Question

A researcher notes that senior corporation executives are not very accurate forecasters of their own annual earnings. He states that his studies of a large number of company executive forecasts "showed that the average estimate missed the mark by \(15 \%\)." a. Suppose the distribution of these forecast errors has a mean of \(15 \%\) and a standard deviation of \(10 \%\). Is it likely that the distribution of forecast errors is approximately normal? b. Suppose the probability is .5 that a corporate executive's forecast error exceeds \(15 \% .\) If you were to sample the forecasts of 100 corporate executives, what is the probability that more than 60 would be in error by more than \(15 \% ?\)

Short answer

What is the probability that more than 60 forecasts out of 100 would have a forecast error greater than 15%? Answer: It is reasonable to assume that the forecast errors follow a normal distribution for the purposes of this exercise. The probability that more than 60 forecasts out of 100 would have a forecast error greater than 15% is approximately 0.0228 or 2.28%.

Step-by-step solution

a. Determining if the distribution is approximately normal

The distribution of forecast errors is given a mean of 15% and a standard deviation of 10%. Based on this information, we cannot conclusively determine if the distribution is approximately normal. However, for the purposes of this exercise and based on the Central Limit Theorem, when the sample size increases, the sampling distribution of the sample means tends to be normal. So, we can assume that the distribution of forecast errors is approximately normal, as it is a reasonable assumption given a large number of company executive forecasts in the study. It is important to note that this assumption might not always hold true in reality, but for this exercise, we will proceed with this assumption.

b. Probability of more than 60 forecasts having errors greater than 15%

First, we will use the given information that the probability is 0.5 that a corporate executive's forecast error exceeds 15%. Next, we will calculate the probability that more than 60 out of 100 forecasts have errors greater than 15%. Let X denote the number of forecasts with errors greater than 15% in a sample of 100 forecasts. Since the probability of a single executive's forecast error exceeding 15% is 0.5, we can model X using a binomial distribution with parameters n = 100 (the number of trials) and p = 0.5 (the probability of "success", or having an error greater than 15%). To find the desired probability P(X > 60), we will use the complementary probability method: P(X > 60) = 1 - P(X ≤ 60). To find P(X ≤ 60), we can use the cumulative distribution function (CDF) of the binomial distribution: P(X ≤ 60) = Σ P(X = x) for x = 0 to 60 Next, as we assumed the distribution of forecast errors is approximately normal (from part a), we can use the normal approximation to the binomial distribution to make the calculation easier. To do this, we need the mean and standard deviation of X. For a binomial distribution: μ = np = 100 * 0.5 = 50 σ = √(np(1-p)) = √(100 * 0.5 * 0.5) ≈ 5 Under the normal approximation, we can replace X with a z-score: z = (x - μ) / σ Now, we can find P(X > 60) using the z-score table (also known as the standard normal table): P(X > 60) = P(z > (60 - 50) / 5) ≈ P(z > 2) Using a z-score table, we find the probability P(z > 2) ≈ 0.0228. Therefore, the probability that more than 60 forecasts out of 100 would have a forecast error greater than 15% is approximately 0.0228, or 2.28%.