Question

Life Span of Tires: A certain brand of automobile tires has a mean life span of 35,000 miles and a standard deviation or of 2,250 miles. (Assume a bell-shape distribution).

1. The span of three randomly selected tires are 34,000 miles, 37,000 miles, and 31,000 miles. Find the z-scores that correspond to each life span. Would any of these tires be considered unusual?
2. The life spans of three randomly selected tires are 30,500 miles, 37,250 miles, and 35,000 miles. Using the Empirical Rule, find the percentile that corresponds to each life span.

Answer 1

a)

for 34,000 miles

Z =(X - µ ) / σ = (   34000   -   35000   ) /    2250
Z =    -0.444              

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for 37000 miles

Z =(X - µ ) / σ = (   37000   -   35000   ) /    2250
Z =    0.889              
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for 31000 miles

Z =(X - µ ) / σ = (   31000   -   35000   ) /    2250
Z =    -1.778              

No, any of these tires would not be considered unusual because z score is within (-2 , 2)

b)

for 30500

Z =(X - µ ) / σ = (   30500   -   35000   ) /    2250
Z =    -2.000              

95% of data lie within 2 std dev away mean
Using the Empirical Rule, percentile that corresponds to life span=(100-95)/2=5th percentile

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for 37250

Z =(X - µ ) / σ = (   37250   -   35000   ) /    2250
Z =    1.000              
68% of data lie within 1 std dev away mean
Using the Empirical Rule, percentile that corresponds to life span=(100-68)/2 + 68%=84th percentile

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Z =(X - µ ) / σ = (   35000   -   35000   ) /    2250
Z =    0.000              

Using the Empirical Rule, percentile that corresponds to life span=50th percentile