A consensus forecast is the average of a large number of individual analysts' forecasts. Suppose the individual forecasts for a particular interest rate are normally distributed with a mean of 3 percent and a standard deviation of 2.0 percent. A single analyst is randomly selected. Find the probability that his/her forecast is

(a) At least 3.9 percent. (Round the z value to 2 decimal places. Round your answer to 4 decimal places.)

To solve this problem, we will use the properties of the **normal distribution**. Specifically, we need to calculate the probability that a randomly selected forecast is at least 3.9 percent, given that the distribution of forecasts is normally distributed with a mean of 3 percent and a standard deviation of 2 percent.

The formula to calculate the z-score is:

\[
z = \frac{x - \mu}{\sigma}
\]

where:
- \( x \) is the value we are interested in (3.9 percent),
- \( \mu \) is the mean (3 percent),
- \( \sigma \) is the standard deviation (2 percent).

### Step 1: Compute the z-score

\[
z = \frac{3.9 - 3}{2} = \frac{0.9}{2} = 0.45
\]

### Step 2: Look up the probability for z = 0.45

Using the standard normal distribution table or a calculator, the probability corresponding to \( z = 0.45 \) is approximately 0.6736.

### Step 3: Find the probability of the forecast being at least 3.9 percent

The question asks for the probability that the forecast is **at least** 3.9 percent, which is the complement of the probability of it being **less than** 3.9 percent. So, we subtract the value from 1:

\[
P(X \geq 3.9) = 1 - P(X \leq 3.9) = 1 - 0.6736 = 0.3264
\]

Thus, the probability that a randomly selected forecast is at least 3.9 percent is **0.3264**.

Would you like any further clarification or details?

Here are some related questions:
1. What is the probability that the forecast is less than 2 percent?
2. How would the probability change if the mean forecast were 4 percent instead of 3 percent?
3. What is the probability that the forecast is between 2.5 percent and 4.5 percent?
4. How do changes in standard deviation affect the forecast probabilities?
5. How can we interpret this result in terms of real-world forecasting?

**Tip**: When working with the normal distribution, always visualize the problem by sketching the bell curve to better understand which area you are calculating.

A consensus forecast is the average of a large number of individual analysts' forecasts. Suppose the individual forecasts for a particular interest rate are normally distributed with a mean of 3 percent and a standard deviation of 2.0 percent. A single analyst is randomly selected. Find the probability that his/her forecast is at least 3.9 percent. (Round the z value to 2 decimal places. Round your answer to 4 decimal places.)

Learn how to calculate the probability of an interest rate forecast being at least 3.9 percent using the properties of a normal distribution with a mean of 3 percent and a standard deviation of 2 percent. Includes step-by-step calculations of the z-score and probability.

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