Tires have a normal distribution with a mean of 70,000 miles and a standard deviation of 4,400 miles. What proportion of tires will last at least 75,000 miles?

Let T = the distance, in miles, a tire lasts
T ~ N(70000,4400 ² )
P(T ≥ 75000) = P(Z ≥ 4400 75000 − 70000 ​ )
= P(Z ≥ 22 25 ​ )
≈ P(Z ≥ 1.14) = 1 - P(Z < 1.14) ≈ 1 - 0.8729 = 0.1271

Answered by superkoopasirf | 2024-06-10

Approximately 12.71% of tires are expected to last at least 75,000 miles, based on the normal distribution with a mean of 70,000 miles and a standard deviation of 4,400 miles. We calculated this by standardizing 75,000 miles to find its Z-score and then determining the corresponding probabilities. Finally, we subtracted the left-tail probability from 1 to find the proportion in the right tail.
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Answered by superkoopasirf | 2024-12-26