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Suppose Abby tracks her travel time to work for 60 days and determines that her mean travel time, in minutes, is 35.6. Assume the travel times are normally distributed with a standard deviation of 10.3 min. Determine the travel time such that 26.11% of the 60 days have a travel time that is longer. (Give your answer rounded to two decimal places.)​

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Final answer:

To find the travel time that has 26.11% of the days being longer, we use the z-score for the 73.89% cumulative probability and the formula for converting a z-score to an original value, which results in approximately 41.99 minutes.

Step-by-step explanation:

The student is asking to find a specific travel time that divides the upper 26.11% of her travel times from the lower 73.89%, given that the mean travel time is 35.6 minutes and the standard deviation is 10.3 minutes with a normal distribution. This value corresponds to a certain z-score. First, we find the z-score that corresponds to the cumulative probability of 73.89% (which is 1 - 26.11%). Using a z-table or a calculator, we find that the z-score for 73.89% is approximately 0.62. Next, we use the formula for converting a z-score to the original value, which is X = μ + (z * σ), where μ is the mean, σ is the standard deviation, and z is the z-score. Substituting the given values, we calculate X as follows: X = 35.6 + (0.62 * 10.3) ≈ 35.6 + 6.386 ≈ 41.986. Therefore, the travel time such that 26.11% of the 60 days have a travel time that is longer is approximately 41.99 minutes when rounded to two decimal places.

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