To find the probability that more than 178 calls will be received in a given hour, we need to use the normal distribution with a mean of 170 calls and a standard deviation of 5 calls.

Step 1: Standardize the value

We first convert the raw score (178 calls) to a z-score using the formula:

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

Where:

• $X = 178$ calls,
• $\mu = 170$ calls,
• $\sigma = 5$ calls.

$z = \frac{178 - 170}{5} = \frac{8}{5} = 1.6$

Step 2: Find the cumulative probability corresponding to the z-score

Using the standard normal distribution table or a calculator, we find the cumulative probability for $z = 1.6$.

• The cumulative probability for $z = 1.6$ is approximately 0.9452.

Step 3: Calculate the probability of more than 178 calls

Since we want the probability of having more than 178 calls, we need to find the complement of the cumulative probability:

$P(X > 178) = 1 - P(z \leq 1.6) = 1 - 0.9452 = 0.0548$

Thus, the probability that more than 178 calls will be received in a given hour is approximately 0.055 (to the nearest thousandth).

Would you like more details or have any questions about this solution?

Here are five related questions:

1. What is the probability that fewer than 165 calls will be received in an hour?
2. What is the probability that between 168 and 175 calls will be received in an hour?
3. How would the probability change if the mean number of calls increased to 175 while keeping the same standard deviation?
4. What is the z-score for receiving exactly 170 calls in an hour?
5. How would you interpret a z-score of 2.0 in this context?

Tip: When working with normal distributions, always remember that the total area under the curve equals 1, representing the entire range of possible outcomes. This helps in calculating the probabilities for different scenarios.