Question

6. The mileage that car owners get with a certain kind of radial tire is normally distributed with a mean of 70,000 miles.

1. If 97.725% of these tires lasted less than 78,000 miles, please calculate the standard deviation for the tire mileage. [Note: Please round your result for the standard deviation to the nearest integer.]                                                               [2 points]
2. What is the probability a randomly selected tire of this type will last 65,000 miles? Please justify your answer.                                                                        [1 point]
3. Calculate the probability that a randomly selected tire of this type will last more than 75,000 miles. Show the necessary steps.                                                          [ 2 points]
4. Calculate the probability that a randomly selected tire of this type will last less than 60,000 miles. Show the necessary steps.                                                   [2 points]
5. Calculate the probability that a randomly selected tire of this type will last between 62,000 and 69,000 miles. Show the necessary steps.                       [2 points]
6. What is the tire mileage above which 75 percent of these tires will last? Show the necessary steps.                                                                                          [2 points]
7. What is the tire mileage below which 40 percent of these tires will last? Show the necessary steps.                                                                              [2 point]
8. If the worst performing 3% of these tires are eligible for free-replacement under the company’s product warrantee policy, what is the cutoff level of tire duration for free-replacement? Please show how you arrived at your answer.                    [3 points]

Answer 1

a)

Z= (X - mean)/sd

P(X < 78000) = 0.97725

P(Z < z) = 0.97725

z = 2

mean + 2* sd = 78000

70000 + 2*sd = 78000

sd = 4000

b)

P(X > 65000)

= P(Z > (65 - 70)/4)

= P(Z > -1.25 )

= 0.8944

c)

P(X > 75000)

= 0.1056

d)

P(X < 60000)

= 0.0062

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By HomeworkLib answering guidelines, we have to answer only first 4 sub-parts in multiple sub-parts

Answer 2

To solve this problem, we can use the standard normal distribution.

Step 1: Find the z-score corresponding to the given percentage. From the standard normal distribution table, we find that the z-score corresponding to the 97.725th percentile is 2.15.

Step 2: Use the formula z = (x - mu) / sigma to solve for the standard deviation. Rearranging the formula, we get sigma = (x - mu) / z. Plugging in the given values, we get: sigma = (78000 - 70000) / 2.15 ≈ 3721

Therefore, the standard deviation for the tire mileage is approximately 3721.

To answer the other questions:

The probability that a randomly selected tire of this type will last 65,000 miles is calculated as follows: z = (65000 - 70000) / 3721 ≈ -1.35 Using the standard normal distribution table, we find that the probability is P(Z < -1.35) ≈ 0.088.

The probability that a randomly selected tire of this type will last more than 75,000 miles is calculated as follows: z = (75000 - 70000) / 3721 ≈ 1.34 Using the standard normal distribution table, we find that the probability is P(Z > 1.34) ≈ 0.0918.

The probability that a randomly selected tire of this type will last less than 60,000 miles is calculated as follows: z = (60000 - 70000) / 3721 ≈ -2.69 Using the standard normal distribution table, we find that the probability is P(Z < -2.69) ≈ 0.0036.

The probability that a randomly selected tire of this type will last between 62,000 and 69,000 miles is calculated as follows: z1 = (62000 - 70000) / 3721 ≈ -2.15 z2 = (69000 - 70000) / 3721 ≈ -0.27 Using the standard normal distribution table, we find that the probability is P(-2.15 < Z < -0.27) ≈ 0.6042.

To find the tire mileage above which 75 percent of these tires will last, we need to find the z-score corresponding to the 75th percentile from the standard normal distribution table, which is approximately 0.67. Then, we use the formula z = (x - mu) / sigma to solve for x: 0.67 = (x - 70000) / 3721 x ≈ 71423 Therefore, 75 percent of the tires will last above 71423 miles.

To find the tire mileage below which 40 percent of these tires will last, we need to find the z-score corresponding to the 40th percentile from the standard normal distribution table, which is approximately -0.25. Then, we use the formula z = (x - mu) / sigma to solve for x: -0.25 = (x - 70000) / 3721 x ≈ 68936 Therefore, 40 percent of the tires will last below 68936 miles.

To find the cutoff level of tire duration for free-replacement, we need to find the z-score corresponding to the 3rd percentile from the standard normal distribution table, which is approximately -1.88. Then, we use the formula z = (x - mu) / sigma to solve for x: -1.88 = (x - 70000) / 372