We are given that the time customers spend waiting at the cashier follows a normal distribution with a mean of 15 minutes and a standard deviation of 2 minutes. We are tasked with finding the probabilities for three different scenarios.

Part (i): Probability that the waiting time is more than 20 minutes

To calculate this, we use the z-score formula:

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

Where:

• $x$ is the value of interest (20 minutes),
• $\mu = 15$ minutes (mean),
• $\sigma = 2$ minutes (standard deviation).

$z = \frac{20 - 15}{2} = \frac{5}{2} = 2.5$

Now, using the standard normal distribution table or a calculator, the probability $P(Z > 2.5)$ is approximately 0.0062.

Thus, the probability of waiting more than 20 minutes is 0.0062, or 0.62%.

Part (ii): Probability that the waiting time is less than 11 minutes

We use the same z-score formula for $x = 11$ minutes:

$z = \frac{11 - 15}{2} = \frac{-4}{2} = -2$

From the standard normal distribution table, $P(Z < -2)$ is approximately 0.0228.

Thus, the probability of waiting less than 11 minutes is 0.0228, or 2.28%.

Part (iii): Probability that the waiting time is between 10 and 18 minutes

We calculate the z-scores for both 10 and 18 minutes.

For $x = 10$ minutes:

$z = \frac{10 - 15}{2} = -2.5$

For $x = 18$ minutes:

$z = \frac{18 - 15}{2} = 1.5$

Using the standard normal table:

• $P(Z < -2.5)$ is approximately 0.0062,
• $P(Z < 1.5)$ is approximately 0.9332.

The probability of being between 10 and 18 minutes is the difference:

$P(10 < X < 18) = P(Z < 1.5) - P(Z < -2.5) = 0.9332 - 0.0062 = 0.9270$

Thus, the probability is 0.9270, or 92.7%.

Would you like further details or any clarification on these steps?

Related Questions:

1. What is the z-score for a waiting time of 12 minutes?
2. How would the probabilities change if the standard deviation was 3 minutes?
3. What is the 95th percentile of waiting times in this scenario?
4. How do we calculate the probability for a range of values in a normal distribution?
5. How does changing the mean impact the probabilities calculated above?

Tip: Z-scores help standardize values, making it easier to find probabilities in a normal distribution.