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One line of specialty tires has a wear-out life that can be modeled using a normal distribution with a mean of 25,000 km and a standard deviation of 2,000 km. Determine each of the following:

a. The percentage of tires that can be expected to wear out within ±2,000 km of the average (i.e., between 23,000 km and 27,000 km).

b. The percentage of tires that can be expected to fail between 26,000 km and 29,000 km.

c. For what tire life would you expect 4 percent of the tires to have worn out?

Answer :

Answer:

Approximately 68.26% of tires wear out between 23,000 km and 27,000 km, 28.57% wear out between 26,000 km and 29,000 km, and 4% wear out by 21,498 km.

Step-by-step explanation:

Given: Normal distribution with Mean (μ) = 25,000 km and Standard Deviation (σ) = 2,000 km.

a. Percentage between 23,000 km and 27,000 km (μ ± 1σ):

Z-scores: -1.00 and +1.00

Percentage: 68.26%

b. Percentage between 26,000 km and 29,000 km:

Z-scores: 0.50 and 2.00

Percentage: P(0.50 ≤ Z ≤ 2.00) = P(Z ≤ 2.00) - P(Z ≤ 0.50) ≈ 0.9772 - 0.6915 = 0.2857

Percentage: 28.57%

c. Tire life (X) for which 4% have worn out (P(Tire Life ≤ X) = 0.04):

Z-score for 0.04 cumulative probability ≈ -1.751

X = μ + Zσ = 25,000 + (-1.751 * 2,000) ≈ 21,498 km

Tire Life: Approximately 21,498 km

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