the symmetricnature of the curve. As a result, a z-score of -2 correspondsto the 2.5% Similarly, with a z score of 1, (1 - 68%) / 2 of the outcomes lieto the right. This means that 1 - (1- 68%)/2 lie to theleft. This means that a z-score of 1 corresponds to the84%. A z score of 0 puts you at the mean. Since the curve issymmetric, this means you are also at the median or 50%.

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