Question

1. A tire company produces a tire that has an average life span of 500 miles with a standard deviation of 250. The distribution of the life spans of the tires is normal. What is the probability that the tires lasts between 530 and 375 miles? (Round three decimal places)

2. A tire company produces a tire that has an average life span of 480 miles with a standard deviation of 30. The distribution of the life spans of the tires is normal. What is the probability that the tires lasts less than 430 miles? Round your answer to three decimal places

3. A tire company produces a tire that has an average life span of 480 miles with a standard deviation of 25. The distribution of the life spans of the tires is normal. What is the probability that the tires lasts greater than 498 miles? (Round three decimal places)

Answer 1

1)

Here, μ = 500, σ = 250, x1 = 375 and x2 = 530. We need to compute P(375<= X <= 530). The corresponding z-value is calculated using Central Limit Theorem

z = (x - μ)/σ
z1 = (375 - 500)/250 = -0.5
z2 = (530 - 500)/250 = 0.12

Therefore, we get
P(375 <= X <= 530) = P((530 - 500)/250) <= z <= (530 - 500)/250)
= P(-0.5 <= z <= 0.12) = P(z <= 0.12) - P(z <= -0.5)
= 0.548 - 0.309
= 0.239

2)

Here, μ = 480, σ = 30 and x = 430. We need to compute P(X <= 430). The corresponding z-value is calculated using Central Limit Theorem

z = (x - μ)/σ
z = (430 - 480)/30 = -1.67

Therefore,
P(X <= 430) = P(z <= (430 - 480)/30)
= P(z <= -1.67)
= 0.047

3)

Here, μ = 480, σ = 25 and x = 498. We need to compute P(X >= 498). The corresponding z-value is calculated using Central Limit Theorem

z = (x - μ)/σ
z = (498 - 480)/25 = 0.72

Therefore,
P(X >= 498) = P(z <= (498 - 480)/25)
= P(z >= 0.72)
= 1 - 0.764 = 0.236