The life span of three randomly selected tires are 34,700 miles, 43,000 miles, 39,000 miles. Using the empirical rule, find the percentile that corresponds to each life span.
What empirical rule? Percentile of what?
To find the percentile that corresponds to each life span using the empirical rule, we need to calculate the z-score for each tire and then use a standard normal distribution table to determine the corresponding percentile.
The empirical rule, also known as the 68-95-99.7 rule, states that for a normal distribution:
- Approximately 68% of the data falls within one standard deviation of the mean
- Approximately 95% of the data falls within two standard deviations of the mean
- Approximately 99.7% of the data falls within three standard deviations of the mean
Step 1: Calculate the mean and standard deviation:
- To find the mean, sum up all the tire life spans and divide by the total number of tires: (34,700 + 43,000 + 39,000) / 3 = 38,900 miles
- To find the standard deviation, we need to calculate the variance first. Variance is the average of the squared differences from the mean. Using the formula, we get:
Variance = [(34,700 - 38,900)^2 + (43,000 - 38,900)^2 + (39,000 - 38,900)^2] / 3
= (16,000,000 + 16,000,000 + 100,000) / 3
= 32,700,000 / 3
≈ 10,900,000
- Now, take the square root of the variance to find the standard deviation: √10,900,000 ≈ 3,302.77 miles
Step 2: Calculate the z-scores for each tire:
- The z-score is a measure of how many standard deviations an observation is from the mean. It tells us how relative the tire's life span is compared to the others.
- The formula for the z-score is: z = (x - μ) / σ, where x is the data value, μ is the mean, and σ is the standard deviation.
- For the first tire, with a life span of 34,700 miles: z1 = (34,700 - 38,900) / 3,302.77 ≈ -1.28
- For the second tire, with a life span of 43,000 miles: z2 = (43,000 - 38,900) / 3,302.77 ≈ 1.24
- For the third tire, with a life span of 39,000 miles: z3 = (39,000 - 38,900) / 3,302.77 ≈ 0.03
Step 3: Determine the percentile using the standard normal distribution table:
- We need to find the area under the standard normal curve to the left of each z-score. This area represents the percentile value.
- For the first tire, with a z-score of -1.28, we find the corresponding percentile in the standard normal distribution table. This percentile will be the approximate percentage of tire life spans that are less than 34,700 miles.
- Using the table, we find that the area to the left of -1.28 is approximately 0.1003, or 10.03%.
- So, the first tire corresponds to the 10.03 percentile.
- For the second tire, with a z-score of 1.24, we find the corresponding percentile in the standard normal distribution table.
- The area to the left of 1.24 is approximately 0.8925, or 89.25%.
- So, the second tire corresponds to the 89.25 percentile.
- For the third tire, with a z-score of 0.03, we find the corresponding percentile in the standard normal distribution table.
- The area to the left of 0.03 is approximately 0.5120, or 51.20%.
- So, the third tire corresponds to the 51.20 percentile.
Therefore, the percentiles that correspond to the life spans of the three randomly selected tires are approximately 10.03%, 89.25%, and 51.20%, respectively.