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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.

In a supermarket, the time customers spend waiting for their turn at the cashier follows a normal distribution with a mean of 15 minutes and standard deviation 2 minutes. Find the probability that the waiting time is (i) More than 20 minutes, (ii) Less than 11 minutes, (iii) Between 10 and 18 minutes.

Solve a normal distribution problem involving customer waiting times at a cashier. The problem asks to calculate the probability of waiting more than 20 minutes, less than 11 minutes, and between 10 to 18 minutes using a mean of 15 minutes and a standard deviation of 2 minutes. Suitable for Grades 10-12.

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