To find the probability that a randomly selected commuter spends more than 30 minutes commuting, we need to standardize the data using the Z-score formula. The Z-score helps to convert a raw value (in this case, 30 minutes) into a standard normal distribution.

The Z-score formula is:

$Z = \frac{X - \mu}{\sigma}$

Where:

• $X = 30$ (the value we're interested in),
• $\mu = 25$ (the mean of the commute time),
• $\sigma = 6.1$ (the standard deviation).

Step 1: Calculate the Z-score

$Z = \frac{30 - 25}{6.1} = \frac{5}{6.1} \approx 0.8197$

Step 2: Find the corresponding probability

Now that we have the Z-score of approximately $0.8197$, we need to find the probability that $Z$ is greater than this value. Using standard normal distribution tables (or a calculator), we find the cumulative probability for $Z = 0.8197$.

The cumulative probability $P(Z < 0.8197) \approx 0.7939$.

Since we are interested in the probability of spending more than 30 minutes commuting (i.e., $P(Z > 0.8197)$):

$P(Z > 0.8197) = 1 - 0.7939 = 0.2061$

Final Answer:

The probability that a randomly selected commuter spends more than 30 minutes commuting one way is approximately 0.2061 (rounded to four decimal places).

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Would you like further details or have any questions? Here are some related questions for you to explore:

1. What is the probability that a commuter spends less than 20 minutes commuting one way?
2. How long do the top 5% of commuters spend commuting one way?
3. If a commuter spends 40 minutes commuting, what is their Z-score?
4. What is the probability that a commuter spends between 20 and 30 minutes commuting?
5. How does increasing the standard deviation affect the probability of commuting longer than 30 minutes?

Tip: Z-scores are a powerful way to compare values from different normal distributions!