To solve this problem, we need to find the standard deviation, $x$, given the following conditions:

• The flight times follow a normal distribution with a mean $\mu = 75$ minutes.
• 2% of the flight times are longer than 82 minutes.

This can be expressed as:

$P(X > 82) = 0.02$

In a normal distribution, we can use the z-score formula to standardize this problem:

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

Here, $X = 82$, $\mu = 75$, and $\sigma = x$. Substituting these values in, we have:

$z = \frac{82 - 75}{x}$

$z = \frac{7}{x}$

We know that $P(X > 82) = 0.02$ corresponds to a z-score such that 2% of the data lies to the right of it. Using a standard normal distribution table or a calculator, we find that the z-score corresponding to the 98th percentile (since 1 - 0.02 = 0.98) is approximately:

$z \approx 2.054$

Setting up the equation:

$\frac{7}{x} = 2.054$

Solving for $x$:

$x = \frac{7}{2.054} \approx 3.41$

Therefore, the standard deviation is approximately $\sigma = 3.41$ minutes.

Would you like more details on any part of this solution?

---

Related Questions:

1. How would the problem change if the mean flight time were 80 minutes?
2. What would be the standard deviation if only 1% of flight times were longer than 82 minutes?
3. How can we use the empirical rule to approximate the probability of flight times?
4. What is the z-score for a flight time of 70 minutes with this standard deviation?
5. How would the z-score change if the mean flight time were 70 minutes?

Tip: For normal distributions, knowing the mean and standard deviation allows you to calculate probabilities for any range of values.