To solve this problem, we need to calculate the probability that the number of calls is less than 149, given that the number of calls is normally distributed with a mean ($\mu$) of 130 calls and a standard deviation ($\sigma$) of 15 calls.

We can solve this by finding the Z-score for 149 calls and then using the standard normal distribution table (or a calculator) to find the corresponding probability.

Step 1: Calculate the Z-score

The Z-score is calculated using the formula:

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

Where:

• $X = 149$ (the value we're interested in),
• $\mu = 130$ (the mean),
• $\sigma = 15$ (the standard deviation).

Substitute the values into the formula:

$Z = \frac{149 - 130}{15} = \frac{19}{15} \approx 1.267$

Step 2: Find the Probability

Next, we use the Z-score of 1.267 to find the probability that a value is less than 149 in the standard normal distribution.

Using a standard normal distribution table or calculator, we find that the probability corresponding to a Z-score of 1.267 is approximately 0.897.

Final Answer:

The probability that there will be less than 149 calls in a given hour is approximately 0.897, or 89.7%.

Would you like further details on this, or do you have any questions?

Here are some related questions to expand your understanding:

1. What is the probability of receiving more than 149 calls in a given hour?
2. What is the probability of receiving between 120 and 149 calls in a given hour?
3. How would the probability change if the standard deviation was larger, say 20 calls per hour?
4. How many calls would correspond to the 95th percentile in this distribution?
5. What is the Z-score for 100 calls in this distribution?

Tip: When calculating probabilities for normal distributions, it's often helpful to visualize the bell curve to better understand where your value falls.