the lytton rubber company wishes set a minimum mileage guarantee on its new Mx 100 tire. Tests reveal the mean mileage is 67,900 with the standard deviation of 2,050 miles and that the deviation follows normal distribution. They want to set the minimum mileage guarantee so that no more than 4 percent of tires will have to be replaced. Due to the costly nature of the test the company decided to take a sample of 200 tires.
A.What average guaranteed mileage should Lytton announce ?
B. Assume that if tires mileage is below 67,500 the company is obliged to return back the used tire ? What is the probability of returning back? How many tires retune buck out of the sample ?
C. Assume that the company wants to know the 95% confidence interval for the average mileage of its tires. What would be the interval?
D. If the company asks you to state the appropriate sample size to determine 99% confidence interval of the average mileage with the error not more than 1000 miles what would be your recommendation ?
Statistics
first question answer
statistics
Lytton rubber company
566385
A. Z = (score-mean)/SD
Find table in the back of your statistics text labeled something like "areas under normal distribution" to find the proportion/probability (.04) and its Z score. Insert Z into above equation and solve for score.
A. To determine the average guaranteed mileage that Lytton should announce, we need to find the minimum mileage value that corresponds to the top 4% of the distribution. Since the distribution is normal, we can use the Z-score formula to find this value.
1. First, we need to find the Z-score for which the area to the right is 4%. This can be done using a Z-score table or a statistical calculator. In this case, the Z-score corresponding to an area of 0.04 is approximately 1.75.
2. Next, we need to convert the Z-score back to the original mileage scale. We can do this using the formula: X = mean + (Z-score * standard deviation).
X = 67,900 + (1.75 * 2,050) = 67,900 + 3,587.5 ≈ 71,487.5
Therefore, Lytton should announce an average guaranteed mileage of approximately 71,487.5.
B. To determine the probability of returning a tire with mileage below 67,500, we need to find the area to the left of this value in the distribution. Again, we can use the Z-score formula to calculate this probability.
1. Convert the mileage value of 67,500 to a Z-score using the formula: Z = (X - mean) / standard deviation.
Z = (67,500 - 67,900) / 2,050 ≈ -0.195
2. Find the corresponding area to the left of this Z-score value using a Z-score table or a statistical calculator. The area is approximately 0.4251.
Therefore, the probability of returning a tire with mileage below 67,500 is approximately 0.4251 or 42.51%.
To calculate the number of tires that would be returned out of the sample of 200, we can multiply the probability by the sample size:
Number of tires returned = 0.4251 * 200 ≈ 85.02
Since we cannot have a fraction of a tire, we would round this value down to the nearest whole number. Therefore, approximately 85 tires would be returned out of the sample.
C. To determine the 95% confidence interval for the average mileage of the tires, we can use the formula:
Confidence Interval = mean ± (Z * (standard deviation / sqrt(sample size)))
1. Find the Z-score corresponding to a 95% confidence level. From a Z-score table or a statistical calculator, the Z-score is approximately 1.96 for a 95% confidence level.
2. Calculate the confidence interval:
Confidence Interval = 67,900 ± (1.96 * (2,050 / sqrt(200)))
Simplifying further:
Confidence Interval = 67,900 ± (1.96 * (2,050 / 14.14))
Confidence Interval ≈ 67,900 ± 283.78
Confidence Interval ≈ (67,616.22, 68,183.78)
Therefore, the 95% confidence interval for the average tire mileage would be approximately (67,616.22, 68,183.78).
D. To determine the appropriate sample size for a 99% confidence interval with an error not exceeding 1000 miles, we can use the following formula:
Sample size = (Z^2 * standard deviation^2) / (error^2)
1. Find the Z-score corresponding to a 99% confidence level. From a Z-score table or a statistical calculator, the Z-score is approximately 2.58 for a 99% confidence level.
2. Plug in the values into the formula:
Sample size = (2.58^2 * 2,050^2) / (1,000^2)
Sample size ≈ (6.6464 * 4,202,500) / 1,000,000
Sample size ≈ 27,868.74 / 1,000
Sample size ≈ 27.87
Therefore, the recommended sample size to determine a 99% confidence interval with an error not exceeding 1000 miles would be approximately 28 (rounded up to the nearest whole number).